AD Validation Guide Vol I 2017 R2 En
Short Description
Advance Design Validation Guide Vol I 2017 R2 En...
Description
VALIDATION GUIDE
2017 R2
Advance Design Validation Guide
Version: 2017 R2 Tests passed on: 29 August 2016 Number of tests: 610
INTRODUCTION Before being officially released, each version of GRAITEC software, including Advance Design, undergoes a series of validation tests. This validation is performed in parallel and in addition to manual testing and beta testing, in order to obtain the "operational version" status. This document contains a description of the automatic tests, highlighting the theoretical background and the results we have obtained using the current software release. Usually, a test is made of a reference (independent from the specific software version tested), a transformation (a calculation or a data processing scenario), a result (given by the specific software version tested) and a difference usually measured in percentage as a drift from a set of reference values. Depending on the cases, the used reference is either a theoretical calculation done manually, a sample taken from the technical literature, or the result of a previous version considered as good by experience. Starting with version 2012, Graitec Advance has made significant steps forward in term of quality management by extending the scope and automating the testing process. While in previous versions, the tests were always about the calculation results which were compared to a reference set, starting with version 2012, tests have been extended to user interface behavior, import/export procedures, etc. The next major improvement is the capacity to pass the tests automatically. These current tests have obviously been passed on the “operational version”, but they are actually passed on a daily basis during the development process, which helps improve the daily quality by solving potential issues, immediately after they have been introduced in the code. In the field of structural analysis and design, software users must keep in mind that the results highly depend on the modeling (especially when dealing with finite elements) and on the settings of the numerous assumptions and options available in the software. A software package cannot replace engineers experience and analysis. Despite all our efforts in term of quality management, we cannot guaranty the correct behavior and the validity of the results issued by Advance Design in any situation. With this validation guide, we are providing a set of concrete test cases showing the behavior of Advance Design in various areas and various conditions. The tests cover a wide field of expertise: modeling, climatic load generation according to Eurocode 1, combinations management, meshing, finite element calculation, reinforced concrete design according to Eurocode 2, steel member design according to Eurocode 3, steel connection design according to Eurocode 3, timber member design according to Eurocode 5, seismic analysis according to Eurocode 8, report generation, import / export procedures and user interface behavior. We hope that this guide will highly contribute to the knowledge and the confidence you are placing in Advance Design.
Manuel LIEDOT Chief Operating Officer
ADVANCE DESIGN VALIDATION GUIDE
Table of Contents 1
FINITE ELEMENT METHOD ................................................................................................. 23 1.1
Thin lozengeshaped plate fixed on one side (alpha = 15 °) (010008SDLSB_FEM) ...................................24
1.2
Thin lozengeshaped plate fixed on one side (alpha = 30 °) (010009SDLSB_FEM) ...................................27
1.3
Circular plate under uniform load (010003SSLSB_FEM) ............................................................................30
1.4
Vibration mode of a thin piping elbow in plane (case 2) (010012SDLLB_FEM) ..........................................33
1.5
Vibration mode of a thin piping elbow in plane (case 3) (010013SDLLB_FEM) ..........................................36
1.6
Thin lozengeshaped plate fixed on one side (alpha = 0 °) (010007SDLSB_FEM) .....................................39
1.7
Vibration mode of a thin piping elbow in plane (case 1) (010011SDLLB_FEM) ..........................................42
1.8
System of two bars with three hinges (010002SSLLB_FEM) ......................................................................45
1.9
Thin circular ring fixed in two points (010006SDLLB_FEM) ........................................................................48
1.10
Thin lozengeshaped plate fixed on one side (alpha = 45 °) (010010SDLSB_FEM) ..................................52
1.11
Double fixed beam (010016SDLLB_FEM) .................................................................................................55
1.12
Short beam on simple supports (on the neutral axis) (010017SDLLB_FEM) .............................................59
1.13
Rectangular thin plate simply supported on its perimeter (010020SDLSB_FEM) ......................................63
1.14
Cantilever beam in Eulerian buckling (010021SFLLB_FEM) .....................................................................67
1.15
Double fixed beam with a spring at mid span (010015SSLLB_FEM) .........................................................69
1.16
Thin square plate fixed on one side (010019SDLSB_FEM) .......................................................................72
1.17
Bending effects of a symmetrical portal frame (010023SDLLB_FEM) .......................................................76
1.18
Slender beam on two fixed supports (010024SSLLB_FEM) ......................................................................80
1.19
Slender beam on three supports (010025SSLLB_FEM) ............................................................................86
1.20
Thin circular ring hanged on an elastic element (010014SDLLB_FEM) .....................................................90
1.21
Short beam on simple supports (eccentric) (010018SDLLB_FEM) ............................................................94
1.22
Annular thin plate fixed on a hub (repetitive circular structure) (010022SDLSB_FEM) ..............................99
1.23
Fixed thin arc in out of plane bending (010028SSLLB_FEM) ..................................................................102
1.24
Double hinged thin arc in planar bending (010029SSLLB_FEM) .............................................................105
1.25
Beam on elastic soil, free ends (010032SSLLB_FEM) ............................................................................108
1.26
EDF Pylon (010033SFLLA_FEM) ............................................................................................................112
1.27
Fixed thin arc in planar bending (010027SSLLB_FEM) ...........................................................................116
1.28
Truss with hinged bars under a punctual load (010031SSLLB_FEM) ......................................................119
1.29
Simply supported square plate (010036SSLSB_FEM) ............................................................................122
1.30
Caisson beam in torsion (010037SSLSB_FEM) ......................................................................................125
1.31
Thin cylinder under a uniform radial pressure (010038SSLSB_FEM) ......................................................128
1.32
Bimetallic: Fixed beams connected to a stiff element (010026SSLLB_FEM) ...........................................130
1.33
Portal frame with lateral connections (010030SSLLB_FEM)....................................................................133
1.34
Beam on elastic soil, hinged ends (010034SSLLB_FEM) ........................................................................137
1.35
Beam on two supports considering the shear force (010041SSLLB_FEM) .............................................141
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8
1.36
Thin cylinder under a uniform axial load (010042SSLSB_FEM) .............................................................. 144
1.37
Torus with uniform internal pressure (010045SSLSB_FEM) ................................................................... 147
1.38
Spherical shell under internal pressure (010046SSLSB_FEM) ............................................................... 150
1.39
Stiffen membrane (010040SSLSB_FEM) ................................................................................................ 153
1.40
Thin cylinder under its self weight (010044SSLSB_MEF) ....................................................................... 156
1.41
Spherical shell with holes (010049SSLSB_FEM) .................................................................................... 158
1.42
Spherical dome under a uniform external pressure (010050SSLSB_FEM) ............................................. 161
1.43
Simply supported square plate under a uniform load (010051SSLSB_FEM) .......................................... 164
1.44
Square plate under planar stresses (010039SSLSB_FEM) ..................................................................... 166
1.45
Thin cylinder under a hydrostatic pressure (010043SSLSB_FEM) .......................................................... 170
1.46
Pinch cylindrical shell (010048SSLSB_FEM) .......................................................................................... 173
1.47
Simply supported rectangular plate loaded with punctual force and moments (010054SSLSB_FEM) .... 175
1.48
Shear plate perpendicular to the medium surface (010055SSLSB_FEM) ............................................... 177
1.49
A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (010058SSLSB_FEM) ............ 179
1.50
A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (010059SSLSB_FEM)..... 181
1.51
Simply supported rectangular plate under a uniform load (010053SSLSB_FEM) ................................... 183
1.52
A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (010057SSLSB_FEM)..... 185
1.53
A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (010061SSLSB_FEM)....... 187
1.54
A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (010062SSLSB_FEM) ......... 189
1.55
A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (010063SSLSB_FEM) ... 191
1.56
Simply supported rectangular plate under a uniform load (010052SSLSB_FEM) ................................... 193
1.57
Triangulated system with hinged bars (010056SSLLB_FEM).................................................................. 195
1.58
A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (010060SSLSB_FEM)..... 198
1.59
A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (010066SSLSB_FEM) ........... 201
1.60
Vibration mode of a thin piping elbow in space (case 1) (010067SDLLB_FEM) ...................................... 203
1.61
Reactions on supports and bending moments on a 2D portal frame (Rafters) (010077SSLPB_FEM) .... 206
1.62
Reactions on supports and bending moments on a 2D portal frame (Columns) (010078SSLPB_FEM) . 208
1.63
A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (010065SSLSB_FEM) ......... 210
1.64
Vibration mode of a thin piping elbow in space (case 3) (010069SDLLB_FEM) ...................................... 212
1.65
Slender beam of variable rectangular section with fixedfree ends (ß=5) (010085SDLLB_FEM) ............ 215
1.66
Slender beam of variable rectangular section (fixedfixed) (010086SDLLB_FEM) .................................. 220
1.67
Plane portal frame with hinged supports (010089SSLLB_FEM) .............................................................. 223
1.68
A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (010064SSLSB_FEM) ......... 225
1.69
Vibration mode of a thin piping elbow in space (case 2) (010068SDLLB_FEM) ...................................... 228
1.70
Short beam on two hinged supports (010084SSLLB_FEM) .................................................................... 231
1.71
A 3D bar structure with elastic support (010094SSLLB_FEM) ................................................................ 233
1.72
Fixed/free slender beam with centered mass (010095SDLLB_FEM) ...................................................... 240
1.73
Simple supported beam in free vibration (010098SDLLB_FEM) ............................................................. 245
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1.74
Membrane with hot point (010099HSLSB_FEM) .....................................................................................248
1.75
Cantilever beam in Eulerian buckling with thermal load (010092HFLLB_FEM) .......................................251
1.76
Double cross with hinged ends (010097SDLLB_FEM) ............................................................................253
1.77
Beam on 3 supports with T/C (k > infinite) (010101SSNLB_FEM) ..........................................................257
1.78
Beam on 3 supports with T/C (k = 10000 N/m) (010102SSNLB_FEM)...................................................260
1.79
Linear system of truss beams (010103SSLLB_FEM) ..............................................................................263
1.80
Double fixed beam in Eulerian buckling with a thermal load (010091HFLLB_FEM) ................................267
1.81
Fixed/free slender beam with eccentric mass or inertia (010096SDLLB_FEM) .......................................270
1.82
Beam on 3 supports with T/C (k = 0) (010100SSNLB_FEM) ...................................................................274
1.83 BAEL 91 (concrete design)  France: Linear element in combined bending/tension  without compressed reinforcements  Partially tensioned section (020158SSLLB_B91) .....................................................................277 1.84 BAEL 91 (concrete design)  France: Linear element in simple bending  without compressed reinforcement (020162SSLLB_B91) ....................................................................................................................282 1.85 BAEL 91 (concrete design)  France: Design of a concrete floor with an opening (030208SSLLG_BAEL91) ..........................................................................................................................................286 1.86
Cantilever rectangular plate (010001SSLSB_FEM) .................................................................................294
1.87
Tied (subtensioned) beam (010005SSLLB_FEM) ..................................................................................297
1.88
Slender beam with variable section (fixedfree) (010004SDLLB_FEM) ...................................................302
1.89
PS92  France: Study of a mast subjected to an earthquake (020112SMLLB_P92) ................................305
1.90
CM66 (steel design)  France: Design of a 2D portal frame (030207SSLLG_CM66) ...............................311
1.91
Non linear system of truss beams (010104SSNLB_FEM) .......................................................................319
1.92
CM66 (steel design)  France: Design of a Steel Structure (030206SSLLG_CM66) ................................323
1.93
FEM Results  United Kingdom: Simply supported laterally restrained (from P364 Open Sections Example 2)....333
1.94
Verifying results of a steel beam subjected to dynamic temporal loadings (TTAD #14586) ......................335
1.95 Checks the bending moments in the central node of a steel frame with two beams having a rotational stiffness of 42590 kN/m. .......................................................................................................................................339 1.96
Verifying a simply supported concrete slab subjected to temperature variation between top and bottom fibers ....339
1.97
Correct use of symetric steel crosssections (eg. IPE300S) ......................................................................339
1.98
Temperature load: SD frame with elements under tempertature gradient, applied on separate systems .339
1.99
Verifying displacements of a prestressed cable structure with results presented in Tibert, 1999. .............340
1.100
EC8 / NF EN 19981  France: Verifying the level mass center (TTAD #11573, TTAD #12315) .............340
1.101
Verifying Sxx results on beams (TTAD #11599)......................................................................................340
1.102
Verifying forces results on concrete linear elements (TTAD #11647) ......................................................340
1.103
Generating results for Torsors NZ/Group (TTAD #11633) ......................................................................341
1.104
Verifying diagrams for Mf Torsors on divided walls (TTAD #11557) ........................................................341
1.105
Generating planar efforts before and after selecting a saved view (TTAD #11849) ................................341
1.106
Verifying stresses in beam with "extend into wall" property (TTAD #11680) ...........................................341
1.107
Verifying forces for triangular meshing on planar element (TTAD #11723) .............................................341
1.108
Verifying constraints for triangular mesh on planar elements (TTAD #11447) ........................................342
1.109
Verifying the displacement results on linear elements for vertical seism (TTAD #11756) .......................342
ADVANCE DESIGN VALIDATION GUIDE
1.110
Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD #11854) ..... 342
1.111
Verifying tension/compression supports on nonlinear analysis (TTAD #11518) ..................................... 342
1.112
Verifying results on punctual supports (TTAD #11489) ........................................................................... 343
1.113
Generating a report with torsors per level (TTAD #11421)...................................................................... 343
1.114
Verifying nonlinear analysis results for frames with semirigid joints and rigid joints (TTAD #11495) ..... 343
1.115
Verifying the main axes results on a planar element (TTAD #11725) ..................................................... 343
1.116
Verifying tension/compression supports on nonlinear analysis (TTAD #11518) ..................................... 344
1.117
Verifying the display of the forces results on planar supports (TTAD #11728)........................................ 344
1.118 Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontal force (TTAD #13175) .......................................................................................................................................................................344
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1.119
Verifying torsors on a single story coupled walls subjected to horizontal forces ..................................... 344
1.120
Verifying forces on a linear elastic support which is defined in a user workplane (TTAD #11929).......... 344
1.121
Verifying the internal forces results for a simple supported steel beam .................................................. 345
CAD, RENDERING AND VISUALIZATION ........................................................................ 347 2.1
Verifying the annotations dimensions when new analysis is made (TTAD #14825) ................................... 348
2.2
Verifying the default view (TTAD #13248) .................................................................................................. 348
2.3
Verifying the saved view of elements with annotations. (TTAD #13033) .................................................... 348
2.4
Verifying the visualisation of supports with rotational or moving DoFs (TTAD #13891) ............................. 348
2.5
Verifying the annotations of a wind generated load (TTAD #13190) .......................................................... 349
2.6 Verifying the dimensions and position of annotations on selection when new analysis is made (TTAD #12807) ................................................................................................................................................................ 349
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2.7
Verifying the annotation of a variable surface load (TTAD #13618) ........................................................... 349
2.8
Verifying hide/show elements command (TTAD #11753) .......................................................................... 349
2.9
Verifying the grid text position (TTAD #11657) ........................................................................................... 349
2.10
System stability during section cut results verification (TTAD #11752) ..................................................... 349
2.11
Verifying the grid text position (TTAD #11704) ......................................................................................... 350
2.12
Generating combinations (TTAD #11721) ................................................................................................ 350
2.13
Verifying the coordinates system symbol (TTAD #11611) ........................................................................ 350
2.14
Verifying descriptive actors after creating analysis (TTAD #11589) .......................................................... 350
2.15
Creating a camera (TTAD #11526) ........................................................................................................... 350
2.16
Creating a circle (TTAD #11525) .............................................................................................................. 350
2.17
Verifying the local axes of a section cut (TTAD #11681) .......................................................................... 351
2.18
Verifying the snap points behavior during modeling (TTAD #11458) ........................................................ 351
2.19
Verifying the representation of elements with HEA cross section (TTAD #11328).................................... 351
2.20
Verifying the display of elements with compound cross sections (TTAD #11486) .................................... 351
2.21
Verifying holes in horizontal planar elements after changing the level height (TTAD #11490) ................. 351
2.22
Modeling using the tracking snap mode (TTAD #10979) .......................................................................... 351
2.23
Verifying the descriptive model display after post processing results in analysis mode (TTAD #11475) .. 352
2.24
Moving a linear element along with the support (TTAD #12110) .............................................................. 352
ADVANCE DESIGN VALIDATION GUIDE
2.25
Verifying the "ghost display on selection" function for saved views (TTAD #12054) .................................352
2.26
Verifying the "ghost" display after changing the display colors (TTAD #12064) ........................................352
2.27
Turning on/off the "ghost" rendering mode (TTAD #11999) ......................................................................352
2.28
Verifying the fixed load scale function (TTAD #12183). .............................................................................352
2.29
Verifying the steel connections modeling (TTAD #11698) .........................................................................353
2.30
Verifying the dividing of planar elements which contain openings (TTAD #12229) ...................................353
2.31
Verifying the display of punctual loads after changing the load case number (TTAD #11958) ..................353
2.32
Verifying the display of a beam with haunches (TTAD #12299) ...............................................................353
2.33
Verifying the program behavior when trying to create lintel (TTAD #12062) .............................................353
2.34 Verifying the program behavior when launching the analysis on a model with overlapped loads (TTAD #11837) ................................................................................................................................................................353
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2.35
Creating base plate connections for nonvertical columns (TTAD #12170) ...............................................354
2.36
Verifying drawing of joints in yz plan (TTAD #12453) ...............................................................................354
2.37
Verifying rotation for steel beam with joint (TTAD #12592) .......................................................................354
2.38
Verifying annotation on selection (TTAD #12700) .....................................................................................354
2.39
Verifying the saved view of elements by crosssection (TTAD #13197) ....................................................354
CLIMATIC GENERATOR .................................................................................................... 355 3.1 EC1 / NF EN 199113/NA  France: Generating snow loads on 2 side by side single roof compounds with different height (TTAD 13159) ..............................................................................................................................356 3.2 EC1 / NF EN 199113/NA  France: Generating snow loads on 2 side by side single roof compounds with different height (TTAD 13158) ..............................................................................................................................356 3.3 EC1 / NF EN 199114/NA  France: Generating Cf and Cp,net wind loads on an multibay canopy roof (DEV2013#4.3) .....................................................................................................................................................356 3.4 EC1 / NF EN 199113/NA  France: Generating snow loads on a 4 slopes shed with parapets. (TTAD #14578) ................................................................................................................................................................356 3.5 EC1 / CSN EN 199113/NA  Czech Republic: Snow load generation on building with 2 slopes > 60 degrees (TTAD #14235) .......................................................................................................................................357 3.6 EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 almost horizontal slope building (TTAD #13663) ................................................................................................................................................................357 3.7 EC1 / NF EN 199113/NA  France: Generating snow loads on a custom multiple slope building (TTAD #14285) ................................................................................................................................................................357 3.8 EC1 / NF EN 199113/NA  France: Generating 2D snow loads on a 2 slope portal with one lateral parapet (TTAD #14530) .....................................................................................................................................................357 3.9 EC1 / NF EN 199114/NA  France: Generating wind loads on a 4 slopes shed with parapets (TTAD #14179) ................................................................................................................................................................357 3.10
EC1 / NF EN 199114/NA  France: Wind load generation on multibay canopies (TTAD #11668) ..........358
3.11 EC1 / NF EN 199113/NA  France: Generating snow loads on a 2 slope building with parapets. (TTAD #13671) ................................................................................................................................................................358 3.12 EC1 / NF EN 199113/NA  France: Snow load generation on compound with a doubleroof volume close to a singleroof volume (TTAD #13559) ................................................................................................................358 3.13
EC1 / NF EN 199114/NA  France: Wind load generation on portal with CsCd set to auto (TTAD #12823) .. 358
3.14
EC1 / NF EN 199114/NA  France: Generating wind loads on a canopy. (TTAD #13855) .....................358
ADVANCE DESIGN VALIDATION GUIDE
3.15 EC1 / NF EN 199114/NA  France: Generating wind loads on a 35m high structure with CsCd min set to 0.7 and Delta to 0.15. (TTAD #11196) ................................................................................................................. 359 3.16 EC1 / NF EN 199114/NA  France: Generating wind loads on a singleroof volume compound with parapets. (TTAD #13672) ..................................................................................................................................... 359 3.17
EC1 / NF EN 199113/NA  France: Generating snow loads on a shed with parapets. (TTAD #12494) .. 359
3.18 EC1 / NF EN 199113/NA  France: Generating snow loads on a shed with gutters building. (TTAD #13856) ................................................................................................................................................................ 359 3.19
NV2009  France: Verifying wind on a protruding canopy. (TTAD #13880) .............................................. 360
3.20 EC1 / NF EN 199113/NA  France: Generating snow loads on 2 side by side single roof compounds with parapets (TTAD #13992)...................................................................................................................................... 360 3.21 EC1 / NF EN 199113/NA  France: Generating snow loads on a 3 slopes 3D portal frame. (TTAD #13169) ................................................................................................................................................................ 360 3.22 EC1 / NF EN 199113/NA  France: Generating snow loads on 2 side by side single roof compounds (TTAD #13286) .................................................................................................................................................... 360 3.23 EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 slope building with parapets. (TTAD #13669) ................................................................................................................................................................ 360 3.24 EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 slope building with increased height. (TTAD #13759) .................................................................................................................................................... 361 3.25
EC1 / CR 113/2012  Romania: Snow load generation on a 3 compound building (TTAD #13930s) ..... 361
3.26
EC1 / CR 114/2012  Romania: Wind load generation on portal with CsCd set to auto (TTAD #13930w) ..... 361
3.27 EC1 / NF EN 199113/NA  France: Generating snow loads on a 2 slope building with custom pressure values. (TTAD #14004) ........................................................................................................................................ 361 3.28 EC1 / NF EN 199113/NA  France: Generating snow loads on a 2 slope building with gutters and lateral parapets (TTAD #14005)...................................................................................................................................... 361 3.29 EC1 / NF EN 199113/NA  France: Generating snow loads on 2 closed building with gutters. (TTAD #12808) ................................................................................................................................................................ 362 3.30 EC1 / NF EN 199113/NA  France: Generating snow loads on two side by side buildings with gutters (TTAD #12806) .................................................................................................................................................... 362 3.31 EC1 / NF EN 199113/NA  France: Generating snow loads on 2 closed building with gutters. (TTAD #12835) ................................................................................................................................................................ 362 3.32 EC1 / NF EN 199113/NA  France: Generating snow loads on a 4 slopes with gutters building. (TTAD #12719) ................................................................................................................................................................ 362 3.33 EC1 / NF EN 199113/NA  France: Generating snow loads on 2 closed building with gutters. (TTAD #12841) ................................................................................................................................................................ 362 3.34 EC1 / NF EN 199113/NA  France: Generating snow loads on a 2 slope building with gutters and parapets. (TTAD #12878) ..................................................................................................................................... 363 3.35 EC1 / NF EN 199114/NA  France: Wind load generation on a high building with double slope roof using different parameters defined per directions (DEV2013#4.2) ................................................................................ 363 3.36 EC1 / NF EN 199114/NA  France: Wind load generation on a high building with a horizontal roof using different CsCd values for each direction (DEV2013#4.4) ..................................................................................... 363 3.37 EC1 / BS EN 199114  United Kingdom: Wind load generation on a high building with horizontal roof (DEV2013#4.1) (TTAD #12608) ........................................................................................................................... 363 3.38 EC1 / NF EN 199114/NA  France: Generating Cf and Cp,net wind loads on an isolated roof with double slope (DEV2013#4.3) ........................................................................................................................................... 364 3.39 EC1 / NF EN 199114/NA  France: Generating Cf and Cp,net wind loads on an isolated roof with one slope (DEV2013#4.3) ........................................................................................................................................... 364 3.40 EC1 / NF EN 199113/NA  France: Generating snow loads on a single slope with lateral parapets (TTAD #12606) ................................................................................................................................................................ 364 12
ADVANCE DESIGN VALIDATION GUIDE
3.41
NV2009  France: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604) .364
3.42 EC1 / NF EN 199113/NA  France: Generating snow loads on a 4 slopes with gutters building (TTAD #12716) ................................................................................................................................................................365 3.43 EC1 / NF EN 199114/NA  France: Generating wind loads on a square based lattice structure with compound profiles and automatic calculation of "n" (TTAD #12744) ....................................................................365 3.44 EC1 / BS EN 199114  United Kingdom: Generating wind loads on a square based structure (TTAD #12608) ................................................................................................................................................................365 3.45 EC1 / NF EN 199113/NA  France: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ................................................................................................................................................................365 3.46 EC1 / NF EN 199113/NA  France: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ................................................................................................................................................................366 3.47 EC1 / NF EN 199113/NA  France: Generating snow loads on a 3 slopes 3D portal frame with parapets (TTAD #11111) .....................................................................................................................................................366 3.48 EC1 / NF EN 199114/NA  France: Verifying the geometry of wind loads on an irregular shed (TTAD #12233) ................................................................................................................................................................366 3.49 EC1 / NF EN 199114/NA  France: Verifying the wind loads generated on a building with protruding roof (TTAD #12071, #12278) .......................................................................................................................................366 3.50 EC1 / NF EN 199113/NA  France: Generating snow loads on a 2 slopes 3D portal frame (VT : 3.4 Snow  Example A) ...............................................................................................................................................366 3.51 EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 slopes 3D portal frame (VT : 3.3  Wind  Example C) .........................................................................................................................................................367 3.52 EC1 / NF EN 199114/NA  France: Generating wind loads on a 3D portal frame with one slope roof (VT : 3.2  Wind  Example B)........................................................................................................................................367 3.53 EC1 / NF EN 199114/NA  France: Wind loads on a triangular based lattice structure with compound profiles and user defined "n" (TTAD #12276) .......................................................................................................367 3.54 EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 slopes 3D portal frame (VT : 3.1  Wind  Example A) .........................................................................................................................................................367 3.55 EC1 / NF EN 199114/NA  France: Generating wind loads on a triangular based lattice structure with compound profiles and automatic calculation of "n" (TTAD #12276) ....................................................................367 3.56
NV2009  France: Verifying wind and snow reports for a protruding roof (TTAD #11318) ........................368
3.57
EC1 / SR EN 199114/NB  Romania: Generating the description of climatic loads report (TTAD #11688) .... 368
3.58
EC1 / CR 113/2012  Romania: Generating snow loads on a 2 slopes 3D portal frame (TTAD #11570)368
3.59 EC1 / NF EN 199113/NA  France: Generating snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height (TTAD #11943) ...................................................................................368 3.60 EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls (TTAD #11937) ........................................................................................................................369 3.61
EC1 / CR 114/2012  Romania: Generating wind loads on a 2 slopes 3D portal frame (TTAD #11687) 369
3.62
EC1 / CR 113/2012  Romania: Generating snow loads on a 2 slopes 3D portal frame (TTAD #11569)369
3.63
EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 slopes 3D portal frame (TTAD #11699) .. 369
3.64
EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 slopes 3D portal frame (TTAD #11531) ...370
3.65 NV2009  France: Generating wind loads and snow loads on a simple structure with planar support (TTAD #11380) ................................................................................................................................................................370 3.66 EC1 / NF EN 199113/NA  France: Verifying the snow loads generated on a monopitch frame (TTAD #11302) ................................................................................................................................................................370 3.67 EC1 / DIN EN 199113/NA  Germany: Generating snow loads on two side by side roofs with different heights (DEV2012 #3.13) .....................................................................................................................................370
ADVANCE DESIGN VALIDATION GUIDE
3.68
EC1 / DIN EN 199113/NA  Germany: Generating wind loads on a 55m high structure (DEV2012 #3.12) .... 371
3.69 EC1 / DIN EN 199113/NA  Germany: Generating snow loads on monopitch multispan roofs (DEV2012 #3.13) 371 3.70 EC1 / NF EN 199113/NA  France: Generating snow loads on a 2 slopes 3D portal frame with gutter (TTAD #11113) .................................................................................................................................................... 371 3.71 EC1 / DIN EN 199113/NA  Germany: Generating snow loads on duopitch multispan roofs (DEV2012 #3.13) 371 3.72 EC1 / CSN EN 199114/NA  Czech Republic: Generating wind loads on double slope 3D portal frame (DEV2012 #3.18).................................................................................................................................................. 372 3.73 EC1 / CSN EN 199113/NA  Czech Republic: Generating snow loads on two close roofs with different heights (DEV2012 #3.18) ..................................................................................................................................... 372 3.74 EC1 / NF EN 199113/NA  France: Snow on a 3D portal frame with horizontal roof and parapet with height reduction (TTAD #11191) .......................................................................................................................... 372 3.75 EC1 / NF EN 199113/NA  France: Generating snow loads on a 3D portal frame with a roof which has a small span (< 5m) and a parapet (TTAD #11735) ................................................................................................ 372 3.76 EC1 / EN 199114  General: Wind load generation on simple 3D portal frame with 4 slopes roof (TTAD #11604) ................................................................................................................................................................ 373 3.77 EC1 / NF EN 199114/NA  France: Generating wind loads on double slope 3D portal frame with a fully opened face (DEV2012 #1.6) ............................................................................................................................... 373 3.78 EC1 / NF EN 199114/NA  France: Generating wind loads on duopitch multispan roofs with pitch < 5 degrees (TTAD #11852) ....................................................................................................................................... 373 3.79
EC1 / EN 199114  General: Wind load generation on a signboard ....................................................... 373
3.80 EC1 / NF EN 199114/NA  France: Generating wind loads on a 3D portal frame with 2 slopes roof (TTAD #11932) ................................................................................................................................................................ 374 3.81
EC1 / EN 199114  General: Wind load generation on a high building with horizontal roof .................... 374
3.82
EC1 / EN 199114  General: Wind load generation on a building with multispan roofs .......................... 374
3.83
EC1 / EN 199114  General: Wind load generation on a simple 3D structure with horizontal roof ......... 374
3.84
CR113/2012  Romania  2D Climatic generator on a portal frame with big slopes ................................ 374
3.85 EC1 / NF EN 199114/NA  France: Generating wind loads on an isolated roof with two slopes (TTAD #11695) ................................................................................................................................................................ 375 3.86
NTC 2008: Wind load generation on a double slope 3D portal frame (DEV2015#6) (TTAD #15660) ....... 375
3.87 EC1 / EN 199114  General: Wind load generation on a simple 3D portal frame with 2 slopes roof (TTAD #11602) ................................................................................................................................................................ 375 3.88
NF EN 199114/NA: Wind generation on a 2 slope building with one awning (TTAD #13999) ................ 375
3.89 NTC 2008: Wind and snow load generation on a one slope compound next to a higher single slope compound (DEV2015#6) (TTAD #15425) ............................................................................................................ 375 3.90
CR113/2012  Romania  Snow with adjacent roof, big slope................................................................. 376
3.91
EC1 / DIN EN 199113/NA  Germany: Generating wind loads on an antenna tower (TTAD #15493) .... 376
3.92 NF EN 199113/NA: Snow generation on a slope compound next to a higher single slope compound (TTAD #15923) .................................................................................................................................................... 376
14
3.93
CR113/2012  Romania  Snow accumulation on multiple roofs............................................................. 376
3.94
CR114/2012  Romania  Wind on signboards ....................................................................................... 377
3.95
EC1 / NF EN 199114/NA  France: Generating 2D wind loads on a 2 slope portal (TTAD #14531)....... 377
3.96
NTC 2008: Wind load generation on a higher double slope 3D portal frame (DEV2015#6) (TTAD #15698) ....... 377
3.97
CR113/2012  Romania: Snow loads values generated on a symetrical duopitch roof ......................... 377
ADVANCE DESIGN VALIDATION GUIDE
3.98
CR114/2012  Romania  The wind load position for cannopy roofs, one slope (TTAD #16230) ............377
3.99 EC1 / NF EN 199114/NA  France: Generating wind loads on a double slope with 5 degrees (TTAD #15307) ................................................................................................................................................................378 3.100
EC1 / NF EN 199114/NA  France: Generating 2D wind loads on a multiple roof portal (TTAD #15140) .... 378
3.101 EC1 / NF EN 199114/NA  France: Generating wind loads on a 2 horizontal slopes building one higher that the other (TTAD #13320) ...............................................................................................................................378 3.102 EC1 / NF EN 199113/NA  France: Generating 2D snow loads on a one horizontal slope portal (TTAD #14975) 378 3.103
EC1 / NF EN 199114/NA  France: Generating 2D wind loads on a 2 slope isolated roof (TTAD #14985).. 378
3.104 EC1 / NF EN 199114/NA + NF EN 199113/NA  France: Generating 2D wind and snow loads on a 2 opposite slopes portal with Z down axis (TTAD #15094) ......................................................................................379 3.105 EC1 / NF EN 199114/NA + NF EN 199113/NA  France: Generating 2D wind and snow loads on a 4 slope shed next to a higher one slope compound (TTAD #15047) .......................................................................379 3.106
EC1 / NF EN 199114/NA  France: Generating wind loads on a 3 compound building (TTAD #12883)379
3.107 EC1 / NF EN 199114/NA  France: Generating 2D wind loads on a double slope roof with an opening (TTAD #15328) .....................................................................................................................................................379 3.108
EC1 / NF EN 199113/NA  France: Generating snow loads on a flat roof .............................................380
3.109 EC1 / NF EN 199113/NA  France: Snow load generation on double compound with gutters and parapets on all sides (TTAD #13717) ...................................................................................................................383 3.110 EC1 / NF EN 199113/NA  France: Generating snow loads on a 3D portal frame with horizontal roof and gutter (TTAD #11113) ...........................................................................................................................................383
4
COMBINATIONS ................................................................................................................. 385 4.1 EC0 / NF EN 1990  France: Generating the concomitance matrix after adding a new dead load case (TTAD #11361) .....................................................................................................................................................386 4.2
NF EN 1990/NA  France: Generating a set of combinations with seismic group of loads (TTAD #11889) 386
4.3 EC0 / EN 1990  General: Generating load combinations with unfavorable and favorable/unfavorable predominant action (TTAD #11357) .....................................................................................................................386 4.4
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.10 (DEV2012 #1.7) ...386
4.5
CR 02012  Romania  Automatic combinations check according with the romanian code .......................387
4.6
Verifying the combinations description report (TTAD #11632) ...................................................................387
4.7
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.8 (DEV2012 #1.7) .....387
4.8
EC0 / EN 1990  General: Generating combinations (TTAD #11673).........................................................387
4.9
EC0 / EN 1990  General: Defining concomitance rules for two case families (TTAD #11355) ..................388
4.10 EC0 / EN 1990  General: Generating load combinations after changing the load case number (TTAD #11359) ................................................................................................................................................................388 4.11
EC0 / EN 1990  General: Generating combinations for NEWEC8.cbn (TTAD #11431) ...........................388
4.12 NF EN 1990/NA  France: Generating the concomitance matrix after switching back the effect for live load (TTAD #11806) .....................................................................................................................................................388 4.13
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.9 (DEV2012 #1.7) ...389
4.14
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.7 (DEV2012 #1.7) ... 389
4.15
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.3 (DEV2012 #1.7) ... 390
4.16
NF EN 1990/NA  France: Performing the combinations concomitance standard test no. 5 (DEV2012 #1.7) .. 390
4.17
NF EN 1990/NA  France: Generating a set of combinations with different Q "Base" types (TTAD #11806).... 391
ADVANCE DESIGN VALIDATION GUIDE
5
4.18
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.6 (DEV2012 #1.7) ... 391
4.19
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.2 (DEV2012 #1.7) ... 392
4.20
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.4 (DEV2012 #1.7) .. 392
4.21
NF EN 1990/NA  France: Generating a set of combinations with Q group of loads (TTAD #11960) ............... 392
4.22
NF EN 1990/NA  France: Performing the combinations concomitance standard test no.1 (DEV2010#1.7) .... 393
4.23
CSN EN 1990/NA  Czech Republic: Verifying combinations for CZ localization (TTAD #12542) ............ 393
CONCRETE DESIGN .......................................................................................................... 395 5.1 EC2 / NF EN 199211/NA  France: Verifying the longitudinal reinforcement area of a beam under a linear load  bilinear stressstrain diagram ..................................................................................................................... 396 5.2 EC2 / NF EN 199211/NA  France: Verifying the longitudinal reinforcement area of a beam under a linear load  horizontal level behavior law ...................................................................................................................... 396 5.3 EC2 / NF EN 199211/NA  France: Verifying the transverse reinforcement area for a beam subjected to linear loads ........................................................................................................................................................... 396 5.4 EC2 / NF EN 199211/NA  France: Verifying the longitudinal reinforcement area of a beam under a linear load  inclined stress strain behavior law .............................................................................................................. 396 5.5 EC2 / NF EN 199211/NA  France: Verifying the longitudinal reinforcement area for a beam subjected to point loads ............................................................................................................................................................ 397 5.6
EC2 / NF EN 199211/NA  France: Verifying the minimum reinforcement area for a simply supported beam . 397
5.7
EC2 / NF EN 199211/NA  France: Verifying the longitudinal reinforcement area of a beam under a linear load397
5.8 Testing the punching verification and punching reinforcement results on loaded analysis model (TTAD #14332) ................................................................................................................................................................ 397 5.9 EC2 / NF EN 199211/NA  France: Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular cross section .............................................................................................................................. 398 5.10
Verifying Aty and Atz for a fixed concrete beam (TTAD #11812) .............................................................. 398
5.11
EC2 / NF EN 199211/NA  France: Verifying concrete results for linear elements (TTAD #11556)........ 398
5.12
Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678) ......................... 398
5.13
Verifying the longitudinal reinforcement for linear elements (TTAD #11636) ............................................ 399
5.14
Verifying the reinforcement of concrete columns (TTAD #11635)............................................................. 399
5.15
EC2 / NF EN 199211/NA  France: Verifying concrete results for planar elements (TTAD #11583) ...... 399
5.16
Verifying the reinforced concrete results on a fixed beam (TTAD #11836) ............................................... 399
5.17
Verifying the longitudinal reinforcement for a fixed linear element (TTAD #11700) .................................. 400
5.18
Verifying the reinforced concrete results on a structure with 375 load cases combinations (TTAD #11683) .... 400
5.19
EC2 / NF EN 199211/NA  France: Calculation of a square column in traction (TTAD #11892)............. 400
5.20
Verifying the minimum transverse reinforcement area results for an articulated beam (TTAD #11342) ... 401
5.21 EC2 / NF EN 199211/NA: Verifying the minimum transverse reinforcement area results for articulated beams (TTAD #11342) ......................................................................................................................................... 401 5.22 EC2/NF EN 199211/NA  France: Stresses and cracks verification for a planar element hinged on all edges 401 5.23 NTC 2008  Italy: Verifying the stresses result for a rectangular concrete beam subjected to a uniformly distributed load ..................................................................................................................................................... 402 5.24
EC2 / EN 199211  Punching verification with imposed reinfrocement on slab ...................................... 402
5.25 NTC 2008  Italy: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load ..................................................................................................................................................... 402 16
ADVANCE DESIGN VALIDATION GUIDE
5.26 Verifying the bending moments about X and Y axis on a single story concrete core subjected to horizontal forces and seismic action .....................................................................................................................................402 5.27 plate
EC2 / EN 199211  Germany: Two field beam with rectangular cross section, calculated as a one span 402
5.28 NTC 2008  Italy: Verifying a square concrete column of a multistorey structure subjected to axial compression .........................................................................................................................................................403 5.29 EC2 / NF EN 199211/NA  France: Comparing deflection of a slab hinged supported on 2 edges with the deflection of the equivalent beam approach .........................................................................................................403 5.30
EC2 / EN 199211  Germany: Single beam with a plate crosssection ....................................................403
5.31
EC2 / EN 199211  Germany: Single beam with rectangular cross section ............................................403
5.32 EC2 / EN 199211  Germany: Two field planar system with two span direction calculation based on the Baumann Method .................................................................................................................................................404 5.33 EC2 / EN 199211  Germany: Two field planar system with two span direction calculation based on the Capra Method .......................................................................................................................................................404 5.34 Verifying the peak smoothing influence over mesh, the punching verification and punching reinforcement results when Z down axis is selected (TTAD #14963) ..........................................................................................404 5.35
EC2,EC8 / NF EN 199211/NA  France: Verifying the capacity design results (DEV2013 #8.3).............404
5.36 EC2 / NF EN 199211/NA  France: Column design with “Nominal Stiffness method” square section (TTAD #11625) .....................................................................................................................................................405 5.37 EC2 / NF EN 199211/NA  France: Verifying a rectangular cross section beam made from concrete C25/30 to resist simple bending  Bilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 1) 406 5.38 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcement Bilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 5) ......................................................411 5.39 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcement Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 10) ...................................................416 5.40 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to uniformly distributed load, without compressed reinforcement Bilinear stressstrain diagram (Class XD3) (evaluated by SOCOTEC France  ref. Test 12) .........................................................................................................................422 5.41 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcementBilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 15) ....................................................428 5.42 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcementBilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 16) ....................................................435 5.43 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement Bilinear stressstrain diagram (Class XD1) (evaluated by SOCOTEC France  ref. Test 11) .........................................................................................................................442 5.44 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement Bilinear stressstrain diagram (Class XD1) (evaluated by SOCOTEC France  ref. Test 14) .........................................................................................................................448 5.45 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement  Bilinear stressstrain diagram (Class XD1) (evaluated by SOCOTEC France  ref. Test 18) .........................................................................................................................455 5.46 EC2/NF EN 199211/NA  France: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement  Bilinear stressstrain diagram(Class XD1)(evaluated by SOCOTEC France  ref Test19) .............................................................................................460 5.47 EC2/NF EN 199211/NA  France: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load without compressed reinforcement  Bilinear stressstrain diagram(Class XD1)(evaluated by SOCOTEC France  ref.Test 20) ............................................................................................467 5.48 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam with compressed reinforcement – Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 9) ...........................474
ADVANCE DESIGN VALIDATION GUIDE
5.49 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement  Bilinear stressstrain diagram (Class XD1) (evaluated by SOCOTEC France  ref. Test 13) ......................................................................................................................... 484 5.50 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement  Inclined stressstrain diagram (Class XD1) (evaluated by SOCOTEC France  ref. Test 17) ......................................................................................................................... 490 5.51 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France ref. Test 24) .......................................................................................................................................................... 496 5.52 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with inclined transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France ref. Test 25) .......................................................................................................................................................... 500 5.53 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a T concrete beam  Bilinear stressstrain diagram (Class X0) (evaluated by SOCOTEC France  ref. Test 28) ......................................................... 504 5.54 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement  Inclined stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 29) 508 5.55 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete section Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 23) ............................... 512 5.56 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France ref. Test 27) .......................................................................................................................................................... 517 5.57 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to compression and rotation moment to the top  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 31) ................................................................................................................................................................ 521 5.58 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 32) .................................................................................................. 534 5.59 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to compression by nominal rigidity method Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 33) 544 5.60 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France ref. Test 26) .......................................................................................................................................................... 552 5.61 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 30) 556 5.62 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column using the method based on nominal curvature Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 36) 560 5.63 EC2 / NF EN 199211/NA  France: Verifying a square concrete column using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 37) 567 5.64 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to a small compression force and significant rotation moment to the top  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 40) ......................................................................................................................... 571 5.65 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to a significant compression force and small rotation moment to the top  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 41) .................................................................................................................... 578 5.66 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column subjected to compression to top – Based on nominal rigidity method  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 35)............................................................................................................................................ 591 5.67 EC2 / NF EN 199211/NA  France: Verifying a circular concrete column using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 39) 604
18
ADVANCE DESIGN VALIDATION GUIDE
5.68 EC2/NF EN 199211/NA  France: Verifying a square concrete column subjected to a small rotation moment and significant compression force to the top with Nominal Curvature MethodBilinear stressstrain diagram(Class XC1)(SOCOTEC FranceTest 43) ................................................................................................608 5.69 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to eccentric loading  Bilinear stressstrain diagram (Class X0) (evaluated by SOCOTEC France  ref. Test 44) ................................618 5.70 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam supporting a balcony  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 45) ............................................624 5.71 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column subjected to compression on the top – Method based on nominal stiffness  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 34) .........................................................................................................................631 5.72 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 38) 639 5.73 EC2/NF EN 199211/NAFrance: Verifying a square concrete column subjected to a significant rotation moment and small compression force to the top with Nominal Curvature MethodBilinear stressstrain diagram(Class XC1)(SOCOTEC FranceTest 42) ................................................................................................642 5.74 EC2 / NF EN 199211/NA  France: Verifying a square concrete beam subjected to a normal force of traction  Bilinear stressstrain diagram (Class X0) (evaluated by SOCOTEC France  ref. Test 46 II) ................651 5.75
NTC 2008  Italy: Longitudinal reinforcement on a simply supported beam ..............................................655
5.76
EC2 / CSN EN 199211  Czech Republic: Main reinforcement design for simple supported beam ........657
5.77
EC2 / NF EN199211  France: Verifying a T concrete section  Inclined stressstrain diagram (Class XC3).. 660
5.78 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to Pivot B efforts – Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 4 II) ..................................................662 5.79 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to tension load Bilinear stressstrain diagram (Class XD2) (evaluated by SOCOTEC France  ref. Test 47I) ..............................667 5.80 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a tension distributed load  Bilinear stressstrain diagram (Class XD2) (evaluated by SOCOTEC France  ref. Test 47 II) .673 5.81
EC2 / NF EN 199211  France: Reinforcement calculation for a simple bending beam ..........................676
5.82
A23.304  Canada: Verifying the longitudinal reinforcement value on a oneway reinforced concrete slab ..... 680
5.83 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement  Bilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 2) ..............................................................................................................................................683 5.84 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcementBilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 6) ......................................................690 5.85 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcementBilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 7) ......................................................694 5.86 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam without compressed reinforcement – Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 8) ...........................698 5.87 EC2 / NF EN 199211/NA  France: Verifying a square concrete beam subjected to a normal force of traction  Inclined stressstrain diagram (Class X0) (evaluated by SOCOTEC France  ref. Test 46 I) .................705 5.88 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to uniformly distributed load, with compressed reinforcement Bilinear stressstrain diagram (evaluated by SOCOTEC France ref. Test 3) ............................................................................................................................................................708 5.89 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 4 I) ...................................................718 5.90
EC2 / NF EN 199211  France: Reinforcement calculation on intermediate support on a continuous beam.. 724
5.91 EC3 / CSN EN 199211  Czech Republic: Verification of stresses in steel and concrete on simply supported beam ....................................................................................................................................................726 5.92
EC2 / NF EN 199211  France: Tie reinforcement ..................................................................................731
ADVANCE DESIGN VALIDATION GUIDE
5.93 NTC 2008  Italy: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement ....................................................................................................................................................... 735
6
5.94
NTC 2008  Italy: Verifying a rectangular concrete beam subjected to a uniformly distributed load ......... 738
5.95
EC2 / CSN EN 199211  Czech Republic: Stresses and cracks verification for simple supported beam 742
GENERAL APPLICATIONS ................................................................................................ 749 6.1
Verifying element creation using commas for coordinates (TTAD #11141) ................................................ 750
6.2
Modifying the "Design experts" properties for concrete linear elements (TTAD #12498) ........................... 750
6.3
Verifying the precision of linear and planar concrete covers (TTAD #12525) ............................................. 750
6.4
Defining the reinforced concrete design assumptions (TTAD #12354) ...................................................... 750
6.5
Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238)................................ 750
6.6
Creating system trees using the copy/paste commands (DEV2012 #1.5) .................................................. 750
6.7
Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724) ........................................ 751
6.8
Changing the default material (TTAD #11870) ........................................................................................... 751
6.9
Creating and updating model views and postprocessing views (TTAD #11552) ....................................... 751
6.10
Importing a cross section from the Advance Steel profiles library (TTAD #11487) ................................... 751
6.11
Verifying the synthetic table by type of connection (TTAD #11422) .......................................................... 751
6.12
Verifying mesh, CAD and climatic forces  LPM meeting .......................................................................... 752
6.13
Verifying the appearance of the local x orientation legend (TTAD #11737) .............................................. 752
6.14
Creating system trees using the copy/paste commands (DEV2012 #1.5) ................................................ 752
6.15
Verifying the objects rename function (TTAD #12162) ............................................................................. 752
6.16
Launching the verification of a model containing steel connections (TTAD #12100) ................................ 752
6.17 Creating a new Advance Design file using the "New" command from the "Standard" toolbar (TTAD #12102) ................................................................................................................................................................ 753
7
20
6.18
Verifying geometry properties of elements with compound cross sections (TTAD #11601) ..................... 753
6.19
Verifying material properties for C25/30 (TTAD #11617) .......................................................................... 753
6.20
Verifying the overlapping when a planar element is built in a hole (TTAD #13772) .................................. 753
IMPORT / EXPORT ............................................................................................................. 755 7.1
Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952) ........................................ 756
7.2
Exporting an Advance Design model to DO4 format (DEV2012 #1.10) ..................................................... 756
7.3
Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172) ................. 756
7.4
Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ........................................................ 756
7.5
Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ........................................................ 756
7.6
Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197) ... 757
7.7
System stability when importing AE files with invalid geometry (TTAD #12232) ........................................ 757
7.8
Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197) ... 757
7.9
Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9) ........... 757
7.10
Exporting linear elements to IFC format (TTAD #10561) .......................................................................... 757
7.11
Verifying the load case properties from models imported as GTC files (TTAD #12306) ........................... 758
ADVANCE DESIGN VALIDATION GUIDE
7.12
Importing IFC files containing continuous foundations (TTAD #12410) .....................................................758
7.13 Verifying the releases option of the planar elements edges after the model was exported and imported via GTC format (TTAD #12137) .................................................................................................................................758
8
7.14
Importing GTC files containing "PH.RDC" system (TTAD #12055) ...........................................................758
7.15
Exporting a meshed model to GTC (TTAD #12550) ..................................................................................758
JOINT DESIGN .................................................................................................................... 759 8.1
Deleting a welded tube connection  1 gusset bar (TTAD #12630).............................................................760
8.2
Creating connections groups (TTAD #11797) ............................................................................................760
1 Finite element method
ADVANCE DESIGN VALIDATION GUIDE
1.1
Thin lozengeshaped plate fixed on one side (alpha = 15 °) (010008SDLSB_FEM) Test ID: 2440 Test status: Passed
1.1.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozengeshaped plate fixed on one side, subjected to its own weight only.
1.1.2 Background 1.1.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozengeshaped plate fixed on one side
Scale =1/10 010008SDLSB_FEM
Units I. S. Geometry
24
■ ■
Thickness: t = 0.01 m, Side: a = 1 m,
■ ■
α = 15° Points coordinates: ► A(0;0;0) ► B(a;0;0) ► C ( 0.259a ; 0.966a ; 0 ) ► D ( 1.259a ; 0.966a ; 0 )
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: AB side fixed, Inner: None.
Loading ■ ■
1.1.2.2
External: None, Internal: None.
Eigen modes frequencies function by α angle
Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies: fj =
1 ⋅ λi2 2 2π ⋅ a
Et 2 12ρ(1 − ν 2 )
where i = 1,2, or λi2 = g(α).
α = 15° λ12 3.601 λ22 8.872 M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
Eigen mode shapes
25
ADVANCE DESIGN VALIDATION GUIDE
1.1.2.3
Theoretical results
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
8.999
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
22.1714
1.1.3 Calculated results Result name
26
Result description
Value
Error
Eigen mode 1 frequency [Hz]
8.95 Hz
0.54%
Eigen mode 2 frequency [Hz]
21.69 Hz
2.17%
ADVANCE DESIGN VALIDATION GUIDE
1.2
Thin lozengeshaped plate fixed on one side (alpha = 30 °) (010009SDLSB_FEM) Test ID: 2441 Test status: Passed
1.2.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozengeshaped plate fixed on one side, subjected to its own weight only.
1.2.2 Background 1.2.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozengeshaped plate fixed on one side
Scale =1/10 010009SDLSB_FEM
Units I. S. Geometry ■ ■
Thickness: t = 0.01 m, Side: a = 1 m,
■ ■
α = 30° Points coordinates: ► A(0;0;0) ► B(a;0;0) ►
3 C ( 0.5a ; 2 a ; 0 ) A
A
EA
E
A
E
►
3 D ( 1.5a ; 2 a ; 0 ) A
A
EA
E
A
E
27
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: AB side fixed, Inner: None.
Loading ■ ■
1.2.2.2
External: None, Internal: None.
Eigen mode frequencies relative to the α angle
Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies: fj =
1 ⋅ λi2 2 2π ⋅ a
Et 2 12ρ(1 − ν 2 )
where i = 1,2, or λi2 = g(α).
α = 30° λ12 3.961 λ22 10.19 M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
Eigen mode shapes
28
ADVANCE DESIGN VALIDATION GUIDE
1.2.2.3
Theoretical results
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
9.8987
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
25.4651
1.2.3 Calculated results Result name
Result description
Value
Error
Eigen mode 1 frequency [Hz]
9.82 Hz
0.80%
Eigen mode 2 frequency [Hz]
23.44 Hz
7.95%
29
ADVANCE DESIGN VALIDATION GUIDE
1.3
Circular plate under uniform load (010003SSLSB_FEM) Test ID: 2435 Test status: Passed
1.3.1 Description On a circular plate of 5 mm thickness and 2 m diameter, an uniform load, perpendicular on the plan of the plate, is applied. The vertical displacement on the plate center is verified.
1.3.2 Background 1.3.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 03/89; Analysis type: linear static; Element type: planar. Circular plate under uniform load 010003SSLSB_FEM
Units I. S. Geometry ■ ■
Circular plate radius: r = 1m, Circular plate thickness: h = 0.005 m.
Materials properties
30
■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Scale =1/10
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
Outer: Plate fixed on the side (in all points of its perimeter), For the modeling, we consider only a quarter of the plate and we impose symmetry conditions on some nodes (see the following model; yz plane symmetry condition):translation restrained nodes along x and rotation restrained nodes along y and z: translation restrained nodes along x and rotation restrained nodes along y and z:
■
Inner: None.
Loading ■ ■
1.3.2.2
External: Uniform loads perpendicular on the plate: pZ = 1000 Pa, Internal: None.
Vertical displacement of the model at the center of the plate
Reference solution Circular plates form: pr4 1000 x 14 u = 64D = 64 x 2404 =  6.50 x 103 m A
E
A
A
E
A
with the plate radius coefficient: D =
Eh3 2.1 x 1011 x 0.0053 2 = 12(10.32) 12(1ν ) A
E
A
A
E
D = 2404 Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 70 nodes, 58 planar elements. Circular plate under uniform load Meshing
Scale =1.5
010003SSLSB_FEM
31
ADVANCE DESIGN VALIDATION GUIDE
Deformed shape Circular plate under uniform load Deformed
1.3.2.3
Scale =1.5
010003SSLSB_FEM
Theoretical results
Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement on the plate center [mm]
6.50
1.3.3 Calculated results
32
Result name
Result description
Value
Error
DZ
Vertical displacement on the plate center [mm]
6.47032 mm
0.46%
ADVANCE DESIGN VALIDATION GUIDE
1.4
Vibration mode of a thin piping elbow in plane (case 2) (010012SDLLB_FEM) Test ID: 2444 Test status: Passed
1.4.1 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.
1.4.2 Background 1.4.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (plane problem); Element type: linear. Vibration mode of a thin piping elbow Case 2
Scale = 1/11 010012SDLLB_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 104 m2, 9 4 Flexure moment of inertia relative to the yaxis: Iy = 4.637 x 10 m , 9 4 Flexure moment of inertia relative to zaxis: Iz = 4.637 x 10 m ,
33
ADVANCE DESIGN VALIDATION GUIDE
■ ■
Polar inertia: Ip = 9.274 x 109 m4. Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0) ► C ( L ; R ; 0 ) ► D ( R ; L ; 0 )
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■
Outer: Fixed at points C and D At A: translation restraint along y and z, ► At B: translation restraint along x and z, Inner: None. ► ►
■
Loading ■ ■
1.4.2.2
External: None, Internal: None.
Eigen mode frequencies
Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =
in plane bending:
λi
2
2π ⋅ R
2
EI z ρA
where i = 1,2,
Finite elements modeling ■ ■ ■
34
Linear element: beam, 23 nodes, 22 linear elements.
ADVANCE DESIGN VALIDATION GUIDE
Eigen mode shapes
1.4.2.3
Theoretical results
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode frequency in plane 1 [Hz]
94
CM2
Eigen mode
Eigen mode frequency in plane 2 [Hz]
180
1.4.3 Calculated results Result name
Result description
Value
Error
Eigen mode frequency in plane 1 [Hz]
94.62 Hz
0.66%
Eigen mode frequency in plane 2 [Hz]
184.68 Hz
2.60%
35
ADVANCE DESIGN VALIDATION GUIDE
1.5
Vibration mode of a thin piping elbow in plane (case 3) (010013SDLLB_FEM) Test ID: 2445 Test status: Passed
1.5.1 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.
1.5.2 Background 1.5.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (plane problem); Element type: linear. Vibration mode of a thin piping elbow Case 3
Scale = 1/12 010013SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■ ■
36
Average radius of curvature: OA = R = 1 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, 4 2 Section: A = 1.131 x 10 m , Flexure moment of inertia relative to the yaxis: Iy = 4.637 x 109 m4, 9 4 Flexure moment of inertia relative to zaxis: Iz = 4.637 x 10 m , 9 4 Polar inertia: Ip = 9.274 x 10 m . Points coordinates (in m): ► O(0;0;0)
ADVANCE DESIGN VALIDATION GUIDE
A(0;R;0) B(R;0;0) C ( L ; R ; 0 ) D ( R ; L ; 0 )
► ► ► ►
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■
Outer: Fixed at points C and Ds, At A: translation restraint along y and z, ► At B: translation restraint along x and z, Inner: None. ► ►
■
Loading ■ ■
1.5.2.2
External: None, Internal: None.
Eigen mode frequencies
Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =
in plane bending:
λi
2
2π ⋅ R
2
EI z ρA
where i = 1,2,
Finite elements modeling ■ ■ ■
Linear element: beam, 41 nodes, 40 linear elements.
37
ADVANCE DESIGN VALIDATION GUIDE
Eigen mode shapes
1.5.2.3
Theoretical results
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode frequency in plane 1 [Hz]
25.300
CM2
Eigen mode
Eigen mode frequency in plane 2 [Hz]
27.000
1.5.3 Calculated results Result name
38
Result description
Value
Error
Eigen mode frequency in plane 1 [Hz]
24.96 Hz
1.34%
Eigen mode frequency in plane 2 [Hz]
26.71 Hz
1.07%
ADVANCE DESIGN VALIDATION GUIDE
1.6
Thin lozengeshaped plate fixed on one side (alpha = 0 °) (010007SDLSB_FEM) Test ID: 2439 Test status: Passed
1.6.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozengeshaped plate fixed on one side, subjected to its own weight only.
1.6.2 Background 1.6.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozengeshaped plate fixed on one side
Scale =1/10 010007SDLSB_FEM
Units I. S. Geometry ■ ■
Thickness: t = 0.01 m, Side: a = 1 m,
■ ■
α = 0° Points coordinates: ► A(0;0;0) ► B(a;0;0) ► C(0;a;0) ► D(a;a;0)
39
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: AB side fixed, Inner: None.
Loading ■ ■
1.6.2.2
External: None, Internal: None.
Eigen mode frequencies relative to the α angle
Reference solution M. V. Barton formula for a side "a" lozenge, leads to the frequencies: fj =
1 ⋅ λi2 2 2π ⋅ a
Et 2 12ρ(1 − ν 2 )
where i = 1,2, and λi2 = g(α).
α=0 λ12 3.492 λ22 8.525 M.V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 61 nodes, 900 surface quadrangles.
Eigen mode shapes
40
ADVANCE DESIGN VALIDATION GUIDE
1.6.2.3
Theoretical results
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
8.7266
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
21.3042
1.6.3 Calculated results Result name
Result description
Value
Error
Eigen mode 1 frequency [Hz]
8.67 Hz
0.65%
Eigen mode 2 frequency [Hz]
21.21 Hz
0.44%
41
ADVANCE DESIGN VALIDATION GUIDE
1.7
Vibration mode of a thin piping elbow in plane (case 1) (010011SDLLB_FEM) Test ID: 2443 Test status: Passed
1.7.1 Description Verifies the vibration modes of a thin piping elbow (1 m radius) with fixed ends and subjected to its self weight only.
1.7.2 Background 1.7.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (plane problem); Element type: linear. Vibration mode of a thin piping elbow in plane Case 1
Scale = 1/7 010011SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■
42
Average radius of curvature: OA = R = 1 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, Section: A = 1.131 x 104 m2, 9 4 Flexure moment of inertia relative to the yaxis: Iy = 4.637 x 10 m , 9 4 Flexure moment of inertia relative to zaxis: Iz = 4.637 x 10 m , 9 4 Polar inertia: Ip = 9.274 x 10 m .
ADVANCE DESIGN VALIDATION GUIDE
■
Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0)
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: Fixed at points A and B , Inner: None.
Loading ■ ■
1.7.2.2
External: None, Internal: None.
Eigen mode frequencies
Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =
in plane bending:
λi
2
2π ⋅ R
2
EI z ρA
where i = 1,2,
Finite elements modeling ■ ■ ■
Linear element: beam, 11 nodes, 10 linear elements.
Eigen mode shapes
43
ADVANCE DESIGN VALIDATION GUIDE
1.7.2.3
Theoretical results
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode frequency in plane 1 [Hz]
119
CM2
Eigen mode
Eigen mode frequency in plane 2 [Hz]
227
1.7.3 Calculated results Result name
44
Result description
Value
Error
Eigen mode frequency in plane 1 [Hz]
120.09 Hz
0.92%
Eigen mode frequency in plane 2 [Hz]
227.1 Hz
0.04%
ADVANCE DESIGN VALIDATION GUIDE
System of two bars with three hinges (010002SSLLB_FEM) Test ID: 2434 Test status: Passed
1.8.1 Description On a system of two bars (AC and BC) with three hinges, a punctual load in applied in point C. The vertical displacement in point C and the tensile stress on the bars are verified.
1.8.2 Background 1.8.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 09/89; Analysis type: linear static; Element type: linear. System of two bars with three hinges
Scale =1/33
0002SSLLB_FEM
B
A
4.5
00
30°
30°
1.8
F
0 .50
m
m
4
C Y Z
X
Units I. S. Geometry ■ ■ ■
Bars angle relative to horizontal: θ = 30°, Bars length: l = 4.5 m, 4 2 Bar section: A = 3 x 10 m .
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa.
45
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Hinged in A and B, Inner: Hinge on C
Loading ■ ■
1.8.2.2
External: Punctual load in C: F = 21 x 103 N. Internal: None.
Displacement of the model in C
Reference solution uc = 3 x 103 m Finite elements modeling ■ ■ ■
Linear element: beam, imposed mesh, 21 nodes, 20 linear elements.
Displacement shape System of two bars with three hinges Displacement in C
1.8.2.3
0002SSLLB_FEM
Bars stresses
Reference solutions σAC bar = 70 MPa σBC bar = 70 MPa Finite elements modeling ■ ■ ■ 46
Linear element: beam, imposed mesh, 21 nodes, 20 linear elements.
Scale =1/33
ADVANCE DESIGN VALIDATION GUIDE
1.8.2.4
Shape of the stress diagram System of two bars with three hinges 0002SSLLB_FEM
Bars stresses
1.8.2.5
Scale =1/34
Theoretical results
Solver
Result name
Result description
Reference value
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
1.8.3 Calculated results Result name
Result description
Value
Error
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%
47
ADVANCE DESIGN VALIDATION GUIDE
1.9
Thin circular ring fixed in two points (010006SDLLB_FEM) Test ID: 2438 Test status: Passed
1.9.1 Description Verifies the first eigen modes frequencies for a thin circular ring fixed in two points, subjected to its own weight only.
1.9.2 Background 1.9.2.1
Model description
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 12/89; Analysis type: modal analysis, plane problem; Element type: linear. Thin circular ring fixed in two points
Scale =1/2 010006SDLLB_FEM
Units I. S. Geometry ■ ■ ■
■
48
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 105 m2, ► Flexure moment of inertia relative to the vertical axis: I = 1.042 x 1010 m4, Point coordinates: ► O (0 ;0),
ADVANCE DESIGN VALIDATION GUIDE
►
A (0.05 3 ; 0.05),
►
B (0.05 3 ; 0.05).
Materials properties ■
Longitudinal elastic modulus: E = 7.2 x 1010 Pa
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 2700 kg/m3.
Boundary conditions ■ ■
Outer: Fixed at A and B, Inner: None.
Loading ■ ■
1.9.2.2
External: None, Internal: None.
Eigen mode frequencies
Reference solutions The deformation of the fixed ring is calculated from the deformations of the freefree thin ring ■
Symmetrical mode: ►
u’i = i cos(iθ)
►
v’i = sin (iθ)
►
■
1i2 θ’i = Error! Bookmark not defined. R sin (iθ) A
A
E
A
Antisymmetrical mode: ►
u’i = i sin(iθ)
►
v’i = cos (iθ)
►
1i2 θ’i = Error! Bookmark not defined. R cos (iθ) A
A
E
A
From Green’s method results: fj =
h 1 ⋅ λj 2π R 2
E 12ρ
with a support angle of 120°.
i Symmetrical mode Antisymmetrical mode
1 4.8497 1.9832
2 14.7614 9.3204
3 23.6157 11.8490
4 21.5545
Finite elements modeling ■ ■ ■
Linear element: beam, without meshing, 32 nodes, 32 linear elements.
49
ADVANCE DESIGN VALIDATION GUIDE
Eigen mode shapes
50
ADVANCE DESIGN VALIDATION GUIDE
1.9.2.3
Theoretic results
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 frequency  1 antisymmetric 1 [Hz]
235.3
CM2
Eigen mode
Eigen mode 2 frequency  2 symmetric 1 [Hz]
575.3
CM2
Eigen mode
Eigen mode 3 frequency  3 antisymmetric 2 [Hz]
1105.7
CM2
Eigen mode
Eigen mode 4 frequency  4 antisymmetric 3 [Hz]
1405.6
CM2
Eigen mode
Eigen mode 5 frequency  5 symmetric 2 [Hz]
1751.1
CM2
Eigen mode
Eigen mode 6 frequency  6 antisymmetric 4 [Hz]
2557
CM2
Eigen mode
Eigen mode 7 frequency  7 symmetric 3 [Hz]
2801.5
1.9.3 Calculated results Result name
Result description
Value
Error
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 3 frequency  3 antisymmetric 2 [Hz]
1112.54 Hz
0.62%
Eigen mode 4 frequency  4 antisymmetric 3 [Hz]
1414.22 Hz
0.61%
Eigen mode 5 frequency  5 symmetric 2 [Hz]
1760 Hz
0.51%
Eigen mode 6 frequency  6 antisymmetric 4 [Hz]
2569.97 Hz
0.51%
Eigen mode 7 frequency  7 symmetric 3 [Hz]
2777.43 Hz
0.86%
51
ADVANCE DESIGN VALIDATION GUIDE
1.10 Thin lozengeshaped plate fixed on one side (alpha = 45 °) (010010SDLSB_FEM) Test ID: 2442 Test status: Passed
1.10.1
Description
Verifies the eigen modes frequencies for a 10 mm thick lozengeshaped plate fixed on one side, subjected to its own weight only.
1.10.2
Background
1.10.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozengeshaped plate fixed on one side
Scale =1/10 010010SDLSB_FEM
Units I. S. Geometry ■ ■
Thickness: t = 0.01 m, Side: a = 1 m,
■ ■
α = 45° Points coordinates: ► A(0;0;0) ► B(a;0;0) ►
52
C(
2 2 a; a;0) 2 2
ADVANCE DESIGN VALIDATION GUIDE
►
D(
2 2+ 2 a; a;0) 2 2
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: AB side fixed, Inner: None.
Loading ■ ■
External: None, Internal: None.
1.10.2.2 Eigen mode frequencies relative to the α angle Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies: fj =
1 ⋅ λi2 2π ⋅ a 2
Et 2 12ρ(1 − ν 2 )
where i = 1,2, or λi2 = g(α).
α = 45° λ1 4.4502 λ22 10.56 M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. 2
Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
53
ADVANCE DESIGN VALIDATION GUIDE
Eigen mode shapes
1.10.2.3 Theoretical results Solver
Result name
Result description
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
11.1212
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
26.3897
1.10.3
Calculated results
Result name
54
Reference value
Result description
Value
Error
Eigen mode 1 frequency [Hz]
11.28 Hz
1.43%
Eigen mode 2 frequency [Hz]
28.08 Hz
6.41%
ADVANCE DESIGN VALIDATION GUIDE
1.11 Double fixed beam (010016SDLLB_FEM) Test ID: 2448 Test status: Passed
1.11.1
Description
Verifies the eigen modes frequencies and the vertical displacement on the middle of a beam consisting of eight elements of length "l", having identical characteristics. A punctual load of 50000 N is applied.
1.11.2
Background
1.11.2.1 Model description ■ ■ ■
Reference: internal GRAITEC test (beams theory); Analysis type: static linear, modal analysis; Element type: linear.
Units I. S. Geometry ■ ■ ■
Length: l = 16 m, 2 Axial section: S=0.06 m 4 Inertia I = 0.0001 m
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7850 kg/m3
Boundary conditions ■ ■
Outer: Fixed at both ends x = 0 and x = 8 m, Inner: None.
Loading ■ ■
External: Punctual load P = 50000 N at x = 4m, Internal: None.
55
ADVANCE DESIGN VALIDATION GUIDE
1.11.2.2 Displacement of the model in the linear elastic range Reference solution The reference vertical displacement v5, is calculated at the middle of the beam at x = 2 m.
v5 =
Pl 3 50000 × 16 3 = = 0.05079 m 192EI 192 × 2.1E11× 0.0001
Finite elements modeling ■ ■ ■
Linear element: beam, imposed mesh, 9 nodes, 8 elements.
Deformed shape Double fixed beam Deformed
1.11.2.3 Eigen mode frequencies of the model in the linear elastic range Reference solution Knowing that the first four eigen mode frequencies of a double fixed beam are given by the following formula:
fn =
2 n
χ 2.π .L2
χ 12 2 χ 2 E.I where for the first 4 eigen modes frequencies 2 ρ.S χ 3 2 χ 4
Finite elements modeling ■ ■ ■
56
Linear element: beam, imposed mesh, 9 nodes, 8 elements.
= 22.37 → f1 = 2.937 Hz = 61.67 → f 2 = 8.095 Hz = 120.9 → f 3 = 15.871 Hz = 199.8 → f 4 = 26.228 Hz
ADVANCE DESIGN VALIDATION GUIDE
Modal deformations Double fixed beam Mode 1
Double fixed beam Mode 2
Double fixed beam Mode 3
Double fixed beam Mode 4
57
ADVANCE DESIGN VALIDATION GUIDE
1.11.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement on the middle of the beam [m]
0.05079
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
2.937
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
8.095
CM2
Eigen mode
Eigen mode 3 frequency [Hz]
15.870
CM2
Eigen mode
Eigen mode 4 frequency [Hz]
26.228
1.11.3
58
Calculated results
Result name
Result description
Value
Error
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%
Eigen mode 2 frequency [Hz]
8.09 Hz
0.06%
Eigen mode 3 frequency [Hz]
15.79 Hz
0.50%
Eigen mode 4 frequency [Hz]
25.76 Hz
1.78%
ADVANCE DESIGN VALIDATION GUIDE
1.12 Short beam on simple supports (on the neutral axis) (010017SDLLB_FEM) Test ID: 2449 Test status: Passed
1.12.1
Description
Verifies the first eigen mode frequencies of a short beam on simple supports (the supports are located on the neutral axis), subjected to its own weight only.
1.12.2
Background
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; Analysis type: modal analysis (plane problem); Element type: linear. Short beam on simple supports on the neutral axis
Scale = 1/6 010017SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■
Height: h = 0.2 m, Length: l = 1 m, Width: b = 0.1 m, 2 4 Section: A = 2 x 10 m , Flexure moment of inertia relative to zaxis: Iz = 6.667 x 105 m4.
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
59
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
■
Outer: ► Hinged at A (null horizontal and vertical displacements), ► Simple support in B. Inner: None.
Loading ■ ■
External: None. Internal: None.
1.12.2.1 Eigen modes frequencies Reference solution The bending beams equation gives, when superimposing, the effects of simple bending, shear force deformations and rotation inertia, Timoshenko formula. The reference eigen modes frequencies are determined by a numerical simulation of this equation, independent of any software. The eigen frequencies in tensioncompression are given by: fi =
λi ⋅ 2πl
E
ρ
where λi =
(2i − 1) 2
Finite elements modeling ■ ■ ■
60
Linear element: S beam, imposed mesh, 10 nodes, 9 linear elements.
ADVANCE DESIGN VALIDATION GUIDE
Eigen mode shapes
1.12.2.2 Theoretical results Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
431.555
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
1265.924
CM2
Eigen mode
Eigen mode 3 frequency [Hz]
1498.295
CM2
Eigen mode
Eigen mode 4 frequency [Hz]
2870.661
CM2
Eigen mode
Eigen mode 5 frequency [Hz]
3797.773
CM2
Eigen mode
Eigen mode 6 frequency [Hz]
4377.837
61
ADVANCE DESIGN VALIDATION GUIDE
1.12.3
Calculated results
Result name
62
Result description
Value
Error
Eigen mode 1 frequency [Hz]
437.12 Hz
1.29%
Eigen mode 2 frequency [Hz]
1264.32 Hz
0.13%
Eigen mode 3 frequency [Hz]
1537.16 Hz
2.59%
Eigen mode 4 frequency [Hz]
2911.46 Hz
1.42%
Eigen mode 5 frequency [Hz]
3754.54 Hz
1.14%
Eigen mode 6 frequency [Hz]
4281.23 Hz
2.21%
ADVANCE DESIGN VALIDATION GUIDE
1.13 Rectangular thin plate simply supported on its perimeter (010020SDLSB_FEM) Test ID: 2452 Test status: Passed
1.13.1
Description
Verifies the first eigen mode frequencies of a thin rectangular plate simply supported on its perimeter.
1.13.2
Background
1.13.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLS 03/89; Analysis type: modal analysis; Element type: planar. Rectangular thin plate simply supported on its perimeter
Scale = 1/8 010020SDLSB_FEM
Units I. S. Geometry ■ ■ ■ ■
Length: a = 1.5 m, Width: b = 1 m, Thickness: t = 0.01 m, Points coordinates in m: ► A (0 ;0 ;0) ► B (0 ;1.5 ;0) ► C (1 ;1.5 ;0) ► D (1 ;0 ;0)
63
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■
Outer: Simple support on all sides, ► For the modeling: hinged at A, B and D. Inner: None. ►
■
Loading ■ ■
External: None. Internal: None.
1.13.2.2 Eigen modes frequencies Reference solution M. V. Barton formula for a rectangular plate with supports on all four sides, leads to: fij =
j 2 π i 2 [( ) +( )] a b 2
Et 2 12ρ(1 − ν 2 )
where: i = number of halflength of wave along y ( dimension a) j = number of halflength of wave along x ( dimension b) Finite elements modeling ■ ■ ■
64
Planar element: shell, 496 nodes, 450 planar elements.
ADVANCE DESIGN VALIDATION GUIDE
Eigen mode shapes
1.13.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 1; j = 1. [Hz]
35.63
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 2; j = 1. [Hz]
68.51
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 1; j = 2. [Hz]
109.62
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 3; j = 1. [Hz]
123.32
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 2; j = 2. [Hz]
142.51
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 3; j = 2. [Hz]
197.32
65
ADVANCE DESIGN VALIDATION GUIDE
1.13.3
Calculated results
Result name
66
Result description
Value
Error
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 = 2; j = 1 (Mode 2) [Hz]
68.29 Hz
0.32%
Eigen mode "i"  "j" frequency, for i = 1; j = 2 (Mode 3) [Hz]
109.98 Hz
0.33%
Eigen mode "i"  "j" frequency, for i = 3; j = 1 (Mode 4) [Hz]
123.02 Hz
0.24%
Eigen mode "i"  "j" frequency, for i = 2; j = 2 (Mode 5) [Hz]
141.98 Hz
0.37%
Eigen mode "i"  "j" frequency, for i = 3; j = 2 (Mode 6) [Hz]
195.55 Hz
0.90%
ADVANCE DESIGN VALIDATION GUIDE
1.14 Cantilever beam in Eulerian buckling (010021SFLLB_FEM) Test ID: 2453 Test status: Passed
1.14.1
Description
Verifies the critical load result on node 5 of a cantilever beam in Eulerian buckling. A punctual load of 100000 is applied.
1.14.2
Background
1.14.2.1 Model description ■ ■ ■
Reference: internal GRAITEC test (Euler theory); Analysis type: Eulerian buckling; Element type: linear.
Units I. S. Geometry ■ ■ ■
L = 10 m 2 S=0.01 m I = 0.0002 m4
Materials properties ■
Longitudinal elastic modulus: E = 2.0 x 1010 N/m2,
■
Poisson's ratio: ν = 0.1.
Boundary conditions ■ ■
Outer: Fixed at end x = 0, Inner: None.
Loading ■ ■
External: Punctual load P = 100000 N at x = L, Internal: None.
67
ADVANCE DESIGN VALIDATION GUIDE
1.14.2.2 Critical load on node 5 Reference solution The reference critical load established by Euler is:
Pcritique =
98696 π 2EI = 98696 N ⇒ λ = = 0.98696 2 100000 4L
Finite elements modeling ■ ■ ■
Planar element: beam, imposed mesh, 5 nodes, 4 elements.
Deformed shape
1.14.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Fx
Critical load on node 5. [N]
98696
1.14.3
68
Calculated results
Result name
Result description
Value
Error
Fx
Critical load on node 5 (mode 1) [N]
100000 N
1.32%
ADVANCE DESIGN VALIDATION GUIDE
1.15 Double fixed beam with a spring at mid span (010015SSLLB_FEM) Test ID: 2447 Test status: Passed
1.15.1
Description
Verifies the vertical displacement on the middle of a beam consisting of four elements of length "l", having identical characteristics. A punctual load of 10000 N is applied.
1.15.2
Background
1.15.2.1 Model description ■ ■ ■
Reference: internal GRAITEC test; Analysis type: linear static; Element type: linear.
Units I. S. Geometry ■ ■ ■
=1m 2 S = 0.01 m 4 I = 0.0001 m
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: Fixed at ends x = 0 and x = 4 m,
■ ■
Elastic support with k = EI/ rigidity Inner: None.
Loading ■ ■
External: Punctual load P = 10000 N at x = 2m, Internal: None.
69
ADVANCE DESIGN VALIDATION GUIDE
1.15.2.2 Displacement of the model in the linear elastic range Reference solution The reference vertical displacement v3, is calculated at the middle of the beam at x = 2 m. Rigidity matrix of a plane beam:
ES 0 0 [K e ] = ES − 0 0
0
0
12EI 3 6EI 2
6EI 2 4EI �
0
0
−
12EI 3 6EI 2

ES 0 0 ES l
6EI 2 2EI
0
−
0
6EI 2 2EI 0 6EI − 2 4EI
0
0
12EI 3 6EI − 2
−
0 12EI 3 6EI − 2
Given the symmetry / X and load of the structure, it is unnecessary to consider the degrees of freedom associated with normal work (u2, u3, u4). The same symmetry allows the deduction of: ■
v2 = v4
■
β2 = β4
■
β3 = 0
12 3 6 2 12 − 3 6 2 EI
70
6 2 4 6 − 2 2
12 3 6 − 2 24 3
−
0 −
12 3 6 2
6 2 2 0 8 6 − 2 2
12 3 6 − 2 24 1 3 + −
0 −
12 3 6 2
6 2 2 0 8 6 − 2 2
12 3 6 − 2 24 3
−
0 −
12 3 6 2
6 2 2 0 8 6 − 2 2
12 3 6 − 2 12 3 6 − 2
−
v1 R1 β M 1 1 v 2 0 (1) β 2 0 (2) v 3 − P (3 ) = β3 0 (4 ) v 0 (5 ) 4 6 β 4 0 (6 ) 2 2 v 5 R5 β5 M5 6 − 2 4
ADVANCE DESIGN VALIDATION GUIDE
The elementary rigidity matrix of the spring in its local axis system, [k 5 ] = the global axis system by means of the rotation matrix (90° rotation):
0 0 0 1 [K 5 ] = EI 0 0 0 0 0 − 1 0 0 →
0 0 0 −1 0 0 0 0 0 1 0 0
0 (u 3 ) 0 (v 3 ) 0 (β 3 ) 0 (u 6 ) 0 (v 6 ) 0 (β 6 )
6 2 8 3 v3 + β3 + β4 = 0 ⇒ β4 = − v3 2 4
→ −
→
0 0 0 0 0 0
EI 1 − 1 (U 3 ) , must be expressed in − 1 1 (U 6 )
12 6 24 v 3 − 2 β 3 + 3 v 4 = 0 ⇒ 2v 4 = v 3 3
6 2 8 6 2 v + β 2 + β 3 − 2 v 4 + β 4 = 0 ⇒ v 4 = v 2 (usually unnecessary) 2 2
3 (3) → − 12 v 2 − 6 β 2 + 24 + 1 v 3 − 12 v 4 − 6 β 4 = − P ⇒ v 3 = − P� = −0.11905 10 − 03 m 3 2 3 3 2 2 EI 3 + l EI
(
)
Finite elements modeling ■ ■ ■
Linear element: beam, imposed mesh, 6 nodes, 4 linear elements + 1 spring,
Deformed shape Double fixed beam with a spring at mid span Deformed
Note: the displacement is expressed here in µm
1.15.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement on the middle of the beam [mm]
0.11905
1.15.3
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement on the middle of the beam [mm]
0.119048 mm
0.00%
71
ADVANCE DESIGN VALIDATION GUIDE
1.16 Thin square plate fixed on one side (010019SDLSB_FEM) Test ID: 2451 Test status: Passed
1.16.1
Description
Verifies the first eigen modes frequencies of a thin square plate fixed on one side.
1.16.2
Background
1.16.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLS 01/89; Analysis type: modal analysis; Element type: planar. Thin square plate fixed on one side
Scale = 1/6 010019SDLSB_FEM
Units I. S. Geometry ■ ■ ■
72
Side: a = 1 m, Thickness: t = 1 m, Points coordinates in m: ► A (0 ;0 ;0) ► B (1 ;0 ;0) ► C (1 ;1 ;0) ► D (0 ;1 ;0)
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: Edge AD fixed. Inner: None.
Loading ■ ■
External: None. Internal: None.
1.16.2.2 Eigen modes frequencies Reference solution M. V. Barton formula for a square plate with side "a", leads to: fj =
1 λi2 2π ⋅ a 2 i λi
Et 2 12ρ(1 − ν 2 ) 1 3.492
where i = 1,2, . . . 2 8.525
3 21.43
4 27.33
5 31.11
6 54.44
Finite elements modeling ■ ■ ■
Planar element: shell, 959 nodes, 900 planar elements.
73
ADVANCE DESIGN VALIDATION GUIDE
Eigen mode shapes Thin square plate fixed on one side Mode 1
Thin square plate fixed on one side Mode 2
Thin square plate fixed on one side Mode 3
74
ADVANCE DESIGN VALIDATION GUIDE
1.16.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
8.7266
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
21.3042
CM2
Eigen mode
Eigen mode 3 frequency [Hz]
53.5542
CM2
Eigen mode
Eigen mode 4 frequency [Hz]
68.2984
CM2
Eigen mode
Eigen mode 5 frequency [Hz]
77.7448
CM2
Eigen mode
Eigen mode 6 frequency [Hz]
136.0471
1.16.3
Calculated results
Result name
Result description
Value
Error
Eigen mode 1 frequency [Hz]
8.67 Hz
0.65%
Eigen mode 2 frequency [Hz]
21.22 Hz
0.40%
Eigen mode 3 frequency [Hz]
53.13 Hz
0.79%
Eigen mode 4 frequency [Hz]
67.74 Hz
0.82%
Eigen mode 5 frequency [Hz]
77.15 Hz
0.77%
Eigen mode 6 frequency [Hz]
134.65 Hz
1.03%
75
ADVANCE DESIGN VALIDATION GUIDE
1.17 Bending effects of a symmetrical portal frame (010023SDLLB_FEM) Test ID: 2455 Test status: Passed
1.17.1
Description
Verifies the first eigen mode frequencies of a symmetrical portal frame with fixed supports.
1.17.2
Background
1.17.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; Analysis type: modal analysis; Element type: linear. Bending effects of a symmetrical portal frame
Scale = 1/5 010023SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■ ■
Straight rectangular sections for beams and columns: Thickness: h = 0.0048 m, Width: b = 0.029 m, 4 2 Section: A = 1.392 x 10 m , Flexure moment of inertia relative to zaxis: Iz = 2.673 x 1010 m4, Points coordinates in m: x y
76
A 0.30 0
B 0.30 0
C 0.30 0.36
D 0.30 0.36
E 0.30 0.81
F 0.30 0.81
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: Fixed at A and B, Inner: None.
Loading ■ ■
External: None. Internal: None.
1.17.2.2 Eigen modes frequencies Reference solution Dynamic radius method (slender beams theory). Finite elements modeling ■ ■ ■
Linear element: beam, 60 nodes, 60 linear elements.
77
ADVANCE DESIGN VALIDATION GUIDE
Deformed shape Bending effects of a symmetrical portal frame
Scale = 1/7
Mode 13
1.17.2.3 Theoretical results
78
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 antisymmetric frequency [Hz]
8.8
CM2
Eigen mode
Eigen mode 2 antisymmetric frequency [Hz]
29.4
CM2
Eigen mode
Eigen mode 3 symmetric frequency [Hz]
43.8
CM2
Eigen mode
Eigen mode 4 symmetric frequency [Hz]
56.3
CM2
Eigen mode
Eigen mode 5 antisymmetric frequency [Hz]
96.2
CM2
Eigen mode
Eigen mode 6 symmetric frequency [Hz]
102.6
CM2
Eigen mode
Eigen mode 7 antisymmetric frequency [Hz]
147.1
CM2
Eigen mode
Eigen mode 8 symmetric frequency [Hz]
174.8
CM2
Eigen mode
Eigen mode 9 antisymmetric frequency [Hz]
178.8
CM2
Eigen mode
Eigen mode 10 antisymmetric frequency [Hz]
206
CM2
Eigen mode
Eigen mode 11 symmetric frequency [Hz]
266.4
CM2
Eigen mode
Eigen mode 12 antisymmetric frequency [Hz]
320
CM2
Eigen mode
Eigen mode 13 symmetric frequency [Hz]
335
ADVANCE DESIGN VALIDATION GUIDE
1.17.3
Calculated results
Result name
Result description
Value
Error
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 4 symmetric frequency [Hz]
56.3 Hz
0.00%
Eigen mode 5 antisymmetric frequency [Hz]
96.05 Hz
0.16%
Eigen mode 6 symmetric frequency [Hz]
102.7 Hz
0.10%
Eigen mode 7 antisymmetric frequency [Hz]
147.08 Hz
0.01%
Eigen mode 8 symmetric frequency [Hz]
174.96 Hz
0.09%
Eigen mode 9 antisymmetric frequency [Hz]
178.92 Hz
0.07%
Eigen mode 10 antisymmetric frequency [Hz]
206.23 Hz
0.11%
Eigen mode 11 symmetric frequency [Hz]
266.62 Hz
0.08%
Eigen mode 12 antisymmetric frequency [Hz]
319.95 Hz
0.02%
Eigen mode 13 symmetric frequency [Hz]
334.96 Hz
0.01%
79
ADVANCE DESIGN VALIDATION GUIDE
1.18 Slender beam on two fixed supports (010024SSLLB_FEM) Test ID: 2456 Test status: Passed
1.18.1
Description
A straight slender beam with fixed ends is loaded with a uniform load, several punctual loads and a torque. The shear force, bending moment, vertical displacement and horizontal reaction are verified.
1.18.2
Background
1.18.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 01/89; Analysis type: linear static; Element type: linear. Slender beam on two fixed supports
Scale = 1/4 010024SSLLB_FEM
Units I. S. Geometry ■ ■
Length: L = 1 m, Beam inertia: I = 1.7 x 108 m4.
Materials properties ■
80
Longitudinal elastic modulus: E = 2.1 x 1011 Pa.
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Fixed at A and B, Inner: None.
Loading ■
External: Uniformly distributed load from A to B: py = p = 24000 N/m, ► Punctual load at D: Fx = F1 = 30000 N, ► Torque at D: Cz = C = 3000 Nm, ► Punctual load at E: Fx = F2 = 10000 N, ► Punctual load at E: Fy = F = 20000 N. Internal: None. ►
■
1.18.2.2 Shear force at G Reference solution Analytical solution: ■
Shear force at G: VG
VG = 0.216F – 1.26
C L
Finite elements modeling ■ ■ ■
Linear element: beam, 5 nodes, 4 linear elements.
81
ADVANCE DESIGN VALIDATION GUIDE
Results shape Slender beam on two fixed supports
Scale = 1/5 Shear force
1.18.2.3 Bending moment in G Reference solution Analytical solution: ■
Bending moment at G: MG 2 pL MG = 24  0.045LF – 0.3C A
E
A
Finite elements modeling ■ ■ ■
82
Linear element: beam, 5 nodes, 4 linear elements.
ADVANCE DESIGN VALIDATION GUIDE
Results shape Slender beam on two fixed supports
Scale = 1/5 Bending moment
1.18.2.4 Vertical displacement at G Reference solution Analytical solution: ■
Vertical displacement at G: vG pl4 0.003375FL3 0.015CL2 vG = 384EI + + EI EI A
E
A
A
E
A
A
E
Finite elements modeling ■ ■ ■
Linear element: beam, 5 nodes, 4 linear elements.
83
ADVANCE DESIGN VALIDATION GUIDE
Results shape Slender beam on two fixed supports
Scale = 1/4 Deformed
1.18.2.5 Horizontal reaction at A Reference solution Analytical solution: ■ Horizontal reaction at A: HA HA = 0.7F1 –0.3F2 Finite elements modeling ■ ■ ■
Linear element: beam, 5 nodes, 4 linear elements.
1.18.2.6 Theoretical results
84
Solver
Result name
Result description
Reference value
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]
4.90
CM2
Fx
Horizontal reaction in point A. [N]
24000
ADVANCE DESIGN VALIDATION GUIDE
1.18.3
Calculated results
Result name
Result description
Value
Error
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%
85
ADVANCE DESIGN VALIDATION GUIDE
1.19 Slender beam on three supports (010025SSLLB_FEM) Test ID: 2457 Test status: Passed
1.19.1
Description
A straight slender beam on three supports is loaded with two punctual loads. The bending moment, vertical displacement and reaction on the center are verified.
1.19.2
Background
1.19.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 03/89; Analysis type: static (plane problem); Element type: linear. Slender beam on three supports
Scale = 1/49 010025SSLLB_FEM
Units I. S. Geometry ■ ■
86
Length: L = 3 m, Beam inertia: I = 6.3 x 104 m4.
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa. Boundary conditions ■
Outer: Hinged at A, ► Elastic support at B (Ky = 2.1 x 106 N/m), ► Simple support at C. Inner: None. ►
■
Loading ■ ■
External: 2 punctual loads F = Fy = 42000N. Internal: None.
1.19.2.2 Bending moment at B Reference solution The resolution of the hyperstatic system of the slender beam leads to: k= ■
6EI L3Ky Bending moment at B: MB
MB = ±
L ( −6 + 2k )F 2 (8 + k )
Finite elements modeling ■ ■ ■
Linear element: beam, 5 nodes, 4 linear elements.
87
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Results shape Slender beam on three supports
Scale = 1/49 Bending moment
1.19.2.3 Reaction in B Reference solution ■
Compression force in the spring: VB 11F VB = 8 + k A
E
Finite elements modeling ■ ■ ■
Linear element: beam, 5 nodes, 4 linear elements.
1.19.2.4 Vertical displacement at B Reference solution ■
Deflection at the spring location: vB 11F vB = Ky(8 + k) A
E
A
Finite elements modeling ■ ■ ■
88
Linear element: beam, 5 nodes, 4 linear elements.
ADVANCE DESIGN VALIDATION GUIDE
Results shape Slender beam on three supports Deformed
1.19.2.5 Theoretical results Solver
Result name
Result description
Reference value
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
1.19.3
Calculated results
Result name
Result description
Value
Error
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%
89
ADVANCE DESIGN VALIDATION GUIDE
1.20 Thin circular ring hanged on an elastic element (010014SDLLB_FEM) Test ID: 2446 Test status: Passed
1.20.1
Description
Verifies the first eigen modes frequencies of a circular ring hanged on an elastic element, subjected to its self weight only.
1.20.2
Background
1.20.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 13/89; Analysis type: modal analysis, plane problem; Element type: linear. Thin circular ring hang from an elastic element
Scale = 1/1 010014SDLLB_FEM
Units I. S.
90
ADVANCE DESIGN VALIDATION GUIDE
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 105 m2, Flexure moment of relative to the vertical axis: I = 1.042 x 1010 m4, ► Elastic element Thickness: h = 0.003 m,
Width: b = 0.010 m, Section: A = 3 x 105 m2, Flexure moment of inertia relative to the vertical axis: I = 2.25 x 1011 m4, ■
Points coordinates: ► O ( 0 ; 0 ), ► A ( 0 ; 0.0725 ), ► B ( 0 ; 0.1 ).
Materials properties ■
Longitudinal elastic modulus: E = 7.2 x 1010 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 2700 kg/m3.
Boundary conditions ■ ■
Outer: Fixed in A, Inner: None.
Loading ■ ■
External: None, Internal: None.
1.20.2.2 Eigen mode frequencies Reference solutions The reference solution was established from experimental results of a mass manufactured aluminum ring. Finite elements modeling ■ ■ ■
Linear element: beam, 43 nodes, 43 linear elements.
91
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Eigen mode shapes
1.20.2.3 Theoretical results
92
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 Asymmetrical frequency [Hz]
28.80
CM2
Eigen mode
Eigen mode 2 Symmetrical frequency [Hz]
189.30
CM2
Eigen mode
Eigen mode 3 Asymmetrical frequency [Hz]
268.80
CM2
Eigen mode
Eigen mode 4 Asymmetrical frequency [Hz]
641.00
CM2
Eigen mode
Eigen mode 5 Symmetrical frequency [Hz]
682.00
CM2
Eigen mode
Eigen mode 6 Asymmetrical frequency [Hz]
1063.00
ADVANCE DESIGN VALIDATION GUIDE
1.20.3
Calculated results
Result name
Result description
Value
Error
Eigen mode 1 Asymmetrical frequency [Hz]
28.81 Hz
0.03%
Eigen mode 2 Symmetrical frequency [Hz]
189.69 Hz
0.21%
Eigen mode 3 Asymmetrical frequency [Hz]
269.38 Hz
0.22%
Eigen mode 4 Asymmetrical frequency [Hz]
642.15 Hz
0.18%
Eigen mode 5 Symmetrical frequency [Hz]
683.9 Hz
0.28%
Eigen mode 6 Asymmetrical frequency [Hz]
1065.73 Hz
0.26%
93
ADVANCE DESIGN VALIDATION GUIDE
1.21 Short beam on simple supports (eccentric) (010018SDLLB_FEM) Test ID: 2450 Test status: Passed
1.21.1
Description
Verifies the first eigen mode frequencies of a short beam on simple supports (the supports are eccentric relative to the neutral axis).
1.21.2
Background
1.21.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; Analysis type: modal analysis, (plane problem); Element type: linear. Short beam on simple supports (eccentric)
Scale = 1/5 010018SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■
94
Height: h = 0.2m, Length: l = 1 m, Width: b = 0.1 m, 2 4 Section: A = 2 x 10 m , 5 4 Flexure moment of inertia relative to zaxis: Iz = 6.667 x 10 m .
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■
■
Outer: ► Hinged at A (null horizontal and vertical displacements), ► Simple support at B. Inner: None.
Loading ■ ■
External: None. Internal: None.
1.21.2.2 Eigen modes frequencies Reference solution The problem has no analytical solution, the solution is determined by averaging several software: Timoshenko model with shear force deformation effects and rotation inertia. The bending modes and the tractioncompression are coupled. Finite elements modeling ■ ■ ■
Linear element: S beam, imposed mesh, 10 nodes, 9 linear elements.
95
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Eigen modes shape Short beam on simple supports (eccentric) Mode 1
Short beam on simple supports (eccentric) Mode 2
Short beam on simple supports (eccentric) Mode 3
96
ADVANCE DESIGN VALIDATION GUIDE
Short beam on simple supports (eccentric) Mode 4
Short beam on simple supports (eccentric) Mode 5
97
ADVANCE DESIGN VALIDATION GUIDE
1.21.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode 1 frequency [Hz]
392.8
CM2
Eigen mode
Eigen mode 2 frequency [Hz]
902.2
CM2
Eigen mode
Eigen mode 3 frequency [Hz]
1591.9
CM2
Eigen mode
Eigen mode 4 frequency [Hz]
2629.2
CM2
Eigen mode
Eigen mode 5 frequency [Hz]
3126.2
1.21.3
Calculated results
Result name
98
Result description
Value
Error
Eigen mode 1 frequency [Hz]
393.7 Hz
0.23%
Eigen mode 2 frequency [Hz]
945.35 Hz
4.78%
Eigen mode 3 frequency [Hz]
1595.94 Hz
0.25%
Eigen mode 4 frequency [Hz]
2526.22 Hz
3.92%
Eigen mode 5 frequency [Hz]
3118.91 Hz
0.23%
ADVANCE DESIGN VALIDATION GUIDE
1.22 Annular thin plate fixed on a hub (repetitive circular structure) (010022SDLSB_FEM) Test ID: 2454 Test status: Passed
1.22.1
Description
Verifies the eigen mode frequencies of a thin annular plate fixed on a hub.
1.22.2
Background
1.22.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLS 04/89; Analysis type: modal analysis; Element type: planar element. Annular thin plate fixed on a hub (repetitive circular structure)
Scale = 1/3
010022SDLSB_FEM
Units I. S. Geometry ■ ■ ■
Inner radius: Ri = 0.1 m, Outer radius: Re = 0.2 m, Thickness: t = 0.001 m.
Material properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
99
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Boundary conditions ■ ■
Outer: Fixed on a hub at any point r = Ri. Inner: None.
Loading ■ ■
External: None. Internal: None.
1.22.2.2 Eigen modes frequencies Reference solution The solution of determining the frequency based on Bessel functions leads to the following formula: fij = A
1 λ2 2πRe2 ij E
A
EA
Et2 12ρ(1ν2) A
AE
A
where: i = the number of nodal diameters j = the number of nodal circles and j \ i 0 1
λij2
such as: 0 13.0 85.1
1 13.3 86.7
2 14.7 91.7
3 18.5 100
Finite elements modeling ■ ■ ■
Planar element: plate, 360 nodes, 288 planar elements.
1.22.2.3 Theoretical results
100
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 0; j = 0. [Hz]
79.26
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 0; j = 1. [Hz]
518.85
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 1; j = 0. [Hz]
81.09
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 1; j = 1. [Hz]
528.61
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 2; j = 0. [Hz]
89.63
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 2; j = 1. [Hz]
559.09
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 3; j = 0. [Hz]
112.79
CM2
Eigen mode
Eigen mode “i"  “j” frequency, for i = 3; j = 1. [Hz]
609.70
ADVANCE DESIGN VALIDATION GUIDE
1.22.3
Calculated results
Result name
Result description
Value
Error
Eigen mode "i"  "j" frequency, for i = 0; j = 0 (Mode 1) [Hz]
79.05 Hz
0.26%
Eigen mode "i"  "j" frequency, for i = 0; j = 1 (Mode 18) [Hz]
521.84 Hz
0.58%
Eigen mode "i"  "j" frequency, for i = 1; j = 0 (Mode 2) [Hz]
80.52 Hz
0.70%
Eigen mode "i"  "j" frequency, for i = 1; j = 1 (Mode 20) [Hz]
529.49 Hz
0.17%
Eigen mode "i"  "j" frequency, for i = 2; j = 0 (Mode 4) [Hz]
88.43 Hz
1.34%
Eigen mode “i"  “j” frequency, for i = 2; j = 1 (Mode 22) [Hz]
552.43 Hz
1.19%
Eigen mode "i"  "j" frequency, for i = 3; j = 0 (Mode 7) [Hz]
110.27 Hz
2.23%
Eigen mode "i"  "j" frequency, for i = 3; j = 1 (Mode 25) [Hz]
593.83 Hz
2.60%
101
ADVANCE DESIGN VALIDATION GUIDE
1.23 Fixed thin arc in out of plane bending (010028SSLLB_FEM) Test ID: 2460 Test status: Passed
1.23.1
Description
An arc of a circle fixed at one end is loaded with a punctual force at its free end, perpendicular to the plane. The out of plane displacement, torsion moment and bending moment are verified.
1.23.2
Background
1.23.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 07/89; Analysis type: static linear; Element type: linear. Fixed thin arc in out of plane bending 010028SSLLB_FEM
Units I. S. Geometry ■ ■
102
Medium radius: R = 1 m , Circular hollow section: ► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 104 m2, ► Ix = 4.637 x 109 m4.
Scale = 1/6
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: Fixed at A. Inner: None.
Loading ■ ■
External: Punctual force in B perpendicular on the plane: Fz = F = 100 N. Internal: None.
1.23.2.2 Displacements at B Reference solution Displacement out of plane at point B: FR3 π EIx 3π uB = EI [ 4 + K ( 4  2)] x T A
E
A
A
E
A
A
E
A
A
E
A
where KT is the torsional rigidity for a circular section (torsion constant is 2Ix). KT = 2GIx =
EIx 1+ν A
E
A
FR3 π 3π ⇒ uB = EI [ 4 + (1 + ν) ( 4  2)] x A
E
A
A
E
A
A
E
A
Finite elements modeling ■ ■ ■
Linear element: beam, 46 nodes, 45 linear elements.
1.23.2.3 Moments at θ = 15° Reference solution ■
Torsion moment: Mx’ = Mt = FR(1  sinθ)
■
Bending moment: Mz’ = Mf = FRcosθ
Finite elements modeling ■ ■ ■
Linear element: beam, 46 nodes, 45 linear elements.
103
ADVANCE DESIGN VALIDATION GUIDE
1.23.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
D
Displacement out of plane in point B [m]
0.13462
CM2
Mx
Torsion moment in θ = 15° [Nm]
74.1180
CM2
Mz
Bending moment in θ = 15° [Nm]
96.5925
1.23.3
104
Calculated results
Result name
Result description
Value
Error
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%
ADVANCE DESIGN VALIDATION GUIDE
1.24 Double hinged thin arc in planar bending (010029SSLLB_FEM) Test ID: 2461 Test status: Passed
1.24.1
Description
Verifies the rotation about Zaxis, the vertical displacement and the horizontal displacement on several points of a double hinged thin arc in planar bending.
1.24.2
Background
1.24.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 08/89; Analysis type: static linear (plane problem); Element type: linear. Double hinged thin arc in planar bending 010029SSLLB_FEM
Scale = 1/8
Units I. S. Geometry ■ ■
Medium radius: R = 1 m , Circular hollow section: ► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 104 m2, ► Ix = 4.637 x 109 m4.
105
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Hinge at A, At B: allowed rotation along z, vertical displacement restrained along y. Inner: None.
► ►
■
Loading ■ ■
External: Punctual load at C: Fy = F =  100 N. Internal: None.
1.24.2.2 Displacements at A, B and C Reference solution ■
Rotation about zaxis
π FR2 θA =  θB = ( 2  1) 2EI A
■
E
A
A
E
Displacement;
π FR 3π FR3 Vertical at C: vC = 8 EA + ( 4  2) 2EI A
E
A
A
E
A
A
E
A
FR FR3 Horizontal at B: uB = 2EA  2EI A
E
A
A
Finite elements modeling ■ ■ ■
106
Linear element: beam, 37 nodes, 36 linear elements.
E
A
E
ADVANCE DESIGN VALIDATION GUIDE
Displacements shape Fixed thin arc in planar bending
Scale = 1/11 Deformed
1.24.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
RY
Rotation about Zaxis in point A [rad]
0.030774
CM2
RY
Rotation about Zaxis 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
1.24.3
Calculated results
Result name
Result description
Value
Error
RY
Rotation about Zaxis in point A [rad]
0.0307785 Rad
0.01%
RY
Rotation about Zaxis 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%
107
ADVANCE DESIGN VALIDATION GUIDE
1.25 Beam on elastic soil, free ends (010032SSLLB_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 displacement and rotation about zaxis on several points of the beam are verified.
1.25.2
Background
1.25.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 15/89; Analysis type: static linear (plane problem); Element type: linear. Beam on elastic soil, free ends 010032SSLLB_FEM
Units I. S. Geometry ■ ■
108
L = (π 10 )/2, I = 104 m4. A
E
A
Scale = 1/21
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa. Boundary conditions ■
Outer: Free A and B extremities, ► Constant linear stiffness of soil ky = K = 840000 N/m2. Inner: None. ►
■
Loading ■ ■
External: Punctual load at A, C and B: Fy = F =  10000 N. Internal: None.
1.25.2.2 Bending moment and displacement at C Reference solution β=
4 A
K/(4EI) E
ϕ = βL/2 λ = sh (2ϕ) + sin (2ϕ) ■
Bending moment: MC = (F/(4β))(ch(2ϕ)  cos (2ϕ) – 8sh(ϕ)sin(ϕ))/λ
■
Vertical displacement: vC =  (Fβ/(2K))( ch(2ϕ) + cos (2ϕ) + 8ch(ϕ)cos(ϕ) + 2)/λ
Finite elements modeling ■ ■ ■
Linear element: beam, 72 nodes, 71 linear elements.
109
ADVANCE DESIGN VALIDATION GUIDE
Bending moment diagram Beam on elastic soil, free ends
Scale = 1/20 Bending moment
1.25.2.3 Displacements at A Reference solution ■
Vertical displacement: vA = (2Fβ/K)( ch(ϕ)cos(ϕ) + ch(2ϕ) + cos(2ϕ))/λ
■
Rotation about zaxis θA = (2Fβ2/K)( sh(ϕ)cos(ϕ)  sin(ϕ)ch(ϕ) + sh(2ϕ)  sin(2ϕ))/λ
Finite elements modeling ■ ■ ■
Linear element: beam, 72 nodes, 71 linear elements
1.25.2.4 Theoretical results
110
Solver
Result name
Result description
Reference value
CM2
My
Bending moment in point C [Nm]
5759
CM2
Dz
Vertical displacement in point C [m]
0.006844
CM2
Dz
Vertical displacement in point A [m]
0.007854
CM2
RY
Rotation θ about zaxis in point A [rad]
0.000706
ADVANCE DESIGN VALIDATION GUIDE
1.25.3
Calculated results
Result name
Result description
Value
Error
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 zaxis in point A [rad]
0.000707427 Rad
0.20%
111
ADVANCE DESIGN VALIDATION GUIDE
1.26 EDF Pylon (010033SFLLA_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 loads corresponding to wind loads are applied on the main arms, on the upper arm and on the lower horizontal frames of the pylon.
1.26.2
Background
1.26.2.1 Model description ■ ■ ■
Reference: Internal GRAITEC test; Analysis type: static linear, Eulerian buckling; Element type: linear
Units I. S. Geometry
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Hinged support, ► For the modeling, a fixed restraint and 4 beams were added at the pylon supports level. Inner: None. ►
■ 112
ADVANCE DESIGN VALIDATION GUIDE
Loading ■
External: Punctual loads corresponding to a wind load. FX = 165550 N, FY =  1240 N, FZ =  58720 N on the main arms, FX = 50250 N, FY =  1080 N, FZ =  12780 N on the upper arm, ► FX = 11760 N, FY = 0 N, FZ = 0 N on the lower horizontal frames Internal: None. ► ►
■
1.26.2.2 Displacement of the model in the linear elastic range Reference solution
113
ADVANCE DESIGN VALIDATION GUIDE
Software Max deflection (m) λ dominating mode
ANSYS 5.3 0.714 2.77
Finite elements modeling ■ ■ ■
Linear element: beam, imposed mesh, 402 nodes, 1034 elements.
Deformed shape
114
NE/NASTRAN 7.0 0.714 2.77
ADVANCE DESIGN VALIDATION GUIDE
Buckling modal deformation (dominating mode)
1.26.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
D
Displacement at the top of the pylon [m]
0.714
Dominating buckling  critical λ, mode 4 [Hz]
2.77
CM2
1.26.3
Calculated results
Result name
Result description
Value
Error
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%
115
ADVANCE DESIGN VALIDATION GUIDE
1.27 Fixed thin arc in planar bending (010027SSLLB_FEM) Test ID: 2459 Test status: Passed
1.27.1
Description
Arc of a circle fixed at one end, subjected to two punctual loads and a torque at its free end. The horizontal displacement, vertical displacement and rotation about Zaxis are verified.
1.27.2
Background
1.27.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 06/89; Analysis type: static linear (plane problem); Element type: linear. Fixed thin arc in planar bending 010027SSLLB_FEM
Units I. S. Geometry ■ ■
116
Medium radius: R = 3 m , Circular hollow section: ► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 104 m2, ► Ix = 4.637 x 109 m4.
Scale = 1/24
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Longitudinal elastic modulus: E = 2 x 1011 Pa. Boundary conditions ■ ■
Outer: Fixed in A. Inner: None.
Loading ■
External: At B: punctual load F1 = Fx = 10 N, punctual load F2 = Fy = 5 N, ► bending moment about Oz, Mz = 8 Nm. Internal: None. ► ►
■
1.27.2.2 Displacements at B Reference solution At point B: ■
R2 displacement parallel to Ox: u = 4EI [F1πR + 2F2R + 4Mz] A
E
A
2
■
R displacement parallel to Oy: v = 4EI [2F1πR + (3π  8)F2R + 2(π  2)Mz]
■
R rotation around Oz: θ = 4EI [4F1R + 2(π  2)F2R + 2πMz]
A
A
E
E
A
A
Finite elements modeling ■ ■ ■
Linear element: beam, 31 nodes, 30 linear elements.
117
ADVANCE DESIGN VALIDATION GUIDE
Results shape Fixed thin arc in planar bending
Scale = 1/19 Deformed
1.27.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DX
Horizontal displacement in point B [m]
0.3791
CM2
DZ
Vertical displacement in point B [m]
0.2417
CM2
RY
Rotation about Zaxis in point B [rad]
0.1654
1.27.3
118
Calculated results
Result name
Result description
Value
Error
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 Zaxis in point B [rad]
0.165362 Rad
0.02%
ADVANCE DESIGN VALIDATION GUIDE
1.28 Truss with hinged bars under a punctual load (010031SSLLB_FEM) Test ID: 2463 Test status: Passed
1.28.1
Description
Verifies the horizontal and the vertical displacement in several points of a truss with hinged bars, subjected to a punctual load.
1.28.2
Background
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 11/89; Analysis type: static linear (plane problem); Element type: linear.
1.28.2.1 Model description Truss with hinged bars under a punctual load
Scale = 1/10 010031SSLLB_FEM
Units I. S. Geometry Elements AC CB CD BD
Length (m) 0.5 2 0.5 2 2.5 2 A
E
A
E
A
E
A
E
Area (m2) 2 x 104 2 x 104 1 x 104 1 x 104
119
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Longitudinal elastic modulus: E = 1.962 x 1011 Pa. Boundary conditions ■ ■
Outer: Hinge at A and B, Inner: None.
Loading ■ ■
External: Punctual force at D: Fy = F =  9.81 x 103 N. Internal: None.
1.28.2.2 Displacements at C and D Reference solution Displacement method. Finite elements modeling ■ ■ ■
Linear element: beam, 4 nodes, 4 linear elements.
Displacements shape Truss with hinged bars under a punctual load
120
Deformed
Scale = 1/9
ADVANCE DESIGN VALIDATION GUIDE
1.28.2.3 Theoretical results Solver
Result name
Result description
Reference value
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
1.28.3
Calculated results
Result name
Result description
Value
Error
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%
121
ADVANCE DESIGN VALIDATION GUIDE
1.29 Simply supported square plate (010036SSLSB_FEM) Test ID: 2467 Test status: Passed
1.29.1
Description
Verifies the vertical displacement in the center of a simply supported square plate.
1.29.2
Background
1.29.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 02/89; Analysis type: static linear; Element type: planar. Simply supported square plate 010036SSLSB_FEM
Units I. S. Geometry ■ ■
Side = 1 m, Thickness h = 0.01m.
Materials properties
122
■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7950 kg/m3.
Scale = 1/9
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
Outer: Simple support on the plate perimeter, For the modeling, we add a fixed support at B. Inner: None.
► ►
■
Loading ■ ■
External: Self weight (gravity = 9.81 m/s2). Internal: None.
1.29.2.2 Vertical displacement at O Reference solution According to Love Kirchhoff hypothesis, the displacement w at a point (x,y): w(x,y) = Σ wmnsinmπxsinnπy where wmn =
192ρg(1  ν2) mn(m2 + n2)π6Eh2 AE
A
E
A
E
Finite elements modeling ■ ■ ■
Planar element: shell, 441 nodes, 400 planar elements.
Deformed shape Simply supported square plate
Deformed
Scale = 1/6
123
ADVANCE DESIGN VALIDATION GUIDE
1.29.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point O [µm]
0.158
1.29.3
124
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point O [µm]
0.164901 µm
4.37%
ADVANCE DESIGN VALIDATION GUIDE
1.30 Caisson beam in torsion (010037SSLSB_FEM) Test ID: 2468 Test status: Passed
1.30.1
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 Zaxis and the stress are verified.
1.30.2
Background
1.30.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 05/89; Analysis type: static linear; Element type: planar. Caisson beam in torsion
010037SSLSB_FEM
Scale = 1/4
Units I. S. Geometry ■ ■ ■
Length; L = 1m, Square section of side: b = 0.1 m, Thickness = 0.005 m.
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
125
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Beam fixed at end x = 0; Inner: None.
Loading ■ ■
External: Torsion moment M = 10N.m applied to the free end (for modeling, 4 forces of 50 N). Internal: None.
1.30.2.2 Displacement and stress at two points Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. Points coordinates: ■ A (0,0.05,0.5) ■ B (0.05,0,0.8) Note: point O is the origin of the coordinate system (x,y,z). Finite elements modeling ■ ■ ■
Planar element: shell, 90 nodes, 88 planar elements.
Deformed shape Caisson beam in torsion
126
Deformed
Scale = 1/4
ADVANCE DESIGN VALIDATION GUIDE
1.30.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
D
Displacement in point A [m]
0.617 x 106
CM2
D
Displacement in point B [m]
0.987 x 106
CM2
RY
Rotation about Zaxis in point A [rad]
0.123 x 104
CM2
RY
Rotation about Zaxis in point B [rad]
0.197 x 104
CM2
sxy_mid
σxy stress in point A [MPa]
0.11
CM2
sxy_mid
σxy stress in point B [MPa]
0.11
1.30.3
Calculated results
Result name
Result description
Value
Error
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 Zaxis in point A [rad]
1.23211e005 Rad
0.17%
RY
Rotation about Zaxis in point B [rad]
1.97172e005 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%
127
ADVANCE DESIGN VALIDATION GUIDE
1.31 Thin cylinder under a uniform radial pressure (010038SSLSB_FEM) Test ID: 2469 Test status: Passed
1.31.1
Description
Verifies the stress, the radial deformation and the longitudinal deformation of a cylinder loaded with a uniform internal pressure.
1.31.2
Background
1.31.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 06/89; Analysis type: static elastic; Element type: planar. Thin cylinder under a uniform radial pressure 010038SSLSB_FEM
Units I. S. Geometry ■ ■ ■
Length: L = 4 m, Radius: R = 1 m, Thickness: h = 0.02 m.
Materials properties
128
■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Scale = 1/18
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
Outer: Free conditions For the modeling, only ¼ of the cylinder is considered and the symmetry conditions are applied. On the other side, we restrained the displacements at a few nodes in order to make the model stable. Inner: None.
► ►
■
Loading ■ ■
External: Uniform internal pressure: p = 10000 Pa, Internal: None.
1.31.2.2 Stresses in all points Reference solution Stresses in the planar elements coordinate system (x axis is parallel with the length of the cylinder): ■ ■
σxx = 0 pR σyy = h A
E
Finite elements modeling ■ ■ ■
Planar element: shell, 209 nodes, 180 planar elements.
1.31.2.3 Cylinder deformation in all points ■
Radial deformation: pR2 δR = Eh A
■
E
Longitudinal deformation: δL =
pRνL Eh A
E
1.31.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
syy_mid
σyy stress in all points [Pa]
500000.000000
CM2
Dz
δL radial deformation of the cylinder in all points [µm]
2.380000
CM2
DY
δL longitudinal deformation of the cylinder in all points [µm]
2.860000
1.31.3
Calculated results
Result name
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%
129
ADVANCE DESIGN VALIDATION GUIDE
1.32 Bimetallic: Fixed beams connected to a stiff element (010026SSLLB_FEM) Test ID: 2458 Test status: Passed
1.32.1
Description
Two beams fixed at one end and rigidly connected to an undeformable beam is loaded with a punctual load. The deflection, vertical reaction and bending moment are verified in several points.
1.32.2
Background
1.32.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 05/89; Analysis type: linear static; Element type: linear. Fixed beams connected to a stiff element 010026SSLLB_FEM
Units I. S. Geometry ■
■ ■
130
Lengths: ► L = 2 m, ► l = 0.2 m, Beams inertia moment: I = (4/3) x 108 m4, 2 The beam sections are squared, of side: 2 x 10 m.
Scale = 1/10
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2 x 1011 Pa.
Boundary conditions ■ ■
Outer: Fixed in A and C, Inner: The tangents to the deflection of beams AB and CD at B and D remain horizontal; practically, we restraint translations along x and z at nodes B and D.
Loading ■ ■
External: In D: punctual load F = Fy = 1000N. Internal: None.
1.32.2.2 Deflection at B and D Reference solution The theory of slender beams bending (EulerBernouilli formula) leads to a deflection at B and D: The resolution of the hyperstatic system of the slender beam leads to: FL3 vB = vD = 24EI A
E
Finite elements modeling ■ ■ ■
Linear element: beam, 4 nodes, 3 linear elements.
Results shape Fixed beams connected to a stiff element Deformed
Scale = 1/10
131
ADVANCE DESIGN VALIDATION GUIDE
1.32.2.3 Vertical reaction at A and C Reference solution Analytical solution. Finite elements modeling ■ ■ ■
Linear element: beam, 4 nodes, 3 linear elements.
1.32.2.4 Bending moment at A and C Reference solution Analytical solution. Finite elements modeling ■ ■ ■
Linear element: beam, 4 nodes, 3 linear elements
1.32.2.5 Theoretical results Solver
Result name
Result description
Reference value
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]
500
CM2
My
Bending moment in point C [Nm]
500
1.32.3
132
Calculated results
Result name
Result description
Value
Error
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%
ADVANCE DESIGN VALIDATION GUIDE
1.33 Portal frame with lateral connections (010030SSLLB_FEM) Test ID: 2462 Test status: Passed
1.33.1
Description
Verifies the rotation about zaxis and the bending moment on a portal frame with lateral connections.
1.33.2
Background
1.33.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 10/89; Analysis type: static linear; Element type: linear. Portal frame with lateral connections 010030SSLLB_FEM
Scale = 1/21
Units I. S. Geometry Beam AB
Length lAB = 4 m
Moment of inertia 64 IAB = 3 x 108 m4 1 IAC = 12 x 108 m4 1 IAD = 12 x 108 m4 4 IAE = 3 x 108 m4 A
AC AD AE
lAC = 1 m lAD = 1 m lAE = 2 m
E
A
E
A
A
E
A
A
■
G is in the middle of DA.
A
E
A
133
ADVANCE DESIGN VALIDATION GUIDE
■
The beams have square sections: ► AAB = 16 x 104 m ► AAD = 1 x 104 m ► AAC = 1 x 104 m ► AAE = 4 x 104 m
Materials properties Longitudinal elastic modulus: E = 2 x 1011 Pa, Boundary conditions ■
Outer: Fixed at B, D and E, ► Hinge at C, Inner: None. ►
■
Loading ■
External: Punctual force at G: Fy = F =  105 N, ► Distributed load on beam AD: p =  103 N/m. Internal: None. ►
■
1.33.2.2 Displacements at A Reference solution Rotation at A about zaxis: EIAn We say: kAn = l An A
E
where n = B, C, D or E A
3 K = kAB + kAD + kAE + 4 kAC A
E
kAn rAn = K FlAD plAB2 C1 = 8  12 C1 θ = 4K Finite elements modeling ■ ■ ■
134
Linear element: beam, 6 nodes, 5 linear elements.
ADVANCE DESIGN VALIDATION GUIDE
Displacements shape Portal frame with lateral connections Deformed
1.33.2.3 Moments in A Reference solution ■
plAB2 MAB = 12 + rAB x C1
■
FlAD MAD =  8 + rAD x C1
■ ■
MAE = rAE x C1 MAC = rAC x C1
Finite elements modeling ■ ■ ■
Linear element: beam, 6 nodes, 5 linear elements
135
ADVANCE DESIGN VALIDATION GUIDE
1.33.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
RY
Rotation θ about zaxis in point A [rad]
0.227118
CM2
My
Bending moment in point A (MAB) [Nm]
11023.72
CM2
My
Bending moment in point A (MAC) [Nm]
113.559
CM2
My
Bending moment in point A (MAD) [Nm]
12348.588
CM2
My
Bending moment in point A (MAE) [Nm]
1211.2994
1.33.3
136
Calculated results
Result name
Result description
Value
Error
RY
Rotation Theta about zaxis 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%
ADVANCE DESIGN VALIDATION GUIDE
1.34 Beam on elastic soil, hinged ends (010034SSLLB_FEM) Test ID: 2466 Test status: Passed
1.34.1
Description
A beam under a punctual load, a distributed load and two torques lays on a soil of constant linear stiffness. The rotation around zaxis, the vertical reaction, the vertical displacement and the bending moment are verified in several points.
1.34.2
Background
1.34.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 16/89; Analysis type: static linear (plane problem); Element type: linear. Beam on elastic soil, hinged ends 010034SSLLB_FEM
Scale = 1/27
Units I. S. Geometry ■ ■
L = (π 10 )/2, 4 4 I = 10 m .
137
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa. Boundary conditions ■
Outer: Free A and B ends, ► Soil with a constant linear stiffness ky = K = 840000 N/m2. Inner: None. ►
■
Loading ■
External: Punctual force at D: Fy = F =  10000 N, ► Uniformly distributed force from A to B: fy = p =  5000 N/m, ► Torque at A: Cz = C = 15000 Nm, ► Torque at B: Cz = C = 15000 Nm. Internal: None. ►
■
1.34.2.2 Displacement and support reaction at A Reference solution β=
4
K/(4EI)
ϕ = βL/2 λ = ch(2ϕ) + cos(2ϕ) ■
Vertical support reaction: VA = p(sh(2ϕ) + sin(2ϕ))  2βFch(ϕ)cos(ϕ) + 2β2C(sh(2ϕ)  sin(2ϕ)) x
■
Rotation about zaxis: θA = p(sh(2ϕ) – sin(2ϕ)) + 2βFsh(ϕ)sin(ϕ)  2β2C(sh(2ϕ) + sin(2ϕ)) x
Finite elements modeling ■ ■ ■
138
1 2βλ
Linear element: beam, 50 nodes, 49 linear elements.
1 (K/β)λ
ADVANCE DESIGN VALIDATION GUIDE
Deformed shape Beam on elastic soil, hinged ends
Scale = 1/20 Deformed
1.34.2.3 Displacement and bending moment at D Reference solution ■
Vertical displacement: vD = 2p(λ  2ch(ϕ)cos(ϕ)) + βF(sh(2ϕ) – sin(2ϕ))  8β2Csh(ϕ)sin(ϕ) x
■
1 2Kλ
Bending moment: MD = 4psh(ϕ)sin(ϕ) + βF(sh(2ϕ) + sin(2ϕ))  8β2Cch(ϕ)cos(ϕ) x
1 4β2λ
Finite elements modeling ■ ■ ■
Linear element: beam, 50 nodes, 49 linear elements.
139
ADVANCE DESIGN VALIDATION GUIDE
Bending moment diagram Beam on elastic soil, hinged ends
Scale = 1/20 Bending moment
1.34.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
RY
Rotation around zaxis 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
1.34.3
Calculated results
Result name
140
Result description
Value
Error
RY
Rotation around zaxis 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%
ADVANCE DESIGN VALIDATION GUIDE
1.35 Beam on two supports considering the shear force (010041SSLLB_FEM) Test ID: 2472 Test status: Passed
1.35.1
Description
Verifies the vertical displacement on a 300 cm long beam, consisting of an I shaped profile of a total height of 20.04 cm, a 0.96 cm thick web and 20.04 cm wide / 1.46 cm thick flanges.
1.35.2
Background
1.35.2.1 Model description ■ ■ ■
Reference: Internal GRAITEC test; Analysis type: static linear (plane problem); Element type: linear.
Units I. S. Geometry l= h= b= tw = tf = Sx= Iz = Sy =
300 20.04 20.04 1.46 0.96 74.95 5462 16.43
cm cm cm cm cm cm2 cm4 cm2
Materials properties ■ ■
Longitudinal elastic modulus: E = 2285938 daN/cm2, Transverse elastic modulus G = 879207 daN/cm2
■
Poisson's ratio: ν = 0.3.
141
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
Outer: Simple support on node 11, For the modeling, put an hinge at node 1 (instead of a simple support). Inner: None.
► ►
■
Loading ■ ■
External: Vertical punctual load P = 20246 daN at node 6, Internal: None.
1.35.2.2 Vertical displacement of the model in the linear elastic range Reference solution The reference displacement is calculated in the middle of the beam, at node 6. shear flexion 3 − 20246 x3003 − 20246 x300 Pl Pl = −0.912 − 0.105 = −1.017 cm = + + v6 = 48 EI z 4GS y 48 x 2285938 x5462 4 x 2285938 x16.43 2(1 + 0.3)
Finite elements modeling ■ ■ ■
Planar element: S beam, imposed mesh, 11 nodes, 10 linear elements.
Deformed shape
142
ADVANCE DESIGN VALIDATION GUIDE
1.35.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement at node 6 [cm]
1.017
1.35.3
Calculated results
Result name
Result description
Value
Error
DZ
Vertical displacement at node 6 [cm]
1.01722 cm
0.02%
143
ADVANCE DESIGN VALIDATION GUIDE
1.36 Thin cylinder under a uniform axial load (010042SSLSB_FEM) Test ID: 2473 Test status: Passed
1.36.1
Description
Verifies the stress, the longitudinal deformation and the radial deformation of a cylinder under a uniform axial load.
1.36.2
Background
1.36.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 07/89; Analysis type: static elastic; Element type: planar. Thin cylinder under a uniform axial load 010042SSLSB_FEM
Units I. S. Geometry ■ ■ ■
Thickness: h = 0.02 m, Length: L = 4 m, Radius: R = 1 m.
Materials properties
144
■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Scale = 1/19
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
Outer: Null axial displacement at the left end: vz = 0, For the modeling, only a ¼ of the cylinder is considered. Inner: None.
► ►
■
Loading ■ ■
External: Uniform axial load q = 10000 N/m Inner: None.
1.36.2.2 Stress in all points Reference solution x axis of the local coordinate system of planar elements is parallel to the cylinders axis. q σxx = h σyy = 0 Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 697 nodes, 640 surface quadrangles.
1.36.2.3 Cylinder deformation at the free end Reference solution ■
δL longitudinal deformation of the cylinder: qL δL = Eh
■
δR radial deformation of the cylinder: δR =
qνR Eh
Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 697 nodes, 640 surface quadrangles.
145
ADVANCE DESIGN VALIDATION GUIDE
Deformation shape Thin cylinder under a uniform axial load
Scale = 1/22
Deformation shape
1.36.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
sxx_mid
σxx stress at all points [Pa]
5 x 105
CM2
syy_mid
σyy stress at all points [Pa]
0
CM2
DY
δL longitudinal deformation at the free end [m]
CM2
Dz
δR radial deformation at the free end [m]
1.36.3
146
9.52 x 106 7.14 x 107
Calculated results
Result name
Result description
Value
Error
sxx_mid
Sigma xx stress at all points [Pa]
500000 Pa
0.00%
syy_mid
Sigma yy stress at all points [Pa]
1.05305e009 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%
ADVANCE DESIGN VALIDATION GUIDE
1.37 Torus with uniform internal pressure (010045SSLSB_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
1.37.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 10/89; Analysis type: static, linear elastic; Element type: planar.
Torus with uniform internal pressure 010045SSLSB_FEM
Units I. S. Geometry ■ ■ ■
Thickness: h = 0.02 m, Transverse section radius: b = 1 m, Average radius of curvature: a = 2 m.
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3. 147
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: For the modeling, only 1/8 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 pb r + a σ11 = 2h r pb σ22 = 2h Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 361 nodes, 324 surface quadrangles.
1.37.2.3 Cylinder deformation Reference solution ■
δR radial deformation of the torus: pb δR = 2Eh (r  ν(r + a))
Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 361 nodes, 324 surface quadrangles.
1.37.2.4 Theoretical results
148
Solver
Result name
Result description
Reference value
CM2
syy_mid
σ11 stresses for r = a  b [Pa]
7.5 x 105
CM2
syy_mid
σ11 stresses for r = a + b [Pa]
CM2
sxx_mid
σ22 stress for all r [Pa]
CM2
Dz
δL radial deformations of the torus for r = a  b [m]
CM2
Dz
δL radial deformations of the torus for r = a + b [m]
4.17 x 105 2.50 x 105 1.19 x 107 1.79 x 106
ADVANCE DESIGN VALIDATION GUIDE
1.37.3
Calculated results
Result name
Result description
Value
Error
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%
149
ADVANCE DESIGN VALIDATION GUIDE
1.38 Spherical shell under internal pressure (010046SSLSB_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
1.38.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 14/89; Analysis type: static, linear elastic; Element type: planar.
Spherical shell under internal pressure 010046SSLSB_FEM
Units I. S. Geometry
150
■ ■
Thickness: h = 0.02 m, Radius: R2 = 1 m,
■
θ = 90° (hemisphere).
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Simple support (null displacement along vertical displacement) on the shell perimeter. For modeling, we consider only half of the hemisphere, so we impose symmetry conditions (DOF restrains placed in the vertical plane xy in translation along z and in rotation along x and y). In addition, the node at the top of the shell is restrained in translation along x to assure the stability of the structure during calculation).
■
Inner: None.
Loading ■ ■
External: Uniform internal pressure p = 10000 Pa Internal: None.
1.38.2.2 Stresses Reference solution (See stresses description on the first scheme of the overview) If 0° ≤ θ ≤ 90° σ11 = σ22 =
pR22 2h
Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 343 nodes, 324 planar elements.
1.38.2.3 Cylinder deformation Reference solution ■
δR radial deformation of the calotte: δR =
pR22 (1  ν) sin θ 2Eh
Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 343 nodes, 324 planar elements.
151
ADVANCE DESIGN VALIDATION GUIDE
Deformed shape Spherical shell under internal pressure
Scale = 1/11 Deformed
1.38.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
sxx_mid
σ11 stress for all θ [Pa]
2.50 x 105
CM2
syy_mid
σ22 stress for all θ [Pa]
CM2
Dz
δR radial deformations for θ = 90° [m]
1.38.3
152
2.50 x 105 8.33 x 107
Calculated results
Result name
Result description
Value
Error
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%
ADVANCE DESIGN VALIDATION GUIDE
1.39 Stiffen membrane (010040SSLSB_FEM) Test ID: 2471 Test status: Passed
1.39.1
Description
Verifies the horizontal displacement and the stress on a plate (8 x 12 cm) fixed in the middle on 3 supports with a punctual load at its free node.
1.39.2
Background
1.39.2.1 Model description ■ ■ ■
Reference: KlausJürgen Bathe  Finite Element Procedures in Engineering Analysis, Example 5.13; Analysis type: static linear; Element type: planar (membrane).
ξ ,η ∈ [− 1;1] Units I. S. Geometry ■ ■ ■
Thickness: e = 0.1 cm, Length: l = 8 cm, Width: B = 12 cm.
Materials properties ■
Longitudinal elastic modulus: E = 30 x 106 N/cm2,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: Fixed on 3 sides, Inner: None.
153
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: Uniform load Fx = F = 6000 N at A, Internal: None.
1.39.2.2 Results of the model in the linear elastic range Reference solution Point B is the origin of the coordinate system used for the results positions.
uA =
F F 6000 = = = 9.3410− 4 cm 6 6 K eabE ES 2.67410 + 3.7510 2 1 2 + + 2 2 b (1 + ν ) 2a 3 a 1 −ν
(
)
η = 1; σ xx1 = 0 E uA (1 − η ) for η = 0; σ xx1 = 1924 N/cm 2 = 19.24 MPa σ xx1 = 2 1 − ν 2a η = −1; σ = 3849 N/cm 2 = 38.49 MPa xx1 η = 1; σ yy1 = 0 σ yy1 = νσ xx1 for η = 0; σ yy1 = 577 N/cm 2 = 5.77 MPa η = −1; σ = 1155 N/cm 2 = 11.55 MPa yy1 ξ = −1; σ xy1 = 0 E uA (1 + ξ ) for ξ = 0; σ xy1 = −898 N/cm 2 = −8.98 MPa σ xy1 = − 1 + ν 8b ξ = 1; σ = −1796 N/cm 2 = −17.96 MPa xy1 Finite elements modeling ■ ■ ■
Planar element: membrane, imposed mesh, 6 nodes, 2 quadrangle planar elements and 1 bar.
Deformed shape
154
ADVANCE DESIGN VALIDATION GUIDE
1.39.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DX
Horizontal displacement Element 1 in A [cm]
9.340000
CM2
sxx_mid
σxx stress Element 1 in y = 0 cm [MPa]
38.490000
CM2
sxx_mid
σxx stress Element 1 in y = 6 cm [MPa]
0
CM2
syy_mid
σyy stress Element 1 in y = 0 cm [MPa]
11.550000
CM2
syy_mid
σyy stress Element 1 in y = 6 cm [MPa]
0
CM2
sxy_mid
σxy stress Element 1 in x = 0 cm [MPa]
0
CM2
sxy_mid
σxy stress Element 1 in x = 4 cm [MPa]
8.980000
CM2
sxy_mid
σxy stress Element 1 in x = 8 cm [MPa]
17.960000
1.39.3
Calculated results
Result name
Result description
Value
Error
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]
1.81899e015 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]
2.72848e015 MPa
0.00%
sxy_mid
Sigma xy stress Element 1 in x = 0 cm [MPa]
0 MPa
0.00%
sxy_mid
Sigma xy stress Element 1 in x = 4 cm [MPa]
8.98076 MPa
0.01%
sxy_mid
Sigma xy stress Element 1 in x = 8 cm [MPa]
17.9615 MPa
0.01%
155
ADVANCE DESIGN VALIDATION GUIDE
1.40 Thin cylinder under its self weight (010044SSLSB_MEF) Test ID: 2475 Test status: Passed
1.40.1
Description
Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder subjected to its self weight only.
1.40.2
Background
1.40.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 09/89; ■ Analysis type: static, linear elastic; ■ Element type: planar. A cylinder of R radius and L length subject of self weight only. Thin cylinder under its self weight 010044SSLSB_FEM
Units I. S. Geometry ■ ■ ■
Thickness: h = 0.02 m, Length: L = 4 m, Radius: R = 1 m.
Materials properties
156
■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: γ = 7.85 x 104 N/m3.
Scale = 1/24
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
Outer: Null axial displacement at z = 0, For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on the nodes that are parallel with the cylinder’s axis. Inner: None.
► ►
■
Loading ■ ■
External: Cylinder self weight, Internal: None.
1.40.2.2 Stresses Reference solution x axis of the local coordinate system of planar elements is parallel to the cylinders axis. σxx = γz σyy = 0 Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 697 nodes, 640 surface quadrangles.
1.40.2.3 Cylinder deformation Reference solution ■
δL longitudinal deformation of the cylinder: γz2 δL = 2E
■
δR radial deformation of the cylinder: δR =
γνRz E
1.40.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
sxx_mid
σxx stress for z = L [Pa]
314000.000000
CM2
DY
δL longitudinal deformation for z = L [mm]
0.002990
CM2
Dz
δR radial deformation for z = L[mm]
0.000440
* To obtain this result, you must generate a calculation note “Planar elements stresses by load case in neutral fiber" with results on center.
1.40.3
Calculated results
Result name
Result description
Value
Error
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%
157
ADVANCE DESIGN VALIDATION GUIDE
1.41 Spherical shell with holes (010049SSLSB_FEM) Test ID: 2479 Test status: Passed
1.41.1
Description
A spherical shell with holes is subjected to 4 forces, opposite 2 by 2. The horizontal displacement is verified.
1.41.2
Background
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 21/89; Analysis type: static, linear elastic; Element type: planar.
1.41.2.1 Model description
Spherical shell with holes
Units I. S. Geometry
158
■ ■
Radius: R = 10 m Thickness: h = 0.04 m,
■
Opening angle of the hole: ϕ0 = 18°.
010049SSLSB_FEM
ADVANCE DESIGN VALIDATION GUIDE
Materials properties ■
Longitudinal elastic modulus: E = 6.285 x 107 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: For modeling, we consider only a quarter of the shell, so we impose symmetry conditions (nodes in the vertical yz plane are restrained in translation along x and in rotation along y and z. Nodes on the vertical xy plane are restrained in translation along z and in rotation along x and y), Inner: None.
Loading ■ ■
External: Punctual loads F = 1 N, according to the diagram, Internal: None.
1.41.2.2 Horizontal displacement at point A Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution. Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 99 nodes, 80 surface quadrangles.
Deformed shape Spherical shell with holes
Deformed
Scale = 1/79
159
ADVANCE DESIGN VALIDATION GUIDE
1.41.2.3 Theoretical background Solver
Result name
Result description
Reference value
CM2
DX
Horizontal displacement at point A(R,0,0) [mm]
94.0
1.41.3
160
Calculated results
Result name
Result description
Value
Error
DX
Horizontal displacement at point A(R,0,0) [mm]
92.6205 mm
1.47%
ADVANCE DESIGN VALIDATION GUIDE
1.42 Spherical dome under a uniform external pressure (010050SSLSB_FEM) Test ID: 2480 Test status: Passed
1.42.1
Description
A spherical dome of radius (a) is subjected to a uniform external pressure. The horizontal displacement and the external meridian stresses are verified.
1.42.2
Background
1.42.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 22/89; Analysis type: static, linear elastic; Element type: planar.
Spherical dome under a uniform external pressure 010050SSLSB_FEM
Units I. S. Geometry ■
Radius: a = 2.54 m,
161
ADVANCE DESIGN VALIDATION GUIDE
■
Thickness: h = 0.0127 m,
■
Angle: θ = 75°.
Materials properties ■
Longitudinal elastic modulus: E = 6.897 x 1010 Pa,
■
Poisson's ratio: ν = 0.2.
Boundary conditions ■ ■
Outer: Fixed on the dome perimeter, Inner: None.
Loading ■ ■
External: Uniform pressure p = 0.6897 x 106 Pa, Internal: None.
1.42.2.2 Horizontal displacement and exterior meridian stress Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution. Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 401 nodes, 400 planar elements.
Deformed shape
162
ADVANCE DESIGN VALIDATION GUIDE
1.42.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DX
Horizontal displacements in ψ = 15°
1.73 x 103
CM2
DX
Horizontal displacements in ψ = 45°
1.02 x 103
CM2
syy_mid
σyy external meridian stresses in ψ = 15°
74
CM2
sxx_mid
σXX external meridian stresses in ψ = 45°
68
1.42.3
Calculated results
Result name
Result description
Value
Error
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%
163
ADVANCE DESIGN VALIDATION GUIDE
1.43 Simply supported square plate under a uniform load (010051SSLSB_FEM) Test ID: 2481 Test status: Passed
1.43.1
Description
A square plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.
1.43.2
Background
1.43.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 24/89; Analysis type: static, linear elastic; Element type: planar. Simply supported square plate under a uniform load 010051SSLSB_FEM
Units I. S. Geometry ■ ■
Side: a =b = 1 m, Thickness: h = 0.01 m,
Materials properties ■
Longitudinal elastic modulus: E = 1.0 x 107 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ 164
Outer: Simple support on the plate perimeter (null displacement along zaxis),
Scale = 1/9
ADVANCE DESIGN VALIDATION GUIDE
■
Inner: None
Loading ■ ■
External: Normal pressure of plate p = pZ = 1.0 Pa, Internal: None.
1.43.2.2 Vertical displacement and bending moment at the center of the plate Reference solution LoveKirchhoff thin plates theory. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 361 nodes, 324 planar elements.
1.43.2.3 Theoretical result Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement at plate center [m]
4.43 x 103
CM2
Mxx
MX bending moment at plate center [Nm]
0.0479
CM2
Myy
MY bending moment at plate center [Nm]
0.0479
1.43.3
Calculated results
Result name
Result description
Value
Error
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%
165
ADVANCE DESIGN VALIDATION GUIDE
1.44 Square plate under planar stresses (010039SSLSB_FEM) Test ID: 2470 Test status: Passed
1.44.1
Description
Verifies the vertical displacement and the stresses on a square plate of 2 x 2 m, fixed on 3 sides with a uniform surface load on its surface.
1.44.2
Background
1.44.2.1 Model description ■ ■ ■
Reference: Internal GRAITEC test; Analysis type: static linear; Element type: planar (membrane). Square plate under planar stresses
ξ ,η ∈ [− 1;1] Units I. S. Geometry ■ ■
Thickness: e = 1 m, 4 square elements of side h = 1 m.
Materials properties
166
■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Modeling
Scale = 1/19
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Fixed on 3 sides, Inner: None.
Loading ■ ■
External: Uniform load p = 1. 108 N/ml on the upper surface, Internal: None.
1.44.2.2 Displacement of the model in the linear elastic range Reference solution The reference displacements are calculated on nodes 7 and 9. v9 =
6ph(3 + ν)(1  ν2) = 0.1809 x 103 m, E(8(3  ν)2  (3 + ν)2)
v7 =
4(3  ν) v9 = 0.592 x 103 m, 3+ν
For element 1.4: (For the stresses calculated above, the abscissa point (x = 0; y = 0) corresponds to node 8.) ξ = 1 ; σyy = 0 ξ = 0 ; σyy = 47.44 MPa ξ = 1 ; σyy = 94.88 MPa
E (v9  v7) σyy = (1 + ξ) for 1  ν2 2h
σxx = νσyy
ξ = 1 ; σxx = 0 ξ = 0 ; σxx = 14.23 MPa ξ = 1 ; σxx = 28.46 MPa
for
E (v9 + v7) + η(v9  v7) σxy = (1 + ξ) 4h 1+ν
for
η = 1 ; ξ = 0 ; σxy = 47.82 MPa η = 0 ; ξ = 0 ; σxy = 31.21 MPa η= 1 ; ξ = 0 ; σxy = 14.61 MPa
Finite elements modeling ■ ■ ■
Planar element: membrane, imposed mesh, 9 nodes, 4 surface quadrangles.
167
ADVANCE DESIGN VALIDATION GUIDE
Deformed shape
168
ADVANCE DESIGN VALIDATION GUIDE
1.44.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement on node 7 [mm]
0.592
CM2
DZ
Vertical displacement on node 5 [mm]
0.1809
CM2
sxx_mid
σxx stresses on Element 1.4 in x = 0 m [MPa]
0
CM2
sxx_mid
σxx stresses on Element 1.4 in x = 0.5 m [MPa]
14.23
CM2
sxx_mid
σxx stresses on Element 1.4 in x = 1 m [MPa]
28.46
CM2
syy_mid
σyy stresses on Element 1.4 in x = 0 m [MPa]
0
CM2
syy_mid
σyy stresses on Element 1.4 in x = 0.5 m [MPa]
47.44
CM2
syy_mid
σyy stresses on Element 1.4 in x = 1 m [MPa]
94.88
CM2
sxy_mid
σxy stresses on Element 1.4 in y = 0 m [MPa]
14.66
CM2
sxy_mid
σxy stresses on Element 1.4 in y = 1 m [MPa]
47.82
1.44.3
Calculated results
Result name
Result description
Value
Error
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]
7.45058e015 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]
1.49012e014 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%
169
ADVANCE DESIGN VALIDATION GUIDE
1.45 Thin cylinder under a hydrostatic pressure (010043SSLSB_FEM) Test ID: 2474 Test status: Passed
1.45.1
Description
Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder under a hydrostatic pressure.
1.45.2
Background
1.45.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 08/89; Analysis type: static, linear elastic; Element type: planar. Thin cylinder under a hydrostatic pressure 010043SSLSB_FEM
Scale = 1/25
Units I. S. Geometry ■ ■ ■
Thickness: h = 0.02 m, Length: L = 4 m, Radius: R = 1 m.
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
170
Outer: For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on the nodes that are parallel with the cylinder’s axis.
ADVANCE DESIGN VALIDATION GUIDE
■
Inner: None.
Loading ■
z External: Radial internal pressure varies linearly with the "p" height, p = p0 L ,
■
Internal: None.
1.45.2.2 Stresses Reference solution x axis of the local coordinate system of planar elements is parallel to the cylinders axis. σxx = 0 σyy =
p0Rz Lh
Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 209 nodes, 180 surface quadrangles.
1.45.2.3 Cylinder deformation Reference solution ■
δL longitudinal deformation of the cylinder: δL =
■
p0Rνz2 2ELh
δL radial deformation of the cylinder: p0R2z δR = ELh
Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 209 nodes, 180 surface quadrangles.
171
ADVANCE DESIGN VALIDATION GUIDE
Deformation shape Thin cylinder under a hydrostatic pressure Deformed
1.45.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
syy_mid
σyy stress in z = L/2 [Pa]
500000.000000
CM2
DY
δL longitudinal deformation of the cylinder at the inferior extremity [mm]
0.002860
CM2
Dz
δL radial deformation of the cylinder in z = L/2 [mm]
0.002380
1.45.3
172
Calculated results
Result name
Result description
Value
Error
syy_mid
Sigma yy stress in z = L/2 [Pa]
504489 Pa
0.90%
DY
Delta L longitudinal deformation of the cylinder at the inferior extremity [mm]
0.00285442 mm
0.20%
Dz
Delta L radial deformation of the cylinder in z = L/2 [mm]
0.00238372 mm
0.16%
ADVANCE DESIGN VALIDATION GUIDE
1.46 Pinch cylindrical shell (010048SSLSB_FEM) Test ID: 2478 Test status: Passed
1.46.1
Description
A cylinder of length L is pinched by 2 diametrically opposite forces (F). The vertical displacement is verified.
1.46.2
Background
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 20/89; Analysis type: static, linear elastic; Element type: planar.
1.46.2.1 Model description A cylinder of length L is pinched by 2 diametrically opposite forces (F).
Pinch cylindrical shell
010048SSLSB_FEM
Units I. S. Geometry ■ ■
Length: L = 10.35 m (total length), Radius: R = 4.953 m, 173
ADVANCE DESIGN VALIDATION GUIDE
■
Thickness: h = 0.094 m.
Materials properties ■
Longitudinal elastic modulus: E = 10.5 x 106 Pa,
■
Poisson's ratio: ν = 0.3125.
Boundary conditions ■ ■
Outer: For the modeling, we consider only half of the cylinder, so we impose symmetry conditions (nodes in the horizontal xz plane are restrained in translation along y and in rotation along x and z), Inner: None.
Loading ■ ■
External: 2 punctual loads F = 100 N, Internal: None.
1.46.2.2 Vertical displacement at point A Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution. Finite elements modeling ■ ■ ■
Planar element: shell, imposed mesh, 777 nodes, 720 surface quadrangles.
1.46.2.3 Theoretical result Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement in point A [m]
113.9 x 103
1.46.3
174
Calculated results
Result name
Result description
Value
Error
DZ
Vertical displacement in point A [mm]
113.3 mm
0.53%
ADVANCE DESIGN VALIDATION GUIDE
1.47 Simply supported rectangular plate loaded with punctual force and moments (010054SSLSB_FEM) Test ID: 2484 Test status: Passed
1.47.1
Description
A rectangular plate simply supported is subjected to a punctual force and moments. The vertical displacement is verified.
1.47.2
Background
1.47.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 26/89; Analysis type: static linear; Element type: planar. Simply supported rectangular plate loaded with punctual force and moments 010054SSLSB_FEM
Units I. S. Geometry ■ ■ ■
Width: DA = CB = 20 m, Length: AB = DC = 5 m, Thickness: h = 1 m,
Materials properties ■
Longitudinal elastic modulus: E =1000 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: Punctual support at A, B and D (null displacement along zaxis), Inner: None.
175
ADVANCE DESIGN VALIDATION GUIDE
Loading ■
■
External: ► In A: MX = 20 Nm, MY = 10 Nm, ► In B: MX = 20 Nm, MY = 10 Nm, ► In C: FZ = 2 N, MX = 20 Nm, MY = 10 Nm, ► In D: MX = 20 Nm, MY = 10 Nm, Internal: None.
1.47.2.2 Vertical displacement at C Reference solution LoveKirchhoff thin plates theory. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 867 nodes, 800 surface quadrangles.
Deformed shape Simply supported rectangular plate loaded with punctual force and moments Deformed
1.47.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point C [m]
12.480
1.47.3
176
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point C [m]
12.6677 m
1.50%
ADVANCE DESIGN VALIDATION GUIDE
1.48 Shear plate perpendicular to the medium surface (010055SSLSB_FEM) Test ID: 2485 Test status: Passed
1.48.1
Description
Verifies the vertical displacement of a rectangular shear plate fixed at one end, loaded with two forces.
1.48.2
Background
1.48.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 27/89; Analysis type: static; Element type: planar. Shear plate
010055SSLSB_FEM
Scale = 1/50
Units I. S. Geometry ■ ■ ■
Length: L = 12 m, Width: l = 1 m, Thickness: h = 0.05 m,
Materials properties ■
Longitudinal elastic modulus: E = 1.0 x 107 Pa,
■
Poisson's ratio: ν = 0.25.
Boundary conditions ■ ■
Outer: Fixed AD edge, Inner: None. 177
ADVANCE DESIGN VALIDATION GUIDE
Loading ■
■
External: ► At B: Fz = 1.0 N, ► At C: FZ = 1.0 N, Internal: None.
1.48.2.2 Vertical displacement at C Reference solution Analytical solution. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 497 nodes, 420 surface quadrangles.
Deformed shape Shear plate
Deformed
Scale = 1/35
1.48.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point C [m]
35.37 x 103
1.48.3
178
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point C [m]
35.6655 mm
0.84%
ADVANCE DESIGN VALIDATION GUIDE
1.49 A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (010058SSLSB_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 with a uniform pressure.
1.49.2
Background
1.49.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar. Square plate of side "a", for the modeling, only a quarter of the plate is considered.
Units I. S. Geometry ■ ■ ■
Side: a = 1 m, Thickness: h = 0.01333 m, a Slenderness: λ = h = 75.
Materials properties ■ ■
Reinforcement, 11 Longitudinal elastic modulus: E = 2.1 x 10 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Fixed sides: AB and BD, For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),
■
Inner: None.
179
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: 1 MPa uniform pressure, Internal: None.
1.49.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.
1.49.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point C [m]
2.8053 x 102
1.49.3
180
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point C [cm]
2.79502 cm
0.37%
ADVANCE DESIGN VALIDATION GUIDE
1.50 A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (010059SSLSB_FEM) 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 a uniform pressure.
1.50.2
Background
1.50.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.
Units I. S. Geometry ■ ■ ■
Side: a = 1 m, Thickness: h = 0.02 m, a Slenderness: λ = h = 50.
Materials properties ■ ■
Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Fixed edges: AB and BD, For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),
■
Inner: None.
181
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: 1 MPa uniform pressure, Internal: None.
1.50.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.
1.50.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point C [m]
0.83480 x 102
1.50.3
182
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point C [cm]
0.82559 cm
1.10%
ADVANCE DESIGN VALIDATION GUIDE
1.51 Simply supported rectangular plate under a uniform load (010053SSLSB_FEM) Test ID: 2483 Test status: Passed
1.51.1
Description
A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.
1.51.2
Background
1.51.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 24/89; Analysis type: static, linear elastic; Element type: planar. Simply supported rectangular plate under a uniform load 010053SSLSB_FEM
Scale = 1/25
Units I. S. Geometry ■ ■ ■
Width: a = 1 m, Length: b = 5 m, Thickness: h = 0.01 m,
Materials properties ■
Longitudinal elastic modulus: E = 1.0 x 107 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: Simple support on the plate perimeter (null displacement along zaxis), Inner: None. 183
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: Normal pressure of plate p = pZ = 1.0 Pa, Internal: None.
1.51.2.2 Vertical displacement and bending moment at the center of the plate Reference solution LoveKirchhoff thin plates theory. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 793 nodes, 720 surface quadrangles.
1.51.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement at plate center [m]
1.416 x 102
CM2
Mxx
MX bending moment at plate center [Nm]
0.1246
CM2
Myy
MY bending moment at plate center [Nm]
0.0375
1.51.3
184
Calculated results
Result name
Result description
Value
Error
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%
ADVANCE DESIGN VALIDATION GUIDE
1.52 A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (010057SSLSB_FEM) Test ID: 2487 Test status: Passed
1.52.1
Description
Verifies the vertical displacement for a square plate (0.01 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.52.2
Background
1.52.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.
Units I. S. Geometry ■ ■ ■
Side: a = 1 m, Thickness: h = 0.01 m, a Slenderness: λ = h = 100.
Materials properties ■ ■
Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Fixed sides: AB and BD, For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),
■
Inner: None.
185
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: 1 MPa uniform pressure, Internal: None.
1.52.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.
1.52.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point C [m]
6.639 x 102
1.52.3
186
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point C [cm]
6.56563 cm
1.11%
ADVANCE DESIGN VALIDATION GUIDE
1.53 A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (010061SSLSB_FEM) Test ID: 2491 Test status: Passed
1.53.1
Description
Verifies the vertical displacement for a square plate (0.1 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.53.2
Background
1.53.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.
Units I. S. Geometry ■ ■ ■
Side: a = 1 m, Thickness: h = 0.1 m, a Slenderness: λ = h = 10.
Materials properties ■ ■
Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Fixed edges: AB and BD, For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),
■
Inner: None.
187
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: 1 MPa uniform pressure, Internal: None.
1.53.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.
1.53.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point C [m]
0.78661 x 104
1.53.3
188
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point C [mm]
0.0781846 mm
0.61%
ADVANCE DESIGN VALIDATION GUIDE
1.54 A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (010062SSLSB_FEM) Test ID: 2492 Test status: Passed
1.54.1
Description
Verifies the vertical displacement for a square plate (0.01 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.54.2
Background
■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.
1.54.2.1 Model description Square plate of side "a". 0.01 m thick plate fixed on its perimeter
Scale = 1/5
010062SSLSB_FEM
Units I. S. Geometry ■ ■ ■
Side: a = 1 m, Thickness: h = 0.01 m, a Slenderness: λ = h = 100.
Materials properties ■ ■
Reinforcement, 11 Longitudinal elastic modulus: E = 2.1 x 10 Pa,
189
ADVANCE DESIGN VALIDATION GUIDE
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: Fixed edges, Inner: None.
Loading ■ ■
External: Punctual force applied on the center of the plate: FZ = 106 N, Internal: None.
1.54.2.2 Vertical displacement at point C (center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
1.54.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement in point C [m]
0.29579
1.54.3
190
Calculated results
Result name
Result description
Value
Error
DZ
Vertical displacement in point C [m]
0.292146 m
1.23%
ADVANCE DESIGN VALIDATION GUIDE
1.55 A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (010063SSLSB_FEM) Test ID: 2493 Test status: Passed
1.55.1
Description
Verifies the vertical displacement for a square plate (0.01333 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.55.2
Background
1.55.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.01333 m thick plate fixed on its perimeter
Scale = 1/5
010063SSLSB_FEM
Units I. S. Geometry ■ ■ ■
Side: a = 1 m, Thickness: h = 0.01333 m, a Slenderness: λ = h = 75.
Materials properties ■ ■
Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa, 191
ADVANCE DESIGN VALIDATION GUIDE
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■ ■
Outer: Fixed sides, 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 (the center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
1.55.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement in point C [m]
0.12525
1.55.3
192
Calculated results
Result name
Result description
Value
Error
DZ
Vertical displacement in point C [m]
0.124583 m
0.53%
ADVANCE DESIGN VALIDATION GUIDE
1.56 Simply supported rectangular plate under a uniform load (010052SSLSB_FEM) Test ID: 2482 Test status: Passed
1.56.1
Description
A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.
1.56.2
Background
1.56.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLS 24/89; Analysis type: static, linear elastic; Element type: planar. Simply supported rectangular plate under a uniform load 010052SSLSB_FEM
Scale = 1/11
Units I. S. Geometry ■ ■ ■
Width: a = 1 m, Length: b = 2 m, Thickness: h = 0.01 m,
Materials properties ■
Longitudinal elastic modulus: E = 1.0 x 107 Pa,
■
Poisson's ratio: ν = 0.3.
193
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Simple support on the plate perimeter (null displacement along zaxis), Inner: None.
Loading ■ ■
External: Normal pressure of plate p = pZ = 1.0 Pa, Internal: None.
1.56.2.2 Vertical displacement and bending moment at the center of the plate Reference solution LoveKirchhoff thin plates theory. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 435 nodes, 392 surface quadrangles.
1.56.2.3 Theoretical background Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement at plate center [m]
1.1060 x 102
CM2
Mxx
MX bending moment at plate center [Nm]
0.1017
CM2
Myy
MY bending moment at plate center [Nm]
0.0464
1.56.3
194
Calculated results
Result name
Result description
Value
Error
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%
ADVANCE DESIGN VALIDATION GUIDE
1.57 Triangulated system with hinged bars (010056SSLLB_FEM) Test ID: 2486 Test status: Passed
1.57.1
Description
A truss with hinged bars is placed on three punctual supports (subjected to imposed displacements) and is loaded with two punctual forces. A thermal load is applied to all the bars. The traction force and the vertical displacement are verified.
1.57.2
Background
1.57.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 12/89; Analysis type: static (plane problem); Element type: linear.
Units I. S. Geometry ■ ■ ■
θ = 30°, Section A1 = 1.41 x 103 m2, 3 2 Section A2 = 2.82 x 10 m .
Materials properties ■
Longitudinal elastic modulus: E =2.1 x 1011 Pa,
■
Coefficient of linear expansion: α = 105 °C1.
Boundary conditions ■
Outer: Hinge at A (uA = vA = 0), Roller supports at B and C ( uB = v’C = 0), Inner: None.
► ►
■
195
ADVANCE DESIGN VALIDATION GUIDE
Loading ■
■
External: ► Support displacement: vA = 0.02 m ; vB = 0.03 m ; v’C = 0.015 m , ► Punctual loads: FE = 150 KN ; FF = 100 KN, ► Expansion effect on all bars for a temperature variation of 150° in relation with the assembly temperature (specified geometry), Internal: None.
1.57.2.2 Tension force in BD bar Reference solution Determining the hyperstatic unknown with the section cut method. Finite elements modeling ■ ■ ■
Linear element: S beam, automatic mesh, 11 nodes, 17 S beams + 1 rigid S beam for the modeling of the simple support at C.
1.57.2.3 Vertical displacement at D Reference solution vD displacement was determined by several software with implemented finite elements method. Finite elements modeling ■ ■ ■
Linear element: S beam, automatic mesh, 11 nodes, 17 S beams + 1 rigid S beam for the modeling of simple support at C.
Deformed shape Triangulated system with hinged bars
196
010056SSLLB_FEM
ADVANCE DESIGN VALIDATION GUIDE
1.57.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
Fx
FX traction force on BD bar [N]
43633
CM2
DZ
Vertical displacement on point D [m]
0.01618
1.57.3
Calculated results
Result name
Result description
Value
Error
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%
197
ADVANCE DESIGN VALIDATION GUIDE
1.58 A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (010060SSLSB_FEM) Test ID: 2490 Test status: Passed
1.58.1
Description
Verifies the vertical displacement for a square plate (0.05 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.58.2
Background
1.58.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.
Units I. S. Geometry ■ ■ ■
Side: a = 1 m, Thickness: h = 0.05 m, a Slenderness: λ = h = 20.
Materials properties ■ ■
Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Fixed edges: AB and BD, For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),
■
198
Inner: None.
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: 1 MPa uniform pressure, Internal: None.
1.58.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.
199
ADVANCE DESIGN VALIDATION GUIDE
1.58.2.3 heoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Vertical displacement in point C [m]
0.55474 x 103
1.58.3
200
T
Calculated results
Result name
Result description
Value
Error
Dz
Vertical displacement in point C [cm]
0.0549874 cm
0.88%
ADVANCE DESIGN VALIDATION GUIDE
1.59 A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (010066SSLSB_FEM) Test ID: 2496 Test status: Passed
1.59.1
Description
Verifies the vertical displacement for a square plate (0.1 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.59.2
Background
1.59.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.1 m thick plate fixed on its perimeter
Scale = 1/5
010066SSLSB_FEM
Units I. S. Geometry ■ ■
Side: a = 1 m, Thickness: h = 0.1 m,
■
Slenderness: λ = 10.
Materials properties ■ ■
Reinforcement, 11 Longitudinal elastic modulus: E = 2.1 x 10 Pa,
■
Poisson's ratio: ν = 0.3.
201
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Fixed edges, Inner: None.
Loading ■ ■
External: punctual force applied in the center of the plate: FZ = 106 N, Internal: None.
1.59.2.2 Vertical displacement at point C (center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
1.59.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement in point C [m]
0.42995 x 103
1.59.3
202
Calculated results
Result name
Result description
Value
Error
DZ
Vertical displacement in point C [mm]
0.412094 mm
4.15%
ADVANCE DESIGN VALIDATION GUIDE
1.60 Vibration mode of a thin piping elbow in space (case 1) (010067SDLLB_FEM) Test ID: 2497 Test status: Passed
1.60.1
Description
Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, fixed on its ends and subjected to its self weight only.
1.60.2
Background
1.60.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (space problem); Element type: linear. Vibration mode of a thin piping elbow
Scale = 1/7
010067SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■
Average radius of curvature: OA = R = 1 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, 4 2 Section: A = 1.131 x 10 m , Flexure moment of inertia relative to the yaxis: Iy = 4.637 x 109 m4, 9 4 Flexure moment of inertia relative to zaxis: Iz = 4.637 x 10 m , 9 4 Polar inertia: Ip = 9.274 x 10 m .
203
ADVANCE DESIGN VALIDATION GUIDE
■
Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0)
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■ ■
Outer: Fixed at points A and B, Inner: None.
Loading ■ ■
External: None, Internal: None.
1.60.2.2 Eigen modes frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =
transverse bending: µi2 2π R2
GIp ρA
where i = 1,2.
Finite elements modeling ■ ■ ■
Linear element: beam, 11 nodes, 10 linear elements.
Eigen mode shapes
1.60.2.3 Theoretical results
204
Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode Transverse 1 frequency [Hz]
44.23
CM2
Eigen mode
Eigen mode Transverse 2 frequency [Hz]
125
ADVANCE DESIGN VALIDATION GUIDE
1.60.3
Calculated results
Result name
Result description
Value
Error
Eigen mode Transverse 1 frequency [Hz]
44.12 Hz
0.25%
Eigen mode Transverse 2 frequency [Hz]
120.09 Hz
3.93%
205
ADVANCE DESIGN VALIDATION GUIDE
1.61 Reactions on supports and bending moments on a 2D portal frame (Rafters) (010077SSLPB_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
1.61.2.1 Model description ■ ■ ■
206
Reference: Design and calculation of metal structures. Analysis type: static linear; Element type: linear.
ADVANCE DESIGN VALIDATION GUIDE
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:
VA = VE =
qL 2
H A = HE =
MB = MD = −Hh
MC =
qL ² 8h + 5f =H 32 h²(k + 3 ) + f (3h + f ) qL ² − H(h + f ) 8
1.61.2.3 Theoretical results Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the chords Solver
Result name
Result description
Reference value
CM2
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]
332.9
CM2
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
1.61.3
Calculated results
Result name
Result description
Value
Error
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%
207
ADVANCE DESIGN VALIDATION GUIDE
1.62 Reactions on supports and bending moments on a 2D portal frame (Columns) (010078SSLPB_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
1.62.2.1 Model description ■ ■ ■
208
Reference: Design and calculation of metal structures. Analysis type: static linear; Element type: linear.
ADVANCE DESIGN VALIDATION GUIDE
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:
VA = − VE = −
MB =
qh² 2L
HE =
qh ² − HEh 2
qh² 5kh + 6(2h + f ) 16 h²(k + 3 ) + f (3h + f )
H A = HE − qh
qh² − HE (h + f ) 4
MD = −HEh
MC =
1.62.2.3 Theoretical results Solver
Result name
Result description
Reference value
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
1.62.3
Calculated results
Result name
Result description
Value
Error
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%
209
ADVANCE DESIGN VALIDATION GUIDE
1.63 A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (010065SSLSB_FEM) Test ID: 2495 Test status: Passed
1.63.1
Description
Verifies the vertical displacement for a square plate (0.05 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.63.2
Background
1.63.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.05 m thick plate fixed on its perimeter 010065SSLSB_FEM
Units I. S. Geometry ■ ■
Side: a = 1 m, Thickness: h = 0.05 m,
■
Slenderness: λ = 20.
Materials properties
210
■ ■
Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
Scale = 1/5
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Fixed sides, Inner: None.
Loading ■ ■
External: Punctual force applied at the center of the plate: FZ = 106 N, Internal: None.
1.63.2.2 Vertical displacement at point C center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
1.63.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement in point C [m]
0.2595 x 102
1.63.3
Calculated results
Result name
Result description
Value
Error
DZ
Vertical displacement in point C [m]
0.00257232 m
0.86%
211
ADVANCE DESIGN VALIDATION GUIDE
1.64 Vibration mode of a thin piping elbow in space (case 3) (010069SDLLB_FEM) Test ID: 2499 Test status: Passed
1.64.1
Description
Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with two straight elements (2 m long) and subjected to its self weight only.
1.64.1.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (space problem); Element type: linear. Vibration mode of a thin piping elbow
Scale = 1/12
010069SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
212
Average radius of curvature: OA = R = 1 m, L = 2 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, 4 2 Section: A = 1.131 x 10 m , Flexure moment of inertia relative to the yaxis: Iy = 4.637 x 109 m4, 9 4 Flexure moment of inertia relative to zaxis: Iz = 4.637 x 10 m , 9 4 Polar inertia: Ip = 9.274 x 10 m . Points coordinates (in m): ► O(0;0;0) ► A(0;R;0)
ADVANCE DESIGN VALIDATION GUIDE
B(R;0;0) C ( L ; R ; 0 ) D ( R ; L ; 0 )
► ► ►
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■
Outer: Fixed at points C and D At A: translation restraint along y and z, ► At B: translation restraint along x and z, Inner: None. ► ►
■
Loading ■ ■
External: None, Internal: None.
1.64.1.2 Eigen modes frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■
transverse bending: fj =
µi2 2π R2
GIp ρA
where i = 1,2 with i = 1,2:
Finite elements modeling ■ ■ ■
Linear element: beam, 41 nodes, 40 linear elements.
Eigen mode shapes
213
ADVANCE DESIGN VALIDATION GUIDE
1.64.1.3 Theoretical results Reference Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode Transverse 1 frequency [Hz]
17.900
CM2
Eigen mode
Eigen mode Transverse 2 frequency [Hz]
24.800
1.64.2
Calculated results
Result name
214
Result description
Value
Error
Eigen mode Transverse 1 frequency [Hz]
17.65 Hz
1.40%
Eigen mode Transverse 2 frequency [Hz]
24.43 Hz
1.49%
ADVANCE DESIGN VALIDATION GUIDE
1.65 Slender beam of variable rectangular section with fixedfree ends (ß=5) (010085SDLLB_FEM) Test ID: 2503 Test status: Passed
1.65.1
Description
Verifies the eigen modes (bending) for a slender beam with variable rectangular section (fixedfree).
1.65.2
Background
1.65.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 09/89; Analysis type: modal analysis (plane problem); Element type: linear.
Units I. S. Geometry ■ ■
■
Length: L = 1 m, Straight initial section: ► h0 = 0.04 m ► b0 = 0.05 m ► A0 = 2 x 103 m² Straight final section ► h1 = 0.01 m ► b1 = 0.01 m ► A1 = 104 m²
Materials properties ■
E = 2 x 1011 Pa
■
ρ = 7800 kg/m3
Boundary conditions ■
Outer: Fixed at end x = 0, ► Free at end x = 1 Inner: None. ►
■
Loading ■ ■
External: None, Internal: None.
215
ADVANCE DESIGN VALIDATION GUIDE
1.65.2.2 Reference results Calculation method used to obtain the reference solution Precise calculation by numerical integration of the differential equation of beams bending (EulerBernoulli theories):
δ2 δ²ν δ²ν EIz = −ρA δx ² δt ² δx 2 where Iz and A vary with the abscissa. The result is:
hο α = h1 = 4 1 h1 E with fi = λi(α, β) 12ρ 2π l² β = bο = 5 b1 β=5
λ1 24.308
λ2 75.56
λ3 167.21
λ4 301.9
Uncertainty about the reference: analytical solution: Reference values Eigen mode type Flexion
MODE 1
216
1 2 3 4 5
Frequency (Hz) 56.55 175.79 389.01 702.36 1117.63 Scale = 1/4
λ5 480.4
ADVANCE DESIGN VALIDATION GUIDE
MODE 2
MODE 3
Scale = 1/4
Scale = 1/4
217
ADVANCE DESIGN VALIDATION GUIDE
MODE 4
MODE 5
218
Scale = 1/4
Scale = 1/4
ADVANCE DESIGN VALIDATION GUIDE
1.65.2.3 Theoretical results Theoretical Frequency
1.65.3
Result name
Result description
Reference value
Eigen mode
Frequency of eigen mode 1 [Hz]
56.55
Eigen mode
Frequency of eigen mode 2 [Hz]
175.79
Eigen mode
Frequency of eigen mode 3 [Hz]
389.01
Eigen mode
Frequency of eigen mode 4 [Hz]
702.36
Eigen mode
Frequency of eigen mode 5 [Hz]
1117.63
Calculated results
Result name
Result description
Value
Error
Frequency of eigen mode 1 [Hz]
58.49 Hz
3.43%
Frequency of eigen mode 2 [Hz]
177.67 Hz
1.07%
Frequency of eigen mode 3 [Hz]
388.85 Hz
0.04%
Frequency of eigen mode 4 [Hz]
697.38 Hz
0.71%
Frequency of eigen mode 5 [Hz]
1106.31 Hz
1.01%
219
ADVANCE DESIGN VALIDATION GUIDE
1.66 Slender beam of variable rectangular section (fixedfixed) (010086SDLLB_FEM) Test ID: 2504 Test status: Passed
1.66.1
Description
Verifies the eigen modes (flexion) for a slender beam with variable rectangular section (fixedfixed).
1.66.2
BackgroundOverview
1.66.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 10/89; Analysis type: modal analysis (plane problem); Element type: linear.
Units I. S. Geometry ■ ■ ■
■
Length: L = 0.6 m, Constant thickness: h = 0.01 m Initial section: ► b0 = 0.03 m ► A0 = 3 x 104 m² Section variation: ►
with (α = 1)
►
b = b0e2αx
►
A = A0e2αx
Materials properties ■
E = 2 x 1011 Pa
■
ν = 0.3
■
ρ = 7800 kg/m3
Boundary conditions ■
Outer: Fixed at end x = 0, Fixed at end x = 0.6 m. Inner: None.
► ►
■
Loading ■ ■
220
External: None, Internal: None.
ADVANCE DESIGN VALIDATION GUIDE
1.66.2.2 Reference results Calculation method used to obtain the reference solution ωi pulsation is given by the roots of the equation:
1 − cos(rl)ch(sl) +
s² − r ² sh(sl)sin(rl) = 0 2rs
with
λ4i =
ρA 0 ωi2 si ; r = α ² + λ2i ; s = λ2i − α ² → λ2i − α ² > 0 EIzo
(
)
Therefore, the ν translation components of φi(x) mode, are:
cos(rl) − ch(sl) Φ i (x ) = e αx cos(rx ) − ch(sx ) + (s sin(rx ) − rsh(sx )) rsh(sl) − s sin(rl) Uncertainty about the reference: analytical solution: Reference values Eigen mode Frequency (Hz) Eigen mode φi(x)* order x=0 0.1 0.2 0.3 0.4 1 143.303 0 0.237 0.703 1 0.859 2 396.821 0 0.504 0.818 0 0.943 3 779.425 0 0.670 0.210 0.831 0.257 4 1289.577 0 0.670 0.486 0 0.594 * φi(x) eigen modes* standardized to 1 at the point of maximum amplitude.
0.5 0.354 0.752 1 1
0.6 0 0 0 0
Eigen modes
221
ADVANCE DESIGN VALIDATION GUIDE
1.66.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Eigen mode
Frequency of eigen mode 1 [Hz]
143.303
CM2
Eigen mode
Frequency of eigen mode 2 [Hz]
396.821
CM2
Eigen mode
Frequency of eigen mode 3 [Hz]
779.425
CM2
Eigen mode
Frequency of eigen mode 4 [Hz]
1289.577
1.66.3
Calculated results
Result name
222
Result description
Value
Error
Frequency of eigen mode 1 [Hz]
145.88 Hz
1.80%
Frequency of eigen mode 2 [Hz]
400.26 Hz
0.87%
Frequency of eigen mode 3 [Hz]
783.15 Hz
0.48%
Frequency of eigen mode 4 [Hz]
1293.42 Hz
0.30%
ADVANCE DESIGN VALIDATION GUIDE
1.67 Plane portal frame with hinged supports (010089SSLLB_FEM) Test ID: 2505 Test status: Passed
1.67.1
Description
Calculation of support reactions of a 2D portal frame with hinged supports.
1.67.2
Background
1.67.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 14/89; Analysis type: static linear; Element type: linear.
Units I. S. Geometry ■ ■ ■ ■ ■ ■
Length: L = 20 m, 4 4 I1 = 5.0 x 10 m a=4m h=8m b = 10.77 m 4 4 I2 = 2.5 x 10 m
Materials properties ■ ■
Isotropic linear elastic material. 11 E = 2.1 x 10 Pa
223
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions Hinged base plates A and B (uA = vA = 0 ; uB = vB = 0). Loading
■ ■ ■ ■
p = 3 000 N/m F1 = 20 000 N F2 = 10 000 N M = 100 000 Nm
1.67.2.2 Calculation method used to obtain the reference solution ■ ■ ■ ■ ■ ■ ■ ■
K = (I2/b)(h/I1) p = a/h m=1+p B = 2(K + 1) + m C = 1 + 2m N = B + mC VA = 3pl/8 + F1/2 – M/l + F2h/l HA = pl²(3 + 5m)/(32Nh) + (F1l/(4h))(C/N) + F2(1(B + C)/(2N)) + (3M/h)((1 + m)/(2N))
1.67.2.3 Reference values Point A A C
Magnitudes and units V, vertical reaction (N) H, horizontal reaction (N) vc (m)
Value 31 500.0 20 239.4 0.03072
1.67.2.4 Theoretical results Solver
Result name
Result description
Reference value
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
1.67.3
224
Calculated results
Result name
Result description
Value
Error
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%
ADVANCE DESIGN VALIDATION GUIDE
1.68 A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (010064SSLSB_FEM) Test ID: 2494 Test status: Passed
1.68.1
Description
Verifies the vertical displacement for a square plate (0.02 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.68.2
Background
1.68.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.02 m thick plate fixed on its perimeter
Scale = 1/5
010064SSLSB_FEM
Units I. S. Geometry ■ ■
Side: a = 1 m, Thickness: h = 0.02 m,
■
Slenderness: λ = 50.
Materials properties ■ ■
Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
225
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: Fixed edges, Inner: None.
Loading ■ ■
External: punctual force applied in the center of the plate: FZ = 106 N, Internal: None.
1.68.2.2 Vertical displacement at point C (the center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■
226
Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.
ADVANCE DESIGN VALIDATION GUIDE
1.68.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DZ
Vertical displacement in point C [m]
0.037454
1.68.3
Calculated results
Result name
Result description
Value
Error
DZ
Vertical displacement in point C [m]
0.0369818 m
1.26%
227
ADVANCE DESIGN VALIDATION GUIDE
1.69 Vibration mode of a thin piping elbow in space (case 2) (010068SDLLB_FEM) Test ID: 2498 Test status: Passed
1.69.1
Description
Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with two straight elements (0.6 m long) and subjected to its self weight only.
1.69.2
Background
1.69.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (in space); Element type: linear. Vibration mode of a thin piping elbow
Scale = 1/11
010068SDLLB_FEM
Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■ ■
228
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 104 m2, 9 4 Flexure moment of inertia relative to the yaxis: Iy = 4.637 x 10 m , 9 4 Flexure moment of inertia relative to zaxis: Iz = 4.637 x 10 m , 9 4 Polar inertia: Ip = 9.274 x 10 m .
ADVANCE DESIGN VALIDATION GUIDE
■
Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0) ► C ( L ; R ; 0 ) ► D ( R ; L ; 0 )
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Density: ρ = 7800 kg/m3.
Boundary conditions ■
Outer: Fixed at points C and D In A: translation restraint along y and z, ► In B: translation restraint along x and z, Inner: None. ► ►
■
Loading ■ ■
External: None, Internal: None.
1.69.2.2 Eigen modes frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■
transverse bending: fj =
µi2 2π R2
GIp ρA
where i = 1,2.
Finite elements modeling ■ ■ ■
Linear element: beam, 23 nodes, 22 linear elements.
Eigen mode shapes
229
ADVANCE DESIGN VALIDATION GUIDE
1.69.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Eigen mode
Eigen mode Transverse 1 frequency [Hz]
33.4
CM2
Eigen mode
Eigen mode Transverse 2 frequency [Hz]
100
1.69.3
Calculated results
Result name
230
Result description
Value
Error
Eigen mode Transverse 1 frequency [Hz]
33.19 Hz
0.63%
Eigen mode Transverse 2 frequency [Hz]
94.62 Hz
5.38%
ADVANCE DESIGN VALIDATION GUIDE
1.70 Short beam on two hinged supports (010084SSLLB_FEM) Test ID: 2502 Test status: Passed
1.70.1
Description
Verifies the deflection magnitude on a nonslender beam with two hinged supports.
1.70.2
Background
1.70.2.1 Model description ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SSLL 02/89 Analysis type: static linear (plane problem); Element type: linear.
Units I. S. Geometry ■ ■ ■ ■
Length: L = 1.44 m, Area: A = 31 x 104 m² 8 4 Inertia: I = 2810 x 10 m Shearing coefficient: az = 2.42 = A/Ar
Materials properties ■
E = 2 x 1011 Pa
■
ν = 0.3
Boundary conditions ■ ■
Hinge at end x = 0, Hinge at end x = 1.44 m.
231
ADVANCE DESIGN VALIDATION GUIDE
Loading Uniformly distributed force of p = 1. X 105 N/m on beam AB.
1.70.2.2 Reference results Calculation method used to obtain the reference solution The deflection on the middle of a nonslender beam considering the shear force deformations given by the Timoshenko function:
v=
5 pl 4 l 2p + 384 EI 8 A r G
where G =
E 2(1 + υ)
and A r = A
az
where "Ar" is the reduced area and "az" the shear coefficient calculated on the transverse section. Uncertainty about the reference: analytical solution: Reference values Point C
Magnitudes and units V, deflection (m)
Value 1.25926 x 103
1.70.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Dz
Deflection magnitude in point C [m]
0.00125926
1.70.3
232
Calculated results
Result name
Result description
Value
Error
Dz
Deflection magnitude in node C [m]
0.00125926 m
0.00%
ADVANCE DESIGN VALIDATION GUIDE
1.71 A 3D bar structure with elastic support (010094SSLLB_FEM) Test ID: 2508 Test status: Passed
1.71.1
Description
A 3D bar structure with elastic support is subjected to a vertical load of 100 kN. The V2 magnitude on node 5, the normal force magnitude, the reaction magnitude on supports and the action magnitude are verified.
1.71.1.1 Model description ■ ■ ■
Reference: Internal GRAITEC; Analysis type: static linear; Element type: linear.
Units I. S. Geometry For all bars: ■ ■ ■
H=3m B=3m S = 0.02 m2 Element 1 (bar) 2 (bar) 3 (bar) 4 (bar) 5 (spring)
Node i 1 2 3 4 5
Node j 5 5 5 5 6
Materials properties ■ ■
Isotropic linear elastic materials Longitudinal elastic modulus: E = 2.1 E8 N/m2,
233
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Outer: At node 5: K = 50000 kN/m ; Inner: None.
Loading ■ ■
External: Vertical load at node: P = 100 kN, Internal: None.
1.71.1.2 Theoretical results System solution
L= H2 + ■
234
B2 . Also, U1 = V1 = U5 = U6 = V6 = 0 2
Stiffness matrix of bar 1
ADVANCE DESIGN VALIDATION GUIDE
x x 1 1 u ( x) = (1 − ).ui + .u j ⇔ u (ξ ) = (1 − ξ ).ui + (1 + ξ ).u j L L 2 2 2x where ξ = −1 L in the local coordinate system:
[k1 ] = ∫ [B] [H ][B]dVe = ES ∫0 T
L
ve
where
1 2 dξ 2 L
1 1 − 4 dξ = ES 1 − 1 (ui ) = ES 0 1 L − 1 1 (u j ) L − 1 4 0
1 1 2 ES 4 = L −∫1 − 1 4 1 [k 1 ] = ES 0 L −1 0
1 − [B]T [B]dx = ES ∫−1 12 − 1 2 2 1
0 −1 0 0 0 1 0 0
0 (ui ) 0 (vi ) 0 (u j ) 0 (v j )
0 − 1 0 (u1 ) 0 0 0 ( v 1 ) 0 1 0 (u 5 ) 0 0 0 ( v 5 )
[ ]
The elementary matrix k e expressed in the global coordinate system XY is the following: (θ angle allowing the transition from the global base to the local base):
0 0 cos θ sin θ − θ θ sin cos 0 0 T − [K e ] = [R e ] [k e ][R e ] avec [R e ] = 0 0 cos θ sin θ − sin θ cos θ 0 0 cos 2 θ − cos 2 θ − cos θ sin θ cos θ sin θ 2 θ − cos θ sin θ − sin 2 θ sin θ [K e ] = ES cos θ sin L − cos 2 θ − cos θ sin θ cos 2 θ cos θ sin θ 2 − sin θ cos θ sin θ sin 2 θ − cos θ sin θ
Knowing that
cosθ =
B L 2
and sinθ
B2 2 cos θ = 2L2 H H2 , then: sin 2 θ = 2 = L L HB sin θ cos θ = 2L2
B2 2 HB ES H for element 1 nodes 1 → 5, θ = arctan ( ) : [K1 ] = 3 22 B L D − 2 HB − 2
HB 2 H2 −
HB 2
−H2
B2 2 HB − 2 B2 2 HB 2 −
HB 2 (U 1 ) −H2 (V1 ) HB (U 5 ) 2 (V5 ) H2
−
235
ADVANCE DESIGN VALIDATION GUIDE
■
Stiffness matrix of spring support 5
We say K ′ =
K 4
1 0 1 − 1 (ui ) ′ in the local coordinate system : [k 5 ] = K ′ = K − 1 − 1 1 (u j ) 0
0 0 for element 5 nodes 5 → 6, θ = 90° : [K 5 ] = K ' 0 0
0 − 1 0 (ui ) 0 0 0 (vi ) 0 1 0 (u j ) 0 0 0 (v j )
0 (U 5 ) 1 0 − 1 (V5 ) 0 0 0 (U 6 ) 1 0 − 1 (V6 ) 0 0
[ ]
System K {Q} = {F}
■
ES B 2 3 L 2 ES HB L3 2 2 − ES B 3 L 2 − ES HB L3 2 0 0
ES HB L3 2 ES 2 H L3 ES HB − 3 L 2 ES 2 − 3 H L 0 0
ES B 2 L3 2 ES HB − 3 L 2 ES B 2 L3 2 ES HB L3 2 0 0
−
ES HB L3 2 ES − 3 H2 L ES HB L3 2 ES 2 H + K′ L3 0 − K′
−
0
0 0 0 0 0
0 U R X1 1 0 V R Y 1 1 R U 5 X5 0 = V5 −P − K ′ U 6 R X 6 V R 6 Y6 0 K ′
If U1 = V1 = U5 = U6 = V6 = 0, then:
V5 =
And
4
ES 2 H + K′ L3
=
−P
4 = −0.001885 m ES 2 K H + 4 L3
ES HB ES HB V5 = 1015 N R X5 = 3 V5 = −1015 N R X6 = 0 L3 2 L 2 ES K R Y 6 = − V5 = 23563 N = − 3 H 2 V5 = 1436 N 4 L
R X1 = − R Y1
Note:
236
−P
■ ■
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
ADVANCE DESIGN VALIDATION GUIDE
Effort in bar 1:
B L 2 u1 H v − 1 L = u5 0 v5 0
H L B L 2
0 0 B
0
L 2 H − L
0
u1 cosθ v − sin θ 1 = u5 0 v5 0
0 U1 0 V 1 and ES 1 − 1 u1 = N1 = 1759 H U 5 L − 1 1 u5 N 5 − 1759 L V 5 B L 2
sin θ cosθ
0 0
0
cosθ
0
− sin θ
0 U1 0 V1 ES 1 − 1 u1 N1 and = sin θ U 5 L − 1 1 u5 N 5 cosθ V5
Reference values
Point 5 All bars Supports 1, 3, 4 and 5 Supports 1, 3, 4 and 5
Magnitude V2 Normal force Fz action Action Fx=Fy
Units m N N N
Support 6
Fz action
N
Value 1.885 103 1759 1436
1015 / 2 = ±718 23563 x 4=94253
Finite elements modeling ■ ■ ■
Linear element: beam, automatic mesh, 5 nodes, 4 linear elements.
237
ADVANCE DESIGN VALIDATION GUIDE
Deformed shape
Normal forces diagram
238
ADVANCE DESIGN VALIDATION GUIDE
1.71.1.3 Reference values Solver
Result name
Result description
Reference value
CM2
D
V2 magnitude on node 5 [m]
1.885 103
CM2
Fx
Normal force magnitude on bar 1 [N]
1759
CM2
Fx
Normal force magnitude on bar 2 [N]
1759
CM2
Fx
Normal force magnitude on bar 3 [N]
1759
CM2
Fx
Normal force magnitude on bar 4 [N]
1759
CM2
Fz
Fz reaction magnitude on support 1 [N]
1436
CM2
Fz
Fz reaction magnitude on support 3 [N]
1436
CM2
Fz
Fz reaction magnitude on support 4 [N]
1436
CM2
Fz
Fz reaction magnitude on support 5 [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]
718
CM2
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
1.71.2
Calculated results
Result name
Result description
Value
Error
D
Displacement on node 5 [mm]
1.88508 mm
0.00%
Fx
Normal force magnitude on bar 1 [N]
1759.4 N
0.02%
Fx
Normal force magnitude on bar 2 [N]
1759.4 N
0.02%
Fx
Normal force magnitude on bar 3 [N]
1759.4 N
0.02%
Fx
Normal force magnitude on bar 4 [N]
1759.4 N
0.02%
Fy
Fz reaction magnitude on support 1 [N]
1436.55 N
0.04%
Fy
Fz reaction magnitude on support 2 [N]
1436.55 N
0.04%
Fy
Fz reaction magnitude on support 3 [N]
1436.55 N
0.04%
Fy
Fz reaction magnitude on support 4 [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%
239
ADVANCE DESIGN VALIDATION GUIDE
1.72 Fixed/free slender beam with centered mass (010095SDLLB_FEM) Test ID: 2509 Test status: Passed
1.72.1
Description
Fixed/free slender beam with centered mass. Tested functions: Eigen mode frequencies, straight slender beam, combined bendingtorsion, plane bending, transverse bending, punctual mass.
1.72.2
Background
■ ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 15/89; Analysis type: modal analysis; Element type: linear. Tested functions: Eigen mode frequencies, straight slender beam, combined bendingtorsion, plane bending, transverse bending, punctual mass.
1.72.2.1 Test data
Units I. S. Geometry ■ ■ ■ ■ ■ ■
240
Outer diameter Inner diameter: Beam length: Area: Polar inertia: Inertia:
de = 0.35 m, di = 0.32 m, l = 10 m, A = 1.57865 x 102 m2 4 4 IP = 4.43798 x 10 m 4 4 Iy = Iz = 2.21899 x 10 m
ADVANCE DESIGN VALIDATION GUIDE
■ ■
Punctual mass: Beam selfweight:
mc = 1000 kg M
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 Pa,
■
Density: ρ = 7800 kg/m3
■
Poisson's ratio: ν=0.3 (this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the more appropriate to obtain the correct frequency value of mode No. 8 with NE/NASTRAN)
Boundary conditions ■ ■
Outer: Fixed at point A, x = 0, Inner: none
Loading None for the modal analysis
1.72.2.2 Reference results Reference frequency For the first mode, the Rayleigh method gives the approximation formula
f1 = 1 / 2Π x
3 EI z 3
I (m c + 0.24M)
Mode 1 2 3 4 5 6 7 8 9 10 Comment:
Shape Flexion Flexion Flexion Flexion Flexion Flexion Traction Torsion Flexion Flexion
Units Hz Hz Hz Hz Hz Hz Hz Hz Hz Hz
Reference 1.65 1.65 16.07 16.07 50.02 50.02 76.47 80.47 103.2 103.2
The mass matrix associated with the beam torsion on two nodes, is expressed as:
ρ × l × IP 1 1/ 2 × 3 1/ 2 1 And to the extent that Advance Design uses a condensed mass matrix, the value of the torsion mass inertia introduced in the model is set to:
ρ × l × Ip 3
Uncertainty about the reference frequencies ■
Analytical solution mode 1
■
Other modes: ± 1%
Finite elements modeling ■ ■
Linear element AB: Beam Beam meshing: 20 elements.
241
ADVANCE DESIGN VALIDATION GUIDE
Modal deformations
242
ADVANCE DESIGN VALIDATION GUIDE
Observation: the deformed shape of mode No. 8 that does not really correspond to a torsion deformation, is actually the display result of the translations and not of the rotations. This is confirmed by the rotation values of the corresponding mode.
Node 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
DX 3.336e033 5.030e013 1.005e012 1.505e012 2.002e012 2.495e012 2.983e012 3.464e012 3.939e012 4.406e012 4.863e012 5.310e012 5.746e012 6.169e012 6.580e012 6.976e012 7.357e012 7.723e012 8.072e012 8.403e012 8.717e012
DY 6.479e031 1.575e008 6.185e008 1.365e007 2.381e007 3.648e007 5.149e007 6.867e007 8.785e007 1.088e006 1.315e006 1.556e006 1.811e006 2.077e006 2.353e006 2.637e006 2.928e006 3.224e006 3.524e006 3.826e006 4.130e006
Eigen modes vector 8 DZ RX 6.316e031 1.055e022 1.520e008 1.472e002 5.966e008 2.944e002 1.317e007 4.416e002 2.296e007 5.887e002 3.517e007 7.359e002 4.963e007 8.831e002 6.618e007 1.030e001 8.464e007 1.177e001 1.049e006 1.325e001 1.267e006 1.472e001 1.499e006 1.619e001 1.744e006 1.766e001 2.000e006 1.913e001 2.265e006 2.061e001 2.539e006 2.208e001 2.819e006 2.355e001 3.104e006 2.502e001 3.393e006 2.649e001 3.685e006 2.797e001 3.977e006 2.944e001
RY 5.770e028 6.022e008 1.171e007 1.705e007 2.206e007 2.673e007 3.106e007 3.506e007 3.873e007 4.207e007 4.508e007 4.777e007 5.015e007 5.221e007 5.396e007 5.541e007 5.658e007 5.746e007 5.808e007 5.844e007 5.856e007
RZ 5.980e028 6.243e008 1.214e007 1.769e007 2.289e007 2.774e007 3.225e007 3.641e007 4.023e007 4.371e007 4.684e007 4.964e007 5.210e007 5.423e007 5.605e007 5.755e007 5.874e007 5.965e007 6.028e007 6.065e007 6.077e007
With NE/NASTRAN, the results associated with mode No. 8, are:
243
ADVANCE DESIGN VALIDATION GUIDE
1.72.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Eigen mode 1 frequency [Hz]
1.65
CM2
Eigen mode 2 frequency [Hz]
1.65
CM2
Eigen mode 3 frequency [Hz]
16.07
CM2
Eigen mode 4 frequency [Hz]
16.07
CM2
Eigen mode 5 frequency [Hz]
50.02
CM2
Eigen mode 6 frequency [Hz]
50.02
CM2
Eigen mode 7 frequency [Hz]
76.47
CM2
Eigen mode 9 frequency [Hz]
103.20
CM2
Eigen mode 10 frequency [Hz]
103.20
Comment: The difference between the reference frequency of torsion mode (mode No. 8) and the one found by Advance Design may be explained by the fact that Advance Design is using a lumped mass matrix (see the corresponding description sheet).
1.72.3
Calculated results
Result name
244
Result description
Value
Error
Eigen mode 1 frequency [Hz]
1.65 Hz
0.00%
Eigen mode 2 frequency [Hz]
1.65 Hz
0.00%
Eigen mode 3 frequency [Hz]
16.06 Hz
0.06%
Eigen mode 4 frequency [Hz]
16.06 Hz
0.06%
Eigen mode 5 frequency [Hz]
50 Hz
0.04%
Eigen mode 6 frequency [Hz]
50 Hz
0.04%
Eigen mode 7 frequency [Hz]
76.46 Hz
0.01%
Eigen mode 9 frequency [Hz]
103.14 Hz
0.06%
Eigen mode 10 frequency [Hz]
103.14 Hz
0.06%
ADVANCE DESIGN VALIDATION GUIDE
1.73 Simple supported beam in free vibration (010098SDLLB_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 Analysis type: modal analysis; Tested functions: Shear force, eigen frequencies.
1.73.2.1 Problem data
Units I. S. Geometry
Full square section: ■ ■ ■
Dimensions: Area: Inertia:
a x b = 2m x 2 m 2 A = 4m 4 IP = 2.25 m Iy = Iz = 1.333 m4
Materials properties ■
Longitudinal elastic modulus: E = 2 x 1011 Pa,
■
Poisson's ratio: ν = 0.3.
■
Density: ρ = 8000 kg/m3
245
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
Outer:
■
x = y = z = Rx = 0 at A ; y = z =0 at B ; Inner: None. ► ►
Loading None for the modal analysis
1.73.2.2 Reference frequencies Mode 1 2 3 4 5 6 7 8 9
Shape Flexion Flexion Torsion Traction Flexion Flexion Torsion Flexion Flexion
Units Hz Hz Hz Hz Hz Hz Hz Hz Hz
Reference 42.649 42.649 77.542 125.00 148.31 148.31 233.10 284.55 284.55
Comment: Due to the condensed (lumped) nature of the mass matrix of Advance Design, the frequencies values of 3 and 7 modes cannot be found by this software. The same modeling done with NE/NASTRAN gave respectively for mode 3 and 7: 77.2 and 224.1 Hz. Finite elements modeling ■ ■
Straight elements: linear element Imposed mesh: 5 meshes
Modal deformations
246
ADVANCE DESIGN VALIDATION GUIDE
1.73.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Frequency of eigen mode 1 [Hz]
42.649
CM2
Frequency of eigen mode 2 [Hz]
42.649
CM2
Frequency of eigen mode 3 [Hz]
77.542
CM2
Frequency of eigen mode 4 [Hz]
125.00
CM2
Frequency of eigen mode 5 [Hz]
148.31
CM2
Frequency of eigen mode 6 [Hz]
148.31
CM2
Frequency of eigen mode 7 [Hz]
233.10
Comment: The torsion modes No. 3 and 7 that are calculated with NASTRAN cannot be calculated with Advance Design CM2 solver and therefore the mode No. 3 of the Advance Design analysis corresponds to mode No. 4 of the reference. The same problem in the case of No. 7  Advance Design, that corresponds to mode No. 8 of the reference.
1.73.3
Calculated results
Result name
Result description
Value
Error
Frequency of eigen mode 1 [Hz]
43.11 Hz
1.08%
Frequency of eigen mode 2 [Hz]
43.11 Hz
1.08%
Frequency of eigen mode 3 [Hz]
124.49 Hz
0.41%
Frequency of eigen mode 4 [Hz]
149.38 Hz
0.72%
Frequency of eigen mode 5 [Hz]
149.38 Hz
0.72%
Frequency of eigen mode 6 [Hz]
269.55 Hz
5.27%
Frequency of eigen mode 7 [Hz]
269.55 Hz
5.27%
247
ADVANCE DESIGN VALIDATION GUIDE
1.74 Membrane with hot point (010099HSLSB_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, thermoelastic; Tested functions: Stresses.
1.74.2.1 Problem data
Observation: the units system of the initial NAFEMS test, defined in mm, was transposed in m for practical reasons. However, this has no influence on the results values.
248
ADVANCE DESIGN VALIDATION GUIDE
Units I. S. Geometry / meshing A quarter of the structure is modeled by incorporating the terms of symmetries. Thickness: 1 m
Materials properties ■
Longitudinal elastic modulus: E = 1 x 1011 Pa,
■
Poisson's ratio: ν = 0.3,
■
Elongation coefficient α = 0.00001.
Boundary conditions ■
Outer: For all nodes in y = 0, uy =0; For all nodes in x = 0, ux =0; Inner: None.
► ►
■
Loading ■
External: None,
■
Internal: Hot point, thermal load ∆T = 100°C;
249
ADVANCE DESIGN VALIDATION GUIDE
1.74.3
σyy stress at point A:
Reference solution:
Reference value: σyy = 50 MPa in A Finite elements modeling ■ ■ ■
Planar elements: membranes, 28 planar elements, 39 nodes.
1.74.3.1 Theoretical results Solver
Result name
Result description
Reference value
CM2
syy_mid
σyy in A [MPa]
50
Note: This value (50.87) is obtained with a vertical cross section through point A. The value represents σyy at the left end of the diagram. With CM2, it is essential to display the results with the “Smooth results on planar elements” option deactivated.
1.74.4
250
Calculated results
Result name
Result description
Value
Error
syy_mid
Sigma yy in A [MPa]
50.8666 MPa
1.73%
ADVANCE DESIGN VALIDATION GUIDE
1.75 Cantilever beam in Eulerian buckling with thermal load (010092HFLLB_FEM) Test ID: 2507 Test status: Passed
1.75.1
Description
Verifies the vertical displacement and the normal force on a cantilever beam in Eulerian buckling with thermal load.
1.75.2
Background
1.75.2.1 Model description ■ ■ ■
Reference: Euler theory; Analysis type: Eulerian buckling; Element type: linear.
Units I. S. Geometry ■ ■ ■
L = 10 m S=0.01 m2 4 I = 0.0002 m
Materials properties ■
Longitudinal elastic modulus: E = 2.0 x 1010 N/m2,
■
Poisson's ratio: ν = 0.1.
■
Coefficient of thermal expansion: α = 0.00001
Boundary conditions ■ ■
Outer: Fixed at end x = 0, Inner: None.
Loading ■ ■
External: Punctual load P = 100000 N at x = L, Internal: T = 50°C (Contraction equivalent to the compression force) ( ε0 =
−100000 N = = 0.0005 = α∆T = 0.00001 × −50 ) ES 2.10 10 × 0.01
251
ADVANCE DESIGN VALIDATION GUIDE
1.75.2.2 Displacement of the model in the linear elastic range Reference solution The reference critical load established by Euler is:
Pcritical =
π 2 EI 2
4L
= 98696 N ⇒ λ =
98696 = 0.98696 100000
Observation: in this case, the thermal load has no effect over the critical coefficient Finite elements modeling ■ ■ ■
Linear element: beam, imposed mesh, 5 nodes, 4 elements.
Deformed shape
1.75.2.3 Theoretical results Solver
Result name
Result description
Reference value
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
1.75.3
Calculated results
Result name DZ
252
Result description Vertical displacement on Node 5 [cm]
Value 1 cm
Error 0.00%
Fx
Normal Force  Case 101 [N]
100000 N
0.00%
Fx
Normal Force  Case 102 [N]
98699.3 N
0.00%
ADVANCE DESIGN VALIDATION GUIDE
1.76 Double cross with hinged ends (010097SDLLB_FEM) Test ID: 2511 Test status: Passed
1.76.1
Description
Double cross with hinged ends. Tested functions: Eigen frequencies, crossed beams, in plane bending.
1.76.2
Background
■ ■ ■
Reference: NAFEMS, FV2 test Analysis type: modal analysis; Tested functions: Eigen frequencies, Crossed beams, In plane bending.
1.76.2.1 Problem data
Units I. S. Geometry
Full square section: ■ ■ ■ ■
Arm length: Dimensions: Area: Inertia:
L=5m a x b = 0.125 x 0.125 2 2 A = 1.563 10 m 5 4 IP = 3.433 x 10 m
253
ADVANCE DESIGN VALIDATION GUIDE
Iy = Iz = 2.035 x 105m4 Materials properties ■
Longitudinal elastic modulus: E = 2 x 1011 Pa,
■
Density: ρ = 8000 kg/m3
Boundary conditions ■ ■
Outer: A, B, C, D, E, F, G, H points restraint along x and y; Inner: None.
Loading None for the modal analysis
1.76.2.2 Reference frequencies Mode 1 2,3 4 to 8 9 10,11 12 to 16
Units Hz Hz Hz Hz Hz Hz
Finite elements modeling ■ ■
Linear elements type: Beam Imposed mesh: 4 Elements / Arms
Modal deformations
254
Reference 11.336 17.709 17.709 45.345 57.390 57.390
ADVANCE DESIGN VALIDATION GUIDE
1.76.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Frequency
Frequency of Eigen Mode 1 [Hz]
11.336
CM2
Frequency
Frequency of Eigen Mode 2 [Hz]
17.709
CM2
Frequency
Frequency of Eigen Mode 3 [Hz]
17.709
CM2
Frequency
Frequency of Eigen Mode 4 [Hz]
17.709
CM2
Frequency
Frequency of Eigen Mode 5 [Hz]
17.709
CM2
Frequency
Frequency of Eigen Mode 6 [Hz]
17.709
CM2
Frequency
Frequency of Eigen Mode 7 [Hz]
17.709
CM2
Frequency
Frequency of Eigen Mode 8 [Hz]
17.709
CM2
Frequency
Frequency of Eigen Mode 9 [Hz]
45.345
CM2
Frequency
Frequency of Eigen Mode 10 [Hz]
57.390
CM2
Frequency
Frequency of Eigen Mode 11 [Hz]
57.390
CM2
Frequency
Frequency of Eigen Mode 12 [Hz]
57.390
CM2
Frequency
Frequency of Eigen Mode 13 [Hz]
57.390
CM2
Frequency
Frequency of Eigen Mode 14 [Hz]
57.390
CM2
Frequency
Frequency of Eigen Mode 15 [Hz]
57.390
CM2
Frequency
Frequency of Eigen Mode 16 [Hz]
57.390
255
ADVANCE DESIGN VALIDATION GUIDE
1.76.3
Calculated results
Result name
256
Result description
Value
Error
Frequency of Eigen Mode 1 [Hz]
11.33 Hz
0.05%
Frequency of Eigen Mode 2 [Hz]
17.66 Hz
0.28%
Frequency of Eigen Mode 3 [Hz]
17.66 Hz
0.28%
Frequency of Eigen Mode 4 [Hz]
17.69 Hz
0.11%
Frequency of Eigen Mode 5 [Hz]
17.69 Hz
0.11%
Frequency of Eigen Mode 6 [Hz]
17.69 Hz
0.11%
Frequency of Eigen Mode 7 [Hz]
17.69 Hz
0.11%
Frequency of Eigen Mode 8 [Hz]
17.69 Hz
0.11%
Frequency of Eigen Mode 9 [Hz]
45.02 Hz
0.72%
Frequency of Eigen Mode 10 [Hz]
56.06 Hz
2.32%
Frequency of Eigen Mode 11 [Hz]
56.06 Hz
2.32%
Frequency of Eigen Mode 12 [Hz]
56.34 Hz
1.83%
Frequency of Eigen Mode 13 [Hz]
56.34 Hz
1.83%
Frequency of Eigen Mode 14 [Hz]
56.34 Hz
1.83%
Frequency of Eigen Mode 15 [Hz]
56.34 Hz
1.83%
Frequency of Eigen Mode 16 [Hz]
56.34 Hz
1.83%
ADVANCE DESIGN VALIDATION GUIDE
1.77 Beam on 3 supports with T/C (k > infinite) (010101SSNLB_FEM) Test ID: 2515 Test status: Passed
1.77.1
Description
Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k > infinite).
1.77.2
Background
1.77.2.1 Model description ■ ■ ■
Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.
Units I. S. Geometry ■ ■
L = 10 m 4 Section: IPE 200, Iz = 0.00001943 m
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: ► ►
Support at node 1 restrained along x and y (x = 0), Support at node 2 restrained along y (x = 10 m),
T/C stiffness ky → ∞ (1.1030N/m), Inner: None.
►
■
Loading ■ ■
External: Vertical punctual load P = 100 N at x = 5 m, Internal: None.
257
ADVANCE DESIGN VALIDATION GUIDE
1.77.2.2 References solutions ky being infinite, the non linear model behaves the same way as a beam on 3 supports. Displacements
( ) = −0.000115 rad 32EI (3EI + 2k L ) PL (3EI + k L ) =− = 0.000077 rad 16EI (3EI + 2k L ) 3PL2 2EI z + k y L3
β1 =
z
z
2
β2
z
z
v3 =
y
z
3
3
y
3
− 3PL3 =0 16 3EI z + 2k y L3
(
2
β3 =
y
(
)
) = −0.000038 rad + 2k L )
PL − 6EI z + k y L3
(
32EI z 3EI z
y
3
Mz Moments
M z1 = 0 Mz2 =
(
3k y PL4
16 3EI z + 2k y L3
M z ( x = 5m) =
) = −93.75 N.m
PL (M z 2 − M z1 ) + = −203.13 N.m 4 2
Finite elements modeling ■ ■ ■
Linear element: S beam, automatic mesh, 3 nodes, 2 linear elements + 1 T/C.
Deformed shape
258
ADVANCE DESIGN VALIDATION GUIDE
Moment diagram
1.77.2.3 Theoretical results Solver
Result name
Result description
Reference value
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
1.77.3
Calculated results
Result name
Result description
Value
Error
RY
Rotation Ry in node 1 [rad]
0.000115005 Rad
0.00%
RY
Rotation Ry in node 2 [rad]
7.65875e005 Rad
0.54%
DZ
Displacement  node 3 [m]
9.36486e030 m
0.00%
RY
Rotation Ry in node 3 [rad]
3.81695e005 Rad
0.45%
My
Moment M  node 1 [Nm]
1.3145e013 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%
259
ADVANCE DESIGN VALIDATION GUIDE
1.78 Beam on 3 supports with T/C (k = 10000 N/m) (010102SSNLB_FEM) Test ID: 2516 Test status: Passed
1.78.1
Description
Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k = 10000 N/m).
1.78.2
Background
1.78.2.1 Model description ■ ■ ■
Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.
Units I. S. Geometry ■ ■
L = 10 m Section: IPE 200, Iz = 0.00001943 m4
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Support at node 1 restrained along x and y (x = 0), Support at node 2 restrained along y (x = 10 m), ► T/C ky Rigidity = 10000 N/m (the – sign corresponds to an upwards restraint), Inner: None. ► ►
■
Loading ■ ■
260
External: Vertical punctual load P = 100 N at x = 5 m, Internal: None.
ADVANCE DESIGN VALIDATION GUIDE
1.78.2.2 References solutions Displacements
( ) = −0.000129 rad 32EI (3EI + 2k L ) PL (3EI + k L ) =− = 0.000106 rad 16EI (3EI + 2k L ) 3PL2 2EI z + k y L3
β1 =
z
z
2
β2
z
z
v3 =
y
z
3
3
y
3
− 3PL3 = 0.00058 m 16 3EI z + 2k y L3
(
2
β3 =
y
(
)
) = 0.000034 rad + 2k L )
PL − 6EI z + k y L3
(
32EI z 3EI z
y
3
Mz Moments
M z1 = 0 Mz2 =
(
3k y PL4
16 3EI z + 2k y L3
M z ( x = 5m) =
) = −58.15 N.m
PL (M z 2 − M z1 ) + = −220.9 N.m 4 2
Finite elements modeling ■ ■ ■
Linear element: S beam, automatic mesh, 3 nodes, 2 linear elements + 1 T/C.
Deformed shape
261
ADVANCE DESIGN VALIDATION GUIDE
Moment diagram
1.78.2.3 Theoretical results Solver
Result name
Result description
Reference value
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
RY
Rotation Ry in node 3 [rad]
0.000034
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
1.78.3
262
Calculated results
Result name
Result description
Value
Error
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]
3.44295e005 Rad
1.26%
My
M moment  node 1 [Nm]
1.77636e015 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%
ADVANCE DESIGN VALIDATION GUIDE
1.79 Linear system of truss beams (010103SSLLB_FEM) Test ID: 2517 Test status: Passed
1.79.1
Description
Verifies the displacement and the normal force for a bar system containing 4 elements of the same length and 2 diagonals.
1.79.2
Background
1.79.2.1 Model description ■ ■ ■
Reference: internal GRAITEC test; Analysis type: static linear; Element type: linear, bar.
Units I. S. Geometry ■ ■
L=5m 2 Section S = 0.005 m
Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 N/m2. Boundary conditions ■
Outer: Support at node 1 restrained along x and y, Support at node 2 restrained along x and y, Inner: None.
► ►
■
263
ADVANCE DESIGN VALIDATION GUIDE
Loading ■ ■
External: Horizontal punctual load P = 50000 N at node 3, Internal: None.
1.79.2.2 References solutions Displacements
30PL = 0.000649 m 11ES − 6PL v3 = = −0.000129 m 11ES 25PL u4 = = 0.000541 m 11ES 5PL v4 = = 0.000108 m 11ES u3 =
N normal forces
N14 =
N12 = 0 N 23 = − N 43 =
6 P = −27272 N 11
5 P = 22727 N 11
N13 =
5 P = 22727 N 11
6 2 P = 38569 N 11
N 42 = −
5 2 P = −32141 N 11
Finite elements modeling ■ ■ ■
Linear element: bar, without meshing, 4 nodes, 6 linear elements.
Deformed shape
264
ADVANCE DESIGN VALIDATION GUIDE
Normal forces
1.79.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
DX
u3 displacement on Node 3 [m]
0.000649
CM2
DZ
v3 displacement on Node 3 [m]
0.000129
CM2
DX
u4 displacement on Node 4 [m]
0.000541
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
N43 normal force on Element 3 [N]
22727
CM2
Fx
N14 normal force on Element 4 [N]
22727
CM2
Fx
N13 normal effort on Element 5 [N]
38569
CM2
Fx
N42 normal force on Element 6 [N]
32141
265
ADVANCE DESIGN VALIDATION GUIDE
1.79.3
Calculated results
Result name
266
Result description
Value
Error
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]
0N
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%
ADVANCE DESIGN VALIDATION GUIDE
1.80 Double fixed beam in Eulerian buckling with a thermal load (010091HFLLB_FEM) Test ID: 2506 Test status: Passed
1.80.1
Description
Verifies the normal force on the nodes of a double fixed beam in Eulerian buckling with a thermal load.
1.80.2
Background
1.80.2.1 Model description ■ ■ ■
Reference: Euler theory; Analysis type: Eulerian buckling; Element type: linear.
Units I. S. Geometry L = 10 m Cross Section IPE200
Sx m² Vx m3 0.002850 0.00000000
Sy m² V1y m3 0.001400 0.00002850
Sz m² V1z m3 0.001799 0.00019400
Ix m4 V2y m3 0.0000000646 0.00002850
Iy m4 V2z m3 0.0000014200 0.00019400
Iz m4 0.0000194300
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,
■
Poisson's ratio: ν = 0.3.
■
Coefficient of thermal expansion: α = 0.00001
Boundary conditions ■ ■
Outer: Fixed at end x = 0, Inner: None.
267
ADVANCE DESIGN VALIDATION GUIDE
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:
N = ESα∆T = 2.1E11 × 0.00285 × 0.00001 × 5 = 29.925 kN
1.80.2.2 Displacement of the model in the linear elastic range Reference solution The reference critical load established by Euler is:
Pcritical =
π 2 EI L 2
2
= 117.724 kN ⇒ λ =
29.925 = 3.93 117.724
Observation: in this case, the thermal load has no effect over the critical coefficient Finite elements modeling ■ ■ ■
Linear element: beam, imposed mesh, 11 nodes, 10 elements.
Deformed shape of mode 1
268
ADVANCE DESIGN VALIDATION GUIDE
1.80.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
Fx
Normal Force on Node 6  Case 101 [kN]
29.925
CM2
Fx
Normal Force on Node 6  Case 102 [kN]
117.724
1.80.3
Calculated results
Result name
Result description
Value
Error
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%
269
ADVANCE DESIGN VALIDATION GUIDE
1.81 Fixed/free slender beam with eccentric mass or inertia (010096SDLLB_FEM) Test ID: 2510 Test status: Passed
1.81.1
Description
Fixed/free slender beam with eccentric mass or inertia. Tested functions: Eigen mode frequencies, straight slender beam, combined bendingtorsion, plane bending, transverse bending, punctual mass.
1.81.2
Background
1.81.2.1 Model description ■ ■ ■ ■
Reference: Structure Calculation Software Validation Guide, test SDLL 15/89; Analysis type: modal analysis; Element type: linear. Tested functions: Eigen mode frequencies, straight slender beam, combined bendingtorsion, plane bending, transverse bending, punctual mass..
1.81.2.2 Problem data
Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■
270
Outer diameter: Inner diameter: Beam length: Distance BC: Area: Inertia: Polar inertia:
de= 0.35 m, di = 0.32 m, l = 10 m, lBC = 1 m A =1.57865 x 102 m2 Iy = Iz = 2.21899 x 104m4 4 4 Ip = 4.43798 x 10 m
ADVANCE DESIGN VALIDATION GUIDE
■
Punctual mass:
mc = 1000 kg
Materials properties ■
Longitudinal elasticity modulus of AB element: E = 2.1 x 1011 Pa,
■
Density of the linear element AB: ρ = 7800 kg/m3
■ ■
Poisson's ratio ν=0.3(this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the more appropriate to obtain the correct frequency value of modes No. 4 and 5 with NE/NASTRAN: Elastic modulus of BC element: E = 1021 Pa
■
Density of the linear element BC: ρ = 0 kg/m3
Boundary conditions Fixed at point A, x = 0, Loading None for the modal analysis
1.81.2.3 Reference frequencies Reference solutions The different eigen frequencies are determined using a finite elements model of Euler beam (slender beam). fz + t0 = flexion x,z + torsion fy + tr = flexion x,y + traction Mode 1 (fz + t0) 2 (fy + tr) 3 (fy + tr) 4 (fz + t0) 5 (fz + t0) 6 (fy + tr) 7 (fz + t0) 1 (fz + t0)
Units Hz Hz Hz Hz Hz Hz Hz Hz
Reference 1.636 1.642 13.460 13.590 28.900 31.960 61.610 63.930
Uncertainty about the reference solutions The uncertainty about the reference solutions: ± 1% Finite elements modeling ■ ■ ■ ■
Linear element AB: Beam Imposed mesh: 50 elements. Linear element BC: Beam Without meshing
Modal deformations
271
ADVANCE DESIGN VALIDATION GUIDE
1.81.2.4 Theoretical results Solver
Result name
Result description
Reference value
CM2
Frequency
Eigen mode 1 frequency (fz + t0) [Hz]
1.636
CM2
Frequency
Eigen mode 2 frequency (fy + tr) [Hz]
1.642
CM2
Frequency
Eigen mode 3 frequency (fy + tr) [Hz]
13.46
CM2
Frequency
Eigen mode 4 frequency (fz + t0) [Hz]
13.59
CM2
Frequency
Eigen mode 5 frequency (fz + t0) [Hz]
28.90
CM2
Frequency
Eigen mode 6 frequency (fy + tr) [Hz]
31.96
CM2
Frequency
Eigen mode 7 frequency (fz + t0) [Hz]
61.61
CM2
Frequency
Eigen mode 8 frequency (fy + tr) [Hz]
63.93
Note: fz + t0 = flexion x,z + torsion fy + tr = flexion x,y + traction Observation: because the mass matrix of Advance Design is condensed and not consistent, the torsion modes obtained are not taking into account the self rotation mass inertia of the beam.
272
ADVANCE DESIGN VALIDATION GUIDE
1.81.3
Calculated results
Result name
Result description
Value
Error
Eigen mode 1 frequency [Hz]
1.64 Hz
0.24%
Eigen mode 2 frequency [Hz]
1.64 Hz
0.12%
Eigen mode 3 frequency [Hz]
13.45 Hz
0.07%
Eigen mode 4 frequency [Hz]
13.65 Hz
0.44%
Eigen mode 5 frequency [Hz]
29.72 Hz
2.84%
Eigen mode 6 frequency [Hz]
31.96 Hz
0.00%
Eigen mode 7 frequency [Hz]
63.09 Hz
2.40%
Eigen mode 8 frequency [Hz]
63.93 Hz
0.00%
273
ADVANCE DESIGN VALIDATION GUIDE
1.82 Beam on 3 supports with T/C (k = 0) (010100SSNLB_FEM) Test ID: 2514 Test status: Passed
1.82.1
Description
Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k = 0).
1.82.2
Background
1.82.2.1 Model description ■ ■ ■
Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.
Units I. S. Geometry ■ ■
L = 10 m 4 Section: IPE 200, Iz = 0.00001943 m
Materials properties ■
Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,
■
Poisson's ratio: ν = 0.3.
Boundary conditions ■
Outer: Support at node 1 restrained along x and y (x = 0), Support at node 2 restrained along y (x = 10 m), ► T/C stiffness ky = 0, Inner: None. ► ►
■
Loading ■ ■
274
External: Vertical punctual load P = 100 N at x = 5 m, Internal: None.
ADVANCE DESIGN VALIDATION GUIDE
1.82.2.2 References solutions ky being null, the non linear model behaves the same way as the structure without support 3. Displacements
( ) = −0.000153 rad 32EI (3EI + 2k L ) PL (3EI + k L ) =− = 0.000153 rad 16EI (3EI + 2k L ) 3PL2 2EI z + k y L3
β1 =
z
z
2
β2
z
z
v3 =
y
z
3
3
y
3
− 3PL3 = 0.00153 m 16 3EI z + 2k y L3
(
2
β3 =
y
(
)
) = 0.000153 rad + 2k L )
PL − 6EI z + k y L3
(
32EI z 3EI z
y
3
Mz Moments
M z1 = 0 Mz2 =
(
3k y PL4
16 3EI z + 2k y L3
M z ( x = 5m) =
)=0
PL (M z 2 − M z1 ) + = −250 N.m 4 2
Finite elements modeling ■ ■ ■
Linear element: S beam, automatic mesh, 3 nodes, 2 linear elements + 1 T/C.
Deformed shape
275
ADVANCE DESIGN VALIDATION GUIDE
Moment diagrams
1.82.2.3 Theoretical results Solver
Result name
Result description
Reference value
CM2
RY
Rotation Ry in node 1 [rad]
0.000153
CM2
RY
Rotation Ry in node 2 [rad]
0.000153
CM2
DZ
Displacement V in node 3 [m]
0.00153
CM2
RY
Rotation Ry in node 3 [rad]
0.000153
CM2
My
Moment M in node 1 [Nm]
0
CM2
My
Moment M  middle span 1 [Nm]
250
1.82.3
276
Calculated results
Result name
Result description
Value
Error
RY
Rotation Ry in node 1 [rad]
0.000153175 Rad
0.11%
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]
5.05771e014 N*m
0.00%
My
Moment M  middle span 1 [Nm]
250 N*m
0.00%
ADVANCE DESIGN VALIDATION GUIDE
1.83 BAEL 91 (concrete design)  France: Linear element in combined bending/tension  without compressed reinforcements  Partially tensioned section (020158SSLLB_B91) Test ID: 2520 Test status: Passed
1.83.1
Description
Verifies the reinforcement results for a concrete beam with 8 isostatic spans subjects to uniform loads and compression normal forces.
1.83.1.1 Model description ■ ■ ■
Reference: J. Perchat (CHEC) reinforced concrete course Analysis type: static linear; Element type: planar.
Units ■ ■ ■ ■
Forces: kN Moment: kN.m Stresses: MPa Reinforcement density: cm²
Geometry ■ ■
Beam dimensions: 0.2 x 0.5 ht Length: l = 48 m in 8 spans of 6m,
Materials properties ■
Longitudinal elastic modulus: E = 20000 MPa,
■
Poisson's ratio: ν = 0.
Boundary conditions ■
Outer: Hinged at end x = 0, Vertical support at the same level with all other supports Inner: Hinged at each beam end (isostatic)
► ►
■
Loading ■
External: Case 1 (DL):uniform linear load g= 5kN/m (on all spans except 8) Fx = 10 kN at x = 42m: Ng = 10 kN for spans from 6 to 7
►
Fx = 140 kN at x = 32m: Ng = 150 kN for span 5 Fx = 50 kN at x = 24m: Ng = 100 kN for span 4 Fx = 50 kN at x = 18m: Ng = 50 kN for span 3 Fx = 50 kN at x = 12m: Ng = 100 kN for span 2 Fx = 70 kN at x = 6m: Ng = 30 kN for span 1 ►
Case 10 (DL):uniform linear load g = 5 kN/m (span 8) Fx = 10 kN at x = 48m: Ng = 10 kN Fx = 10 kN at x = 42m
►
Case 2 to 8 (LL):uniform linear load q = 9 kN/m (on spans 1, 3 to 7)
277
ADVANCE DESIGN VALIDATION GUIDE
uniform linear load q = 15 kN/m (on span 2) Fx = 30 kN at x = 6m (case 2 span 1) Fx = 50 kN at x = 6m (case 3 span 2) Fx = 50 kN at x = 12m (case 3 span 2) Fx = 40 kN at x = 12m (case 4 span 3) Fx = 40 kN at x = 18m (case 4 span 3) Fx = 100 kN at x = 18m (case 5 span 4) Fx = 100 kN at x = 24m (case 5 span 4) Fx = 150 kN at x = 24m (case 6 span 5) Fx = 150 kN at x = 30m (case 6 span 5) Fx = 8 kN at x = 30m (case 7 span 6) Fx = 8 kN at x = 36m (case 7 span 6) Fx = 8 kN at x = 36m (case 8 span 7 Fx = 8 kN at x = 42m (case 8 span 7) Case 9 (ACC):uniform linear load a = 25 kN/m (on 8th span) Fx = 8 kN at x = 36m (case 9 span 8)
►
Fx = 8 kN at x = 42m (case 9 span 8) Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107) Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102) Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103) Comb BAELS: 1xDL + 1*LL (comb 108 to 114) Comb BAELA: 1xDL + 1xACC with duration of less than 1h (comb 115) ■
Internal: None.
Reinforced concrete calculation hypothesis: All concrete covers are set to 5 cm BAEL 91 calculation (according to 99 revised version)
278
Span 1
Concrete B20
Reinforcement HA fe500
Application D>24h
Concrete No
2
B35
Adx fe235
1h24h D>24h D µ luc = 0.271
therefore the compressed reinforcement is present in the beam section
Reference reinforcement calculation at SLU: The calculation will be divided for theoretical sections: Calculation of the tension steel section (Section A1): The calculation of tensioned steel section must be conducted with the corresponding moment of µ lu :
M Ed 1 = µ lu * bw * d ² * f cd = 0.271 * 0.40 * 0.707² * 16.67 = 0.902 * 10 6 Nm ■
The α value:
[
]
[
]
α lu = 1.25 * 1 − (1 − 2 * µ lu ) = 1.25 * 1 − (1 − 2 * 0.271) = 0.404 ■
Calculation of the lever arm zc:
zlu = d * (1 − 0.4 * α lu ) = 0.707 * (1 − 0.4 * 0.404) = 0.593m ■
Tensioned reinforcement elongation calculation:
ε su =
478
1 − αu
αu
* ε cu 2 =
1 − 0.404 * 3.5 = 5.17 0.404 ‰
ADVANCE DESIGN VALIDATION GUIDE
■
Tensioned reinforcement efforts calculation(S500A):
σ su = 432,71 + 952,38 * ε su ≤ 454MPa
σ su = 432,71 + 952,38 * 0.00517 = 437.63MPa ≤ 454Mpa ■
Calculation of the reinforcement area:
A1 =
M Ed 1 0.902 *10 −6 Nm = = 34.76cm² z lu . f yd 0.593m * 437.63MPa
Compressed steel reinforcement reduction (Section As2): Reduction coefficient: ε sc =
3,5 3.5 (0.404 * 0.707 − 0.045) = 2.957 ‰ (α lu * d − d' ) = 1000 * α lu * d 1000 * 0.404 * 0.707
ε sc = 0.00295 > ε yd = 0.00217 ⇒ σ sc = 432.71 + 952.38 * 0.00295 = 435.52MPa Compressed reinforcement calculation:
As 2 =
M Ed − M Ed 1 1.002 − 0.902 = 3.47cm² = (0.707 − 0.045) * 435.52 (d − d ' )σ sc
The steel reinforcement condition:
A2 = As 2 .
σ sc f yd
= 3.47 *
435.52 = 3.48cm² 437.64
Total area to be implemented: ■ ■
In the lower part: As1=A1+A2=38.24 cm2 (tensioned reinforcement) In the top part: As2=3.47 cm2 (compressed reinforcement)
Reference reinforcement calculation at SLS: The concrete beam design at SLS will be made considering a limitation of the characteristic compressive cylinder strength of concrete at 28 days, at 0.6*fck The assumptions are: ■
The SLS moment: MEcq = 705kNm
■
The equivalence coefficient: α e = 20.94
■
The stress on the concrete will be limited at 0.6*fck=15Mpa and the stress on steel at 0.8*fyk, or 400MPa
479
ADVANCE DESIGN VALIDATION GUIDE
Calculation of the resistance moment MRd for detecting the presence of the compressed reinforcement:
x1 =
αe *σ c 20.94 * 15 *d = * 0.707 = 0.311m 20.94 * 15 + 400 αe *σ c + σ s
1 1 Fc = * bw * x1 * σ c = * 0.40 * 0.311×15 = 0.933 *10 6 N 2 2 zc = d −
x1 0.311 = 0.707 − = 0.603m 3 3
M rb = Fc * zc = 0.933 * 10 6 * 0.603 = 0.563 * 10 6 Nm Therefore the compressed reinforced established earlier was correct. Theoretical section 1 (tensioned reinforcement only)
M 1 = M rb = 0.563 * 10 6 Nm
α1 =
x1 0.311 = = 0.440 d 0.707
zc = d − A1 =
x1 0.311 = 0.707 − = 0.603m 3 3
M1 0.563 = = 23.34cm² z c × σ s 0.603 × 400
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
M 2 = M ser ,cq − M rb = (0.705 − 0.591) * 10 6 = 0.142 * 10 6 Nm Compressed reinforcement stresses:
σ sc = α e * σ c *
α1 * d − d ' α1 * d
σ sc = 20.94 * 15 *
0.440 * 0.707 − 0.045 = 268.66 MPa 0.440 * 0.707
Compressed reinforcement area:
A' =
M2 0.142 = = 7.98cm² (d − d ' ) * σ sc (0.707 − 0.045) * 268.66
Complementary tensioned reinforcement area:
As = A' *
σ sc 268.66 = 7.98 * = 5.36cm² 400 σs
Section area: Tensioned reinforcement: 23.34+5.36=28.7 cm2 Compressed reinforcement: 7.98 cm2 Considering an envelope calculation of ULS and SLS, it will be obtained: Tensioned reinforcement ULS: A=38.24cm2 Compressed reinforcement SLS: A=7.98cm2 480
ADVANCE DESIGN VALIDATION GUIDE
To optimize the reinforcement area, it is preferable a third iteration by recalculating with SLS as a baseline amount of 2 tensioned reinforcement (after ULS: Au=38.24cm ) Reference reinforcement third calculation at SLS: For this third iteration, the calculation will begin considering the section of the tensile reinforcement found when calculating for ULS: Au=38.24cm2 From this value, it will be calculated the stress obtained in the tensioned reinforcement:
σs =
AELS 28.70 *σ s = * 400 = 300.21MPa AELU 38.24
Calculating the moment resistance Mrb for detecting the presence of compressed steel reinforcement:
x1 =
α e *σ c 20.94 *15 * 0.707 = 0.361m *d = 20.94 *15 + 300.21 α e *σ c + σ s
1 1 Fc = * bw * x1 * σ c = * 0.40 * 0.361×15 = 1.083 *10 6 N 2 2 zc = d −
x1 0.361 = 0.707 − = 0.587 m 3 3
M rb = Fc * z c = 1.083 *10 6 * 0.587 = 0.636 *10 6 Nm Therefore the compressed reinforced established earlier was correct. Theoretical section 1 (tensioned reinforcement only)
M 1 = M rb = 0.636 *10 6 Nm
α1 =
x1 0.361 = = 0.511 d 0.707
zc = d − A1 =
x1 0.361 = 0.587 m = 0.707 − 3 3
M1 0.36 = = 36.09cm² z c * σ s 0.578 * 400
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
M 2 = M ser ,cq − M rb = (0.705 − 0.636) *10 6 = 0.069 *10 6 Nm Compressed reinforcement stresses:
σ sc = α e * σ c *
α1 * d − d ' 0.511 * 0.707 − 0.045 = 20.94 * 15 * = 275MPa 0.511 * 0.707 α1 * d
Compressed reinforcement area:
A' =
M2 0.069 = 3.79cm² = (d − d ' ) * σ sc (0.707 − 0.045)* 275
Complementary tensioned reinforcement area:
As = A' *
σ sc 275 = 3.79 * = 3.47cm² 300.21 σs 481
ADVANCE DESIGN VALIDATION GUIDE
Section area: Tensioned reinforcement: 36.09+3.47=39.56 cm2 Compressed reinforcement: 3.79 cm2 Reference solution for minimal reinforcement area The minimum percentage for a rectangular beam in pure bending is:
f ct , eff * bw * d 0.26 * = Max f yk 0.0013 * b * d w
As , min
According to: EC2 Part 1,1 EN 1992112002 Chapter 9.2.1.1(1); Note 2 Where:
f ct ,eff = f ctm = 2.56 MPa
from cracking conditions
Therefore:
As ,min
2.56 * 0.40 * 0.707 = 3.77 * 10 −4 m² 0.26 * = max = 3.77cm² 500 −4 0.0013 * 0.40 * 0.707 = 3.68 * 10 m²
Finite elements modeling ■ ■ ■
Linear element: S beam, 9 nodes, 1 linear element.
ULS and SLS load combinations(kNm) Simply supported beam subjected to bending
ULS (reference value: 1002kNm)
SLS (reference value: 705kNm)
482
ADVANCE DESIGN VALIDATION GUIDE
SLS –Quasipermanent (reference value: 637kNm)
Theoretical reinforcement area(cm2) For Class A reinforcement steel ductility (reference value: A=39.56cm2 and A’=3.79cm2)
2
Minimum reinforcement area(cm ) (reference value: 3.77cm2)
5.48.2.4 Reference results Result name
Result description
Reference value
My,ULS
My corresponding to the 101 combination (ULS) [kNm]
1001 kNm
My,SLS,cq
My corresponding to the 102 combination (SLS) [kNm]
705 kNm
My,SLS,qp
My corresponding to the 102 combination (SLS) [kNm]
637 cm2
Az (Class A)
Theoretical reinforcement area [cm2]
39.56 cm2
2
5.48.3
3.77 cm2
Minimum reinforcement area [cm ]
Amin
Calculated results
Result name
Result description
Value
Error
My My
My USL
1001.92 kN*m
0.0080 %
My SLS cq
704.714 kN*m
0.0406 %
My
My SLS qp
637.304 kN*m
0.0477 %
Az
Az
39.5884 cm²
0.0718 %
Amin
Amin
3.77193 cm²
0.0512 %
483
ADVANCE DESIGN VALIDATION GUIDE
5.49 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement  Bilinear stressstrain diagram (Class XD1) (evaluated by SOCOTEC France  ref. Test 13) Test ID: 4986 Test status: Passed
5.49.1
Description
Simple Bending Design for Service State Limit  Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. The verification of the bending stresses at service limit state is performed.
5.49.2
Background
Simple Bending Design for Service State Limit Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses will be made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. This test was evaluated by the French control office SOCOTEC.
5.49.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■
Loadings from the structure: G = 25 kN/m (including dead load), Exploitation loadings (category A): Q = 30kN/m,
■ ■ ■
Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Quasipermanent combination of actions: CQP = 1.0 x G + 0.3 x Q
Simply supported beam
Units Metric System Geometry Below are described the beam cross section characteristics: ■
484
Height: h = 0.60 m,
ADVANCE DESIGN VALIDATION GUIDE
■ ■ ■ ■ ■
Width: b = 0.25 m, Length: L = 5.80 m, 2 Section area: A = 0.150 m , Concrete cover: c=4.5cm Effective height: d=h(0.6*h+ebz)=0.519m; d’=ebz=0.045m
Materials properties Rectangular solid concrete C25/30 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■
Exposure class XD1 3 Concrete density: 25kN/m Stressstrain law for reinforcement: Bilinear stressstrain diagram The concrete age t0=28 days Humidity RH=50%
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x = 0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.80) restrained in translation along Y, Z and restrained in rotation along X. Inner: None. ►
■
Loading The beam is subjected to the following load combinations: ■
Load combinations: Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q = 25 + 30 = 55 kN/ml Quasipermanent combination of actions: CQP = 1.0 x G + 0.3 x Q = 25 + 0.3*30 = 34 kN/ml
■
Load calculations:
M Ecq =
(25 + 30) * 5.80² = 231.28kNm 8
M Eqp =
(25 + 0.3 * 30) * 5.80² = 142.97kNm 8
485
ADVANCE DESIGN VALIDATION GUIDE
5.49.2.2 Reference results in calculating the concrete final value of creep coefficient
ϕ (∞, t0 ) = ϕ RH β ( f cm ) β (t0 ) Where:
β ( f cm ) = β (t0 ) =
16.8 16.8 = = 2.92MPa 25 + 8 f cm
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
t0 : concrete age t0 = 28 days
ϕ RH
RH 1− 100 * α * α = 1 + 1 2 0.1*3 h0
α1 = α 2 = 1 if f ≤ 35Mpa cm 35 α1 = f cm If not
0.7
35 α 2 = f cm and
0 .2
In this case,
therefore:
α1 = α 2 = 1 In this case: Humidity RH = 50 %
h0 =
ϕ RH
2 Ac 2 * 250 * 600 = = 176mm u 2 * (250 + 600)
50 100 = 1.89 ⇒ ϕ (∞, t ) = ϕ * β ( f ) * β (t ) = 1.89 * 2.92 * 0.488 = 2.70 =1+ RH cm 0 0 0.1 * 3 176 1−
Calculating the equivalence coefficient: The coefficient of equivalence is determined by the following formula:
Es Ecm
αe =
1 + ϕ ( ∞, t 0 ) *
M Eqp M Ecar
Where:
Ecm
f +8 = 22 * ck 10
E s = 200000 MPa
486
0.3
25 + 8 = 22 * 10
0.3
= 31476 MPa
ADVANCE DESIGN VALIDATION GUIDE
1 + ϕ ( ∞, t 0 ) *
Es Ecm
αe =
1 + ϕ ( ∞, t 0 ) *
M Eqp M Ecar
143 = 2.67 = 1 + 2.69 * 231
= M Eqp M Ecar
200000 = 16.97 31476 143 1 + 2.69 * 231
Material characteristics: The maximum compression on the concrete is:
σ bc = 0,6 * f ck = 0,6 *15 = 15Mpa
For the maximum stress on the steel taut, we consider the constraint limit
σ s = 0,8 * f yk = 400Mpa
Neutral axis position calculation: The position of the neutral axis must be determined by calculating maximum stress on the concrete and reinforcement):
α1 =
α1
(position corresponding to the state of
α e *σ c 16.97 *15 = = 0.389 α e * σ c + σ s 16.97 *18 + 400
Moment resistance calculation: Knowing the α1 value, it can be determined the moment resistance of the concrete section, using the following formulas:
x1 = α1 * d = 0.389 * 0.519 = 0.202 m
M rb =
1 0.202 x * bw * x1 * σ c * (d − 1 ) = 0.5 * 0.25 * 0.202 * 15 * 0.519 − = 0.171MNm 2 3 3
Where: Utile height : d = h – (0.06h + ebz) = 0.519m The moment resistance Mrb = 171KNm Because
M Ecq = 231kNm > M rb = 171kNm
, the supposition of having no compressed reinforcement is
incorrect. The calculation of the tension reinforcement theoretical section A1
zc = d −
A1 =
x1 0.202 = 0.519 − = 0.452m 3 3
M rb 0.171 = = 9.46cm² zc * σ s 0.452 * 400
Stress calculation for steel reinforcement σsc:
δ '=
d ' 0.045 = = 0.087 d 0.519
σ sc = α e * σ c *
α1 − δ ' 0.389 − 0.087 = 16.97 * 15 * = 197.75MPa 0.389 α1 487
ADVANCE DESIGN VALIDATION GUIDE
Calculation of the steel compressed reinforcement A’:
A' =
M ser − M rb 0.231 − 0.171 = = 6.44cm² (d − d ' ) * σ sc (0.519 − 0.045) *197.75
Calculation of the steel tensioned reinforcement A2:
A2 = A'*
σ sc 197.75 = 6.44 * = 3.18cm² 400 σs
Calculation of the steel reinforcement :
A = A1 + A2 = 9.46 + 3.18 = 12.64cm²
A' = 6.44cm² Reference solution for minimal reinforcement area The minimum percentage for a rectangular beam in pure bending is:
According to: EC2 Part 1,1 EN 1992112002 Chapter 9.2.1.1(1); Note 2 Where: fct,eff = fctm = 2.896MPa from cracking conditions
Therefore:
2.56 * 0.25 * 0.519 = 1.73 * 10− 4 m² 0.26 * = 1.73cm² As , min = max 500 −4 0.0013 * 0.25 * 0.519 = 1.69 * 10 m² Finite elements modeling ■ ■ ■
Linear element: S beam, 6 nodes, 1 linear element.
ULS and SLS load combinations(kNm) Simply supported beam subjected to bending
SLS (reference value: 231.28kNm)
488
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area(cm2) (reference value: As=12.64cm2; A’=6.44cm2)
Minimum reinforcement area(cm2) (reference value: 1.73cm2)
5.49.2.3 Reference results Result name
Result description
My,SLS
My corresponding to the 102 combination (SLS) [kNm]
231.28 kNm
2
12.64 cm2
Theoretical reinforcement area [cm ]
Az
2
1.73 cm2
Minimum reinforcement area [cm ]
Amin
5.49.3
Reference value
Calculated results
Result name
Result description
Value
Error
My
My SLS
231.275 kN*m
0.0022 %
Az
Az
12.6425 cm²
0.0990 %
Amin
Amin
1.73058 cm²
0.0335 %
489
ADVANCE DESIGN VALIDATION GUIDE
5.50 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement  Inclined stressstrain diagram (Class XD1) (evaluated by SOCOTEC France  ref. Test 17) Test ID: 5000 Test status: Passed
5.50.1
Description
Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C20/25 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section.
5.50.2
Background
Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C20/25 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of compressing stresses in concrete section σc and compressing stresses in the steel reinforcement section σs. This test was evaluated by the French control office SOCOTEC.
5.50.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ Loadings from the structure: G = 9.375 kN/m (including the dead load), ■ Mfq = Mcar = Mqp = 75 kNm ■ Structural class: S4 ■ Characteristic combination of actions: CCQ = 1.0 x G ■ Reinforcement steel ductility: Class B The objective is to verify: ■ ■ ■
490
The stresses results The compressing stresses in concrete section σc The compressing stresses in the steel reinforcement section σs.
ADVANCE DESIGN VALIDATION GUIDE
Simply supported beam
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■
Height: h = 0.50 m, Width: b = 0.20 m, Length: L = 8.00 m, 2 Section area: A = 0.10 m , Concrete cover: c=4.5cm Effective height: d=44cm;
Materials properties Rectangular solid concrete C20/25 is used. The following characteristics are used in relation to this material: ■ ■ ■ ■
Exposure class XD1 Stressstrain law for reinforcement: Inclined stressstrain diagram The concrete age t0=28 days Humidity RH=50%
■
Characteristic compressive cylinder strength of concrete at 28 days:
■
Characteristic yield strength of reinforcement:
■
Ast = 9,42cm ²
f ck = 20Mpa
f yk = 400Mpa
for 3 HA20
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, Support at end point (x = 8) restrained in translation along Y and Z Inner: None.
► ►
■
Loading The beam is subjected to the following load combinations: ■
Characteristic combination of actions: CCQ = 1.0 x G =9.375kN/ml
■
Load calculations: Mfq = Mcar = Mqp = 75 kNm
491
ADVANCE DESIGN VALIDATION GUIDE
5.50.2.2 Reference results in calculating the concrete final value of creep coefficient
ϕ (∞, t0 ) = ϕ RH β ( f cm ) β (t0 ) Where:
β ( f cm ) = β (t0 ) =
16.8 16.8 = = 3.17 MPa f cm 20 + 8
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
t0 : concrete age t0=28days
ϕ RH
RH 1− 100 * α * α = 1 + 1 2 0.1*3 h0
α1 = α 2 = 1 if f ≤ 35Mpa cm 35 If not, α1 = f cm
0.7
35 and α 2 = f cm
0.2
In this case, Therefore
α1 = α 2 = 1
In this case: Humidity RH=50 %
h0 =
ϕ RH
2 Ac 2 * 200 * 500 = = 142.86mm u 2 * (200 + 500)
50 100 = 1.96 ⇒ ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.96 * 3.17 * 0.488 = 3.03 =1+ 3 0.1 * 142.86 1−
The coefficient of equivalence is determined by the following formula:
αe =
Es Ecm 1 + ϕ ( ∞, t 0 ) *
492
M Eqp M Ecar
ADVANCE DESIGN VALIDATION GUIDE
Where:
Ecm
f +8 = 22 * ck 10
0.3
20 + 8 = 22 * 10
0.3
= 29962 MPa
E s = 200000 MPa 1 + ϕ ( ∞, t 0 ) * Es Ecm
αe =
1 + ϕ (∞, t 0 ) *
M Eqp M Ecar
= 1 + 3.03 * 1 = 4.03 =
M Eqp
200000 = 26.90 29962 1 + 3.03 *1
M Ecar
Material characteristics: The maximum compression on the concrete is: σ bc = 0,6fck = 0,6 * 25 = 15Mpa For the maximum stress on the steel taut, we consider the constraint limit σ s = 0,8 * f yk = 320Mpa Checking inertia cracked or not: Before computing the constraints, check whether the section is cracked or not. For this, we determine the cracking moment which corresponds to a tensile stress on the concrete equal to f ctm :
M cr =
f ctm * I v
Where:
I=
b * h3 0,20 * 0,503 = = 0,00208m 4 12 12
v=
h = 0,25m 2
The average stress in concrete is: 2
2
f ctm = 0.30 * f ck3 = 0.30 * 20 3 = 2,21Mpa The critical moment of cracking is therefore:
M cr =
f ctm * I 2,21 * 0,00208 = = 0,018MNm 0,25 v
The servility limit state moment is 0.075MNm therefore the cracking inertia is present.
493
ADVANCE DESIGN VALIDATION GUIDE
Neutral axis position calculation: Neutral axis equation:
x1 =
1 * bw * x1 ² − Ast * α e * (d − x1 ) + Asc * α e * ( x1 − d ' ) = 0 2
− α e * ( Asc + Ast ) + α e2 * ( Asc + Ast )² + 2 * bw * α e * (d '*Asc + d * Ast ) bw
By simplifying the previous equation, by considering
Asc = 0
, it will be obtained:
x1 =
− α e * ( Ast ) + α e2 * Ast ² + 2 * bw * α e * d * Ast ) bw
x1 =
− 26.90 * 9,42 + 26.90² * 9,42² + 2 * 20 * 26,90 * (44 / 9,42) = 23,04cm 20
Calculating the second moment:
b * x3 0,20 * 0,2303 I = w 1 + Ast * α e * (d − x1 )² = + 9,42 * 10− 4 * 26.90 * (0,44 − 0,230)² = 0,001929m 4 3 3
Stresses calculation:
σc =
0,075 M ser * x1 = * 0,230 = 8,94 Mpa ≤ σ c = 12 Mpa 0,001929 I
σ st = α e * σ c *
d − x1 0,44 − 0,230 = 26.90 * 8,94 * = 219.57 Mpa ≤ σ s = 320 Mpa x1 0,230
Finite elements modeling ■ ■ ■
Linear element: S beam, 9 nodes, 1 linear element.
SLS load combinations (kNm) and stresses (MPa) Simply supported beam subjected to bending
SLS (reference value: Mscq=75kNm)
494
ADVANCE DESIGN VALIDATION GUIDE
Compressing stresses in concrete section σc (reference value: σc =8.94MPa )
Compressing stresses in the steel reinforcement section σs (reference value: σs=218.57MPa)
5.50.2.3 Reference results Result name
Result description
Reference value
My,SLS
My corresponding to the 103 combination (SLS) [kNm]
75 kNm
σc
Compressing stresses in concrete section σc (MPa)
8.94 MPa
σs
Compressing stresses in the steel reinforcement section σs (MPa)
219.67 MPa
5.50.3
Calculated results
Result name
Result description
Value
Error
My
My SLS
75 kN*m
0.0000 %
Sc CQ
Sc CQ
8.99322 MPa
0.5953 %
Ss CQ
Ss CQ
219.639 MPa
0.0314 %
495
ADVANCE DESIGN VALIDATION GUIDE
5.51 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 24) Test ID: 5058 Test status: Passed
5.51.1
Description
Verifies a rectangular cross section beam made from concrete C20/25 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the crosssectional area of the shear reinforcement (Asw) calculation.
5.51.2
Background
Verifies a rectangular cross section beam made from concrete C20/25 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the crosssectional area of the shear reinforcement (Asw) calculation. This test was evaluated by the French control office SOCOTEC.
5.51.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete C20/25 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: The dead load will be represented by a punctual load of 105kN
■
Exploitation loadings: The live load will be considered from one point load of 95kN
■ ■ ■
Structural class: S1 Reinforcement steel ductility: Class B The reinforcement will be displayed like in the picture below:
The objective is to verify: ■ ■
496
The shear stresses results The crosssectional area of the shear reinforcement, Asw
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.50 m, Width: b = 0.20 m, Length: L = 3.00 m, Concrete cover: c=3.5cm Effective height: d=h(0.06*h+ebz)=0.435m; d’=ehz=0.035m
■ ■
Stirrup slope: α= 90° Strut slope: θ=30˚
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x = 0) restrained in translation along X, Y and Z, ► Support at end point (x = 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:
a VEd = Pu 1 − l According to EC2 the Pu point load is defined by the next formula: Pu = 1,35*Pg + 1,50*Pq=1.35*105kN=1.50*95kN=284.25kN In this case, (a = 1 m; l = 3 m):
1 VEd = 284,25 * 1 − = 189,5 KN 3
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.
497
ADVANCE DESIGN VALIDATION GUIDE
5.51.2.2 Reference results in calculating the maximum design shear resistance
VRd, max = α cw * ν1 * fcd * z u * b w *
(cotα + cotθ ) 1 + cot 2 θ According to: EC2 Part 1,1 EN 1992112002 Chapter 6.2.3.(3)
Where:
f α cw = 1 coefficient taking account of the state of the stress in the compression chord and v 1 = 0,6 * 1 − ck 250 When the transverse frames are vertical, the above formula simplifies to: v 1 * f cd * z u * b w tgθ + cot θ
VRd,max =
In this case,
θ = 45°
and
α = 90°
v1 strength reduction factor for concrete cracked in shear
20 f = 0.55 v1 = 0,6 * 1 − ck = 0.6 * 1 − 250 250
zu = 0.9 * d = 0.9 * 0.435 = 0.392m VRd , max =
0.55 * 13.33 * 0.392 * 0.20 = 0.288MN 2
VEd = 0.1895MN < VRd , max = 0.288MN Calculation of transversal reinforcement: ■
is determined by considering the transverse reinforcement steel vertical (α=90°) and connecting rods inclined at 45 °, at different points of the beam
■
before the first point load:
■
beyond the point load, the shear force is constant and equal to Rb, therefore:
Vu = − Pu * ■
1 a = 284,25 * = −94,75 KN l 3
it also calculates the required reinforcement area to the right side of the beam:
Finite elements modeling ■ ■ ■
498
Asw VEd . * tgθ 0,18950 * tg 45° = = = 11.13cm² / ml 500 s zu * f ywd 0,392 * 1,15
Linear element: S beam, 3 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
Advance Design gives the following results for Atz (cm2/ml)
5.51.2.3 Reference results Result name
Result description
Reference value
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
2
At,z,1
Theoretical reinforcement area x=1m [cm /ml]
11.13 cm2/ml
At,z,2
Theoretical reinforcement area x=1.01m [cm2/ml]
5.57 cm2/ml
5.51.3
Calculated results
Result name
Result description
Value
Error
Fz
Fz,1
189.5 kN
0.0000 %
Fz
Fz,2
94.75 kN
0.0000 %
Atz
Atz,1
11.1328 cm²
0.1151 %
Atz
Atz,2
5.56641 cm²
0.0645 %
499
ADVANCE DESIGN VALIDATION GUIDE
5.52 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with inclined transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 25) Test ID: 5065 Test status: Passed
5.52.1
Description
Verifies the shear resistance for a rectangular concrete beam C20/25 with inclined transversal reinforcement Bilinear stressstrain diagram (Class XC1). For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) is calculated, along with the crosssectional area of the shear reinforcement (Asw).
5.52.2
Background
Verifies a rectangular cross section beam made from concrete C20/25 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) is calculated, along with the crosssectional area of the shear reinforcement (Asw). This test was evaluated by the French control office SOCOTEC.
5.52.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete C20/25 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: The dead load will be represented by a punctual load of 105kN
■
Exploitation loadings: The live load will be considered from one point load of 95kN
■ ■ ■
Structural class: S1 Reinforcement steel ductility: Class B The reinforcement will be displayed like in the picture below:
The objective is to verify: ■ ■
500
The shear stresses results The crosssectional area of the shear reinforcement, Asw
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.50 m, Width: b = 0.20 m, Length: L = 3.00 m, Concrete cover: c=3.5cm Effective height: d=h(0.06*h+ebz)=0.435m; d’=ehz=0.035m
■ ■
Stirrup slope: α= 45° Strut slope: θ=30˚
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 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:
a VEd = Pu 1 − l According to EC2 the Pu point load is defined by the next formula: Pu = 1,35*Pg + 1,50*Pq=1.35*105kN=1.50*95kN=284.25kN In this case (a=1m; l=3m):
1 VEd = 284,25 * 1 − = 189,5 KN 3
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.
501
ADVANCE DESIGN VALIDATION GUIDE
5.52.2.2 Reference results in calculating the maximum design shear resistance
According to: EC2 Part 1,1 EN 1992112002 Chapter 6.2.3.(3) Where:
α cw = 1 coefficient taking account of the state of the stress in the compression chord and v1 = 0,6 * 1 −
f ck 250
In this case,
θ = 30°
and
α = 45°
v1 strength reduction factor for concrete cracked in shear f 20 v1 = 0,6 * 1 − ck = 0.6 * 1 − = 0.55 250 250
zu = 0.9 * d = 0.9 * 0.435 = 0.392m VRd, max = 0.54 * 13.33 * 0.392 * 0.20 *
(cot 45° + cot30°) = 0.394 1 + cot 2 30°
VEd = 0.1895MN < VRd , max = 0.394MN Calculation of transverse reinforcement: The transverse reinforcement is calculated using the following formula:
Asw VEd 0.18950 = = = 5.76cm² / ml s zu * f ywd * (cot θ + cot α ) * sin α 0.392 * 434.78 * (cot 30 + cot 45) * sin 45 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:
Asw 0.09475 = = 2.88cm² / ml s 0.392 * 434.78 * (cot 30 + cot 45) * sin 45 Finite elements modeling ■ ■ ■
502
Linear element: S beam, 3 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
Advance Design gives the following results for Atz (cm2/ml)
5.52.2.3 Reference results Result name
Result description
Reference value
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
2
Theoretical reinforcement area at x=1.0m [cm /ml]
At,z,1
Theoretical reinforcement area at x=1.01m [cm /ml]
At,z,2
5.52.3
5.76 cm2/ml
2
2.88 cm2/ml
Calculated results
Result name
Result description
Value
Error
Fz
Fz,1
189.5 kN
0.0000 %
Fz
Fz,2
94.75 kN
0.0000 %
Atz
Atz,1
5.76277 cm²
0.0000 %
Atz
Atz,2
2.88139 cm²
0.0000 %
503
ADVANCE DESIGN VALIDATION GUIDE
5.53 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a T concrete beam Bilinear stressstrain diagram (Class X0) (evaluated by SOCOTEC France  ref. Test 28) Test ID: 5083 Test status: Passed
5.53.1
Description
Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the crosssectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value.
5.53.2
Background
Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the crosssectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value. This test was evaluated by the French control office SOCOTEC.
5.53.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete: C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
504
■
Loadings from the structure: The dead load will be represented by a linear load of 13.83kN/m
■
Exploitation loadings: The live load will be considered from one linear load of 26.60kN/m
■ ■ ■ ■
Structural class: S1 Reinforcement steel ductility: Class B Exposure class: X0 Concrete density: 25kN/m3
ADVANCE DESIGN VALIDATION GUIDE
■
The reinforcement will be displayed like in the picture below:
The objective is to: ■ ■ ■ ■
Verify the shear stresses results Verify the transverse reinforcement Verify the transverse reinforcement distribution by the Caqout method Identify the steel sewing
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■
Length: L = 10.00 m, Concrete cover: c=3.5cm Effective height: d=h(0.06*h+ebz)=0.764m; d’=ehz=0.035m
■ ■
Stirrup slope: α= 90° Strut slope: θ=30˚
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, Support at end point (x = 10.00) restrained in translation along Y, Z and rotation along X Inner: None.
► ►
■
505
ADVANCE DESIGN VALIDATION GUIDE
Loading: For a beam under a uniformly distributed load Pu, the shear force is defined by the following equation:
V ( x) = Pu * x −
Pu * l 2
For the beam end, (x=0) the shear force will be:
VEd = −
Pu * l 58,57 * 10 =− = −292,9kN 2 2
In the following calculations, the negative sign of the shear will be neglected, as this has no effect in the calculations.
Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreduced shear, that will be used for further calculations.
5.53.2.2 Reference results in calculating the maximum design shear resistance
VRd , max =
v1 * f cd * zu * bw tgθ + cot θ According to: EC2 Part 1,1 EN 1992112002 Chapter 6.2.3.(3)
In this case:
θ = 30°
and
α = 90°
v1 strength reduction factor for concrete cracked in shear 25 f = 0.54 v1 = 0,6 * 1 − ck = 0.6 * 1 − 250 250
zu = 0.9 * d = 0.9 * 0.764 = 0.688m VRd , max =
0.54 * 16.67 * 0.688 * 0.22 = 0.590MN tg 30 + cot 30
VEd = 0.293MN < VRd , max = 0.590MN Calculation of transverse reinforcement: The transverse reinforcement is calculated using the following formula:
Asw VEd . * tgθ 0.293 * tg 30° = = = 5.66cm² / ml 500 s zu * f ywd 0.688 * 1,15 Finite elements modeling ■ ■ ■ 506
Linear element: S beam, 11 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
Advance Design gives the following results for Atz (cm2/ml)
5.53.2.3 Reference results Result name
Result description
Reference value
Fz
Fz corresponding to the 101 combination (ULS) [kNm]
292.9 kN
At,z
Theoretical reinforcement area [cm2/ml]
5.66 cm2/ml
5.53.3
Calculated results
Result name
Result description
Value
Error
Fz
Fz
292.852 kN
0.0007 %
Atz
Atz
5.65562 cm²
0.0774 %
507
ADVANCE DESIGN VALIDATION GUIDE
5.54 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement  Inclined stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 29) Test ID: 5092 Test status: Passed
5.54.1
Description
Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max., will be determined, along with the crosssectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value. The beam model was provided by Bouygues.
5.54.2
Background
Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with the crosssectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value. The beam model was provided by Bouygues. This test was evaluated by the French control office SOCOTEC.
5.54.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete: C35/40 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
508
■
Loadings from the structure: The dead load will be represented by two linear loads of 22.63kN/m and 47.38kN/m
■
Exploitation loadings: The live load will be considered from one linear load of 80.00kN/m
■ ■ ■
Structural class: S3 Reinforcement steel ductility: Class B Exposure class: XC1
■
Characteristic compressive cylinder strength of concrete at 28 days:
ADVANCE DESIGN VALIDATION GUIDE
■ ■ ■ ■ ■
Partial factor for concrete: Relative humidity: RH=50% Concrete age: t0=28days Design value of concrete compressive strength: fcd=23 MPa Secant modulus of elasticity of concrete: Ecm=34000 MPa
■ ■
Concrete density: Mean value of axial tensile strength of concrete: fctm=3.2 MPa
■
Final value of creep coefficient:
■ ■ ■
Characteristic yield strength of reinforcement: Steel ductility: Class B K coefficient: k=1.08
■ ■
Design yield strength of reinforcement: Design value of modulus of elasticity of reinforcing steel: Es=200000MPa
■
Steel density:
■
Characteristic strain of reinforcement or prestressing steel at maximum load:
■
Strain of reinforcement or prestressing steel at maximum load:
■
Slenderness ratio:
λ = 0.80
The objective is to verify: ■ ■ ■
The of shear stresses results The longitudinal reinforcement corresponding to a 5cm concrete cover The transverse reinforcement corresponding to a 2.7cm concrete cover
Units Metric System Geometry Below are described the beam cross section characteristics: ■
Length: L = 10.00 m,
■ ■
Stirrup slope: α= 90° Strut slope: θ=29.74˚
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, Support at end point (x = 10.00) restrained in translation along Y, Z and rotation along X Inner: None.
► ►
■
509
ADVANCE DESIGN VALIDATION GUIDE
Loading:
Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreduced shear, that will be used for further calculations.
5.54.2.2 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:
5.54.2.3 Reference results in calculating the transversal reinforcement
Asw VEd . * tgθ = s zu * f ywd Where:
zu = 0.9 * d = 0.9 *1.125 = 1.0125m cot θ = 1.75 ⇒ θ = 29.74° α = 90° Asw VEd . * tgθ 1.502 * tg 29.74° = = = 19.50cm² / ml 500 s zu * f ywd 1.0125 * 1,15
510
ADVANCE DESIGN VALIDATION GUIDE
The minimum reinforcement percentage calculation:
0,08 * f ck 0,08 * 35 Asw ≥ ρ w, min * bw * sin α = * bw * sin α = * 0.55 * sin 90° = 5.21cm² / ml 500 s f yk Finite elements modeling ■ Linear element: S beam, ■ 15 nodes, ■ 1 linear element. 2 Advance Design gives the following results for Atz (cm /ml)
5.54.2.4 Reference results Result name
Result description
Reference value
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 cm2/ml
Az(2.7cm cover)
Az longitudinal reinforcement corresp. to a 2.7cm cover [cm2/ml]
105.69 cm2/ml
At,z,1
At,z transversal reinforcement for the beam end [cm /ml]
19.10 cm2/ml
At,z,2
At,z transversal reinforcement for the middle of the beam [cm2/ml]
5.21 cm2/ml
5.54.3
2
Calculated results
Result name
Result description
Value
Error
My
My
5255.58 kN*m
0.0000 %
Fz
Fz
1501.59 kN
0.0000 %
Az
Az 5cm
108.301 cm²
0.0000 %
Az
Az 2.7cm
108.301 cm²
2.4666 %
Atz
Atz end
19.4877 cm²
2.0443 %
Atz
Atz middle
5.20615 cm²
0.0000 %
511
ADVANCE DESIGN VALIDATION GUIDE
5.55 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete section  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 23) Test ID: 5053 Test status: Passed
5.55.1
Description
Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the crosssectional area of the shear reinforcement (Asw) calculation.
5.55.2
Background
Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the crosssectional area of the shear reinforcement (Asw) calculation. This test was evaluated by the French control office SOCOTEC.
5.55.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
512
Loadings from the structure: The dead load family is be considered from three loads: two point loads of 55kN and 65kN and one linear load of 25kN/m, placed along the beam as described in the picture:
ADVANCE DESIGN VALIDATION GUIDE
■
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:
■ ■ ■
Structural class: S1 Reinforcement steel ductility: Class B The reinforcement will be displayed like in the picture below:
The objective is to verify: ■ ■ ■
The shear stresses results The crosssectional area of the shear reinforcement, Asw The theoretical reinforcement value
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.70 m, Width: b = 0.30 m, Length: L = 5.30 m, Concrete cover: c=3.5cm Effective height: d=h(0.06*h+ebz)=0.623m; d’=ehz=0.0335m
■ ■
Stirrup slope: α= 90° Stirrup slope: 45˚
513
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x = 0) restrained in translation along X, Y and Z, Support at end point (x = 5.30) restrained in translation along Y, Z and rotation along X Inner: None.
► ►
■
Loading The maximum shear stresses from the concrete beam: Using the software structure calculation the following values were obtained: x = 0m => VEd = 406 KN x = 0.50m (for the first point load) => VEd = 373.76 KN and VEd = 239.51 KN x = 0.95m (for the second point load) => VEd = 210.82 KN and VEd = 70.57 KN x = 2.06m the shear force is null => VEd = 0 KN
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.
514
ADVANCE DESIGN VALIDATION GUIDE
5.55.2.2 Reference results in calculating the maximum design shear resistance
VRd , max = α cw *ν 1 * f cd * zu * bw *
(cotα + cotθ ) 1 + cot 2θ
According to: EC2 Part 1,1 EN 1992112002 Chapter 6.2.3.(3) Where:
α cw = 1 coefficient taking account of the state of the stress in the compression chord and v1 = 0,6 * 1 −
f ck 250
When the transverse frames are vertical, the above formula simplifies to:
VRd , max =
v1 * f cd * zu * bw tgθ + cot θ
In this case:
θ = 45°
and
α = 90°
v1 strength reduction factor for concrete cracked in shear
25 f = 0.54 v1 = 0,6 * 1 − ck = 0.6 * 1 − 250 250
zu = 0.9 * d = 0.9 * 0.623 = 0.561m VRd , max =
0.54 * 16.67 * 0.561 * 0.30 = 0.757 MN 2
VEd = 0.406 MN < VRd ,max = 0.757 MN Calculation of transversal reinforcement: Is determined by considering the transverse reinforcement steels vertical (α = 90°) and connecting rods inclined at 45 °, at different points of the beam. Before the first point load:
Asw VEd . * tgθ 0,406 * tg 45° = 16,65cm² / ml = = 500 zu * f ywd s 0,561 * 1,15 Between the first and the second point load:
Asw 0,23951 * tg 45° = = 9,82cm² / ml 500 s 0,561 * 1,15 It also calculates the required reinforcement area to the right side of the beam:
Asw 0,20674 * tg 45° = = 8,48cm² / ml 500 s 0,561 * 1,15 Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element. 515
ADVANCE DESIGN VALIDATION GUIDE
Advance Design gives the following results for Atz (cm2/ml)
Note: after the second point load, the minimum transverse reinforcement is set (noted with Atmin in ADVANCE Design) (in cm²/ml):
The reinforcement theoretical value is calculated using the formula:
0,08 * f ck Asw 0,08 * 25 * 0.30 * sin 90 = 2.4cm² / ml ≥ ρ w, min * bw * sin α = * bw * sin α = f yk s 500 5.55.2.3 Reference results Result name
Result description
Reference value
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
Fz
Fz corresponding to 101 combination (ULS) x=0.501m [kN]
239.51 kN
Fz
Fz corresponding to 101 combination (ULS) x=0.95m [kN]
210.82 kN
Fz
Fz corresponding to 101 combination (ULS) x=0.9501m [kN]
70.57 kN
Atz
Transversal reinforcement area x=0m [cm /ml]
16,65 cm2/ml
Atz
Transversal reinforcement area x=0.501m [cm2/ml]
9.82 cm2/ml
Atz
Transversal reinforcement area x=5.3m [cm2/ml]
8.84 cm2/ml
At,min,z
Theoretical reinforcement area [cm2/ml]
2.40 cm2/ml
5.55.3
516
2
Calculated results
Result name
Result description
Value
Error
Fz
Fz,0
405.633 kN
0.0007 %
Fz
Fz,1
373.758 kN
0.0005 %
Fz
Fz,1'
239.445 kN
0.0271 %
Fz
Fz,2
210.821 kN
0.0005 %
Fz
Fz,2'
70.5644 kN
0.0079 %
Atz
Atz,0
16.6391 cm²
0.0547 %
Atz
Atz,1
9.82205 cm²
0.0209 %
Atz
Atz,r
8.48058 cm²
0.0068 %
Atminz
At,min,z
2.4 cm²
0.0000 %
ADVANCE DESIGN VALIDATION GUIDE
5.56 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 27) Test ID: 5076 Test status: Passed
5.56.1
Description
Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram is generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max., will be calculated, along with the crosssectional area of the shear reinforcement, Asw, and the theoretical reinforcement. For the calculation, the reduced shear force values will be used.
5.56.2
Background
Description: Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram is generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max. will be calculated, along with the crosssectional area of the shear reinforcement, Asw, and the theoretical reinforcement. This test was evaluated by the French control office SOCOTEC.
5.56.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: The dead load will be represented by a linear load of 40kN/m
■
Exploitation loadings: The live load will be considered from one linear load of 25kN
■ ■ ■
Structural class: S1 Reinforcement steel ductility: Class B The reinforcement will be displayed like in the picture below:
The objective is to verify: ■ ■
The shear stresses results The theoretical reinforcement value
517
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.70 m, Width: b = 0.35 m, Length: L = 5.75 m, Concrete cover: c=3.5cm Effective height: d=h(0.06*h+ebz)=0.623m; d’=ehz=0.035m
■ ■
Stirrup slope: α= 90° Strut slope: θ=45˚
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.30) restrained in translation along Y, Z and rotation along X Inner: None. ►
■
Loading: For a beam subjected to a point load, Pu, the shear stresses are defined by the formula below:
VEd =
Pu * l 2 According to EC2 the Pu point load is defined by the next formula: Pu = 1,35*Pg + 1,50*Pq=1.35*40kN+1.50*25kN=91.5kN
In this case :
VEd =
5.75 * 91.5 = 263KN 2
5.56.2.2 Reference results in calculating the lever arm zc: The lever arm is calculated using the design formula for pure bending:
Pu = 91.5kN / ml M Ed =
µcu =
518
91.5 * 5.75² = 378.15kNm 8
0.378 M Ed = = 0.167 bw * d ² * f cd 0.35 * 0.623² * 16.67
ADVANCE DESIGN VALIDATION GUIDE
(
)
(
)
α u = 1.25 * 1 − 1 − 2 * µcu = 1.25 * 1 − 1 − 2 * 0.167 = 0.230 zc = d * (1 − 0.4 * α u ) = 0.623 * (1 − 0.4 * 0.230) = 0.566m Calculation of reduced shear force When the shear force curve has no discontinuities, the EC2 allows to consider, as a reduction, the shear force to a horizontal axis x = z. cot θ . In the case of a member subjected to a distributed load, the equation of the shear force is:
V ( x) = Pu * x −
Pu * l 2
Therefore:
x = z * cot θ = 0.566 * cot 45° = 0.566m
VEd , red = 91.5 * 0.566 − 263 = −211kN Warning: Advance Design does not apply the reduction of shear force corresponding to the distributed loads (cutting edge at x = d). Calculation of maximum design shear resistance:
VRd, max = α cw * ν1 * fcd * z u * b w *
(cotα + cotθ ) 1 + cot 2 θ According to: EC2 Part 1,1 EN 1992112002 Chapter 6.2.3.(3)
Where:
α cw = 1
coefficient taking account of the state of the stress in the compression chord and
f v1 = 0,6 * 1 − ck 250 When the transverse frames are vertical, the above formula simplifies to:
VRd , max =
v1 * f cd * zu * bw tgθ + cot θ
In this case:
θ = 45°
and
α = 90°
v1 strength reduction factor for concrete cracked in shear 25 f = 0.54 v1 = 0,6 * 1 − ck = 0.6 * 1 − 250 250
zu = 0.9 * d = 0.9 * 0.623 = 0.56m VRd , max =
0.54 * 16.67 * 0.566 * 0.35 = 0.891MN = 891kN 2
VEd = 236kN < VRd , max = 891kN
519
ADVANCE DESIGN VALIDATION GUIDE
Calculation of transversal reinforcement: Given the vertical transversal reinforcement (α = 90°), the transverse reinforcement is calculated using the following formula:
Asw VEd . * tgθ 0,211 * tg 45 = = = 8,67cm² / ml 500 s zu * f ywd 0,566 * 1,15 Calculation of theoretical reinforcement value: The French national annex indicates the formula:
Asw ≥ ρ w, min * bw * sin α s With:
ρ w, min =
0,08 * f ck 0,08 * 20 = = 7.15 × 10− 4 f yk 500
Asw ≥ ρ w, min * bw * sin α = 7.15 * 10− 4 * 0.2 * sin 90° = 1.43cm² / ml s Finite elements modeling ■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. 2 Advance Design gives the following results for At,min,z (cm /ml)
5.56.2.3 Reference results Result name
Result description
Reference value
Fz
Fz corresponding to the 101 combination (ULS) [kNm]
263 kN
At,min,z
Theoretical reinforcement area [cm2/ml]
1.43 cm2/ml
5.56.3
520
Calculated results
Result name
Result description
Value
Error
Atminz
Atmin,z
1.43108 cm²
0.0755 %
ADVANCE DESIGN VALIDATION GUIDE
5.57 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to compression and rotation moment to the top  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 31) Test ID: 5101 Test status: Passed
5.57.1
Description
Nominal rigidity method. Verifies the adequacy of a rectangular cross section column made from concrete C25/30. The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. Verifies the column to resist simple bending. The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed. The column is considered connected to the ground by a fixed connection and free to the top part.
5.57.2
Background
Nominal rigidity method. Verifies the adequacy of a rectangular cross section made from concrete C25/30. The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.57.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: 15t axial force ► 1.5tm rotation moment applied to the column top ► The selfweight is neglected Exploitation loadings: ► 6.5t axial force ► 0.7tm rotation moment applied to the column top ►
■
■
ψ 2 = 0,3
■ ■ ■ ■ ■ ■ ■
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Concrete cover 5cm Transversal reinforcement spacing a=30cm Concrete C25/30 Steel reinforcement S500B The column is considered isolated and braced
521
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.40 m, Width: b = 0.40 m, Length: L = 6.00 m, Concrete cover: c=5cm
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading The beam is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*15+150*6.5=30t=0.300MN MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm
■
522
e0 =
MEd 0.031 = = 0.10m 0.300 NEd
ADVANCE DESIGN VALIDATION GUIDE
5.57.2.2 Reference results in calculating the concrete column Geometric characteristics of the column: The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * l = 12m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 * 12 = = 104 a 0.40
Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 )
creep coefficient
serviceability first order moment under quasipermanent load combination ULS first order moment (including the geometric imperfections) First order eccentricity evaluation:
e1 = e0 + ei ei = 0.03m The first order moment provided by the quasipermanent loads:
e1 = e0 + ei =
M Eqp 0 N Eqp 0
+ ei =
1.50 + 0.30 * 0.70 + 0.30 = 0.13m 15 + 0.30 * 0.65
N Eqp1 = 1.50 + 0.30 * 0.70 = 16.95t M Eqp1 = N Eqp1 * e1 = 16.95 * 0.13 = 2.20tm = 0.022MNm The first order ULS moment is defined latter in this example:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) = β (t0 ) =
16.8 16.8 = = 2.92 f cm 25 + 8
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0= 28 days concrete age).
523
ADVANCE DESIGN VALIDATION GUIDE
ϕ RH
h0 =
RH 100 =1+ 0.1 * 3 h0 1−
2 * Ac 2 * 400 * 400 = = 200mm ⇒ ϕ RH u 2 * (400 + 400 )
50 100 = 1.85 =1+ 0.1 * 3 200 1−
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.85 * 2.92 * 0.488 = 2.64 The effective creep coefficient calculation:
ϕef = ϕ (∞, t0 ) *
M EQP M Ed
= 2.64 *
0.022 = 1.49 0.039 According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
The second order effects; The buckling calculation: For an isolated column, the slenderness limit check is done using the next formula:
20 * A * B * C n
λlim =
According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1) Where:
n=
0.300 N Ed = = 0.112 Ac * f cd 0.40² * 16.67
A=
1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.49 = 0.77
B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70
λlim =
because the reinforcement ratio in not yet known
because the ratio of the first order moment is not known
20 * 0.77 *1.1* 0.7 = 35.43 0.112
λ = 104 > λlim = 35.43 Therefore, the second order effects most be considered. The second order effects; The buckling calculation: The stresses for the ULS load combination are: NEd= 1.35*15 + 1.50*6.50= 30 t = 0,300MN MEd= 1.35*1.50 + 1.50*0.7= 3.075 t = 0,031MNm Therefore it must be determined: ■ ■
524
The eccentricity of the first order ULS moment, due to the stresses applied The additional eccentricity considered for the geometrical imperfections
ADVANCE DESIGN VALIDATION GUIDE
Initial eccentricity:
e0 =
M Ed 0.031 = = 0.10m N Ed 0.300
Additional eccentricity:
ei =
l0 12 = = 0.03m 400 400
The first order eccentricity: stresses correction: The forces correction, used for the combined flexural calculations:
N Ed = 0.300 MN e1 = e0 + ei = 0.13m M Ed = e1 * N Ed = (0.10 + 0.03) * 0.300 = 0.039MNm M = N Ed * e0
According to Eurocode 2 – EN 199211 2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation: To play the nominal rigidity method a starting section frame is needed. For this it will be used a concrete section considering only the first order effects. Advance Design iterates as many time as necessary. The needed frames will be determined assuming a compound bending with compressive stress. All the results above were obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid of the tensioned steel.
h 0.40 M ua = M G 0 + N * (d − ) = 0.039 + 0.300 * 0.35 − = 0.084MNm 2 2 Verification if the section is partially compressed:
µ BC = 0,8 *
µcu =
h h 0,40 0,40 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,496 d d 0,35 0,35
M ua 0.084 = = 0.103 bw * d ² * f cd 0.40 * 0.35² * 16.67
525
ADVANCE DESIGN VALIDATION GUIDE
µcu = 0.103 < 0.496 = µ BC
therefore the section is partially compressed.
Calculations of steel reinforcement in pure bending:
µcu = 0.103
[
]
α u = 1,25 * 1 − (1 − 2 * 0,103) = 0,136 zc = d * (1 − 0,4 * α u ) = 0,35 * (1 − 0,4 * 0,136) = 0,331m
Calculations of steel reinforcement in combined bending: For the combined bending:
N 0.300 = 5.84 * 10− 4 − = −1.06cm 2 f yd 434.78
A = A'−
The minimum reinforcement percentage:
As , min =
0,10 * N Ed 0.10 * 0.300 = = 0.69cm² ≥ 0.002 * Ac = 3.02cm 2 f yd 434.78
2 The reinforcement will be 4HA10 representing a 3.14cm section
The second order effects calculation: The second order effect will be determined by applying the method of nominal rigidity: Calculation of nominal rigidity: The nominal rigidity of a post or frame member, it is estimated from the following formula:
EI = K c * Ecd * I c + K s * Es * I s According to Eurocode 2 – EN 199211 2004; Chapter 5.8.7.2 (1) With:
Ecd =
Ecm 1.2
f cm = f ck + 8Mpa = 33Mpa Ecm
f = 22000 * cm 10
Ecd = Ic =
33 = 22000 * 10
0.3
= 31476 Mpa
Ecm 31476 = = 26230 Mpa 1.2 1.2
b * h 3 0.40 4 = = 2,133.10 −3 m4 12 12
Es = 200000 Mpa I s : Inertia
526
0.3
(cross section inertia)
ADVANCE DESIGN VALIDATION GUIDE
As 3,14.10−4 ρ= = = 0.002 Ac 0.40 × 0.40 0.002 ≤ ρ = f ck = 20
k1 =
n=
As < 0.01 Ac 25 = 1.12Mpa 20
N Ed 0.300 = = 0.112 Ac * f cd 0.40² *16.67
k2 = n *
104 λ = 0.112 * = 0.069 ≤ 0.20 170 170 2
2
A h 3,14 * 10−4 0.40 Is = 2 * s * − c = 2 * * − 0.05 = 7,06 * 10− 6 m4 2 2 2 2
K s = 1 and
Kc =
k1 * k2 1.12 * 0.069 = = 0.031 1 + ϕef 1 + 1.49
Therefore:
EI = 0.031 * 26230 * 2,133 * 10−3 + 1 * 200000 * 7,06 * 10−6 = 3.15MNm² Corrected stresses: The total moment, including the second order effects is defined as a value and is added to the first order moment value:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.7.3 (1)
M 0 Ed = 0.039 MNm
(time of first order (ULS) taking into account the geometric imperfections, relative to the
center of gravity of concrete).
N Ed = 0.300MN (normal force acting at ULS). And:
β=
π² c0
with
c0 = 8
because the moment is constant (no horizontal force at the top of post). According to Eurocode 2 – EN 199211 2004; Chapter 5.8.7.3 (2)
β=
π² 8
= 1.234
NB = π ² *
EI 3.15 =π²* = 0.216 MN 2 l0 12²
527
ADVANCE DESIGN VALIDATION GUIDE
Therefore the second order moment is:
M Ed
1.234 = −0.133MNm = 0.039 * 1 + 0.216 − 1 0.300
There is a second order moment that is negative because it was critical that the normal force, NB, is less than the applied normal force => instability. A section corresponding to a ratio of 5 ‰ will be considered and the corresponding equivalent stiffness is recalculated.
EI = K c * Ecd * I c + K s * Es * I s With:
Ecd = 26230 Mpa ; I c = 2,133.10−3 m4 Es = 200000 Mpa Is
: Inertia
ρ = 0,005 ⇒ As = 0,005 * Ac = 0,005 * 0,40² = 8cm² It sets up : 4HA16 =>
0.002 ≤ ρ =
ρ = 0,005 ⇒ As = 8,04cm²
As < 0.01 Ac 2
2
As h 8,04 * 10−4 0.40 Is = 2 * * − c = 2 * * − 0.05 = 1,81 * 10− 5 m 4 2 2 2 2
K s = 1 and
Kc =
k1 * k2 1.12 * 0.069 = = 0.031 1 + ϕef 1 + 1.49
Therefore:
EI = 0.031 * 26230 * 2,133.10−3 + 1 * 200000 * 1,81.10−5 = 5.35MNm² The second order effects must be recalculated:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed
β = 1.234 NB = π ² *
M Ed
528
EI 5.35 =π²* = 0.367 MN 2 l0 12²
1.234 = 0.254MNm = 0.039 * 1 + 0.367 − 1 0.300
ADVANCE DESIGN VALIDATION GUIDE
There is thus a second order moment of 0.254MNm. This moment is expressed relative to the center of gravity. Reinforcement calculation for combined bending The necessary frames for the second order stresses can now be determined. A column frames can be calculated from the diagram below:
The input parameters in the diagram are:
µ=
0.254 M Ed = = 0.238 2 b * h * f cd 0.40 * 0.40² *16.67
v=
0.300 N Ed = = 0.112 b * h * f cd 0.40 * 0.40 * 16.67
ω = 0.485 is obtained, which gives:
∑A
s
=
ω * b * h * f cd f yd
=
0.485 * 0.40² * 16.67 = 29.75cm² 434.78
Therefore set up a section 14.87cm ² per side must be set, or 3HA32 per side (by excess) Buckling checking The column in place will be verified without buckling. The new reinforcement area must be considered for the 2 previous calculations: 6HA32 provides As=48.25cm The normal rigidity evaluation: It is estimated nominal rigidity of a post or frame member from the following formula:
EI = K c .E cd . I c + K s .E s . I s With:
Is
: Inertia
ρ=
As 48,25 * 10−4 = = 0.03 Ac 0.40 * 0.40 529
ADVANCE DESIGN VALIDATION GUIDE
2
2
A h 48,25 * 10−4 0.40 Is = 2 * s * − c = 2 * * − 0.05 = 1,086.10− 4 m 4 2 2 2 2
k1 = 1.12 k2 = 0.069 K s = 1 and
Kc =
k1 * k2 1.12 * 0.069 = = 0.031 1 + ϕef 1 + 1.49
The following conditions are not implemented in Advance Design:
ρ= If:
Kc =
As ≥ 0.01 Ac
0,3 1 + 0,5 * ϕef
Then:
EI = 0.031 * 26230 * 2,133 * 10−3 + 200000 * 1,086 * 10−4 = 23.45MNm² Corrected stresses The total moment, including second order effects, is defined as a value plus the moment of the first order:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed
M 0 Ed = 0.039 MNm N Ed = 0.300MN
(normal force acting at ULS)
β = 1.234 NB = π ² *
23.45 EI =π²* = 1.607 MN 2 12² l0
It was therefore a moment of second order which is:
M Ed
1.234 = 0.05MNm = 0.039 * 1 + 1.607 − 1 0.300
Reinforcement calculation:
M Ed = 0.05MNm => µ =
N Ed = 0.300MN => v =
530
0.05 M Ed = = 0.047 2 b * h * f cd 0.40 * 0.40² * 16.67
0.300 N Ed = = 0.112 b * h * f cd 0.40 * 0.40 * 16.67
ADVANCE DESIGN VALIDATION GUIDE
The minimum reinforcement percentage conditions are not satisfied therefore there will be one more iteration. Additional iteration: The additional iteration will be made for a section corresponding to 1%; As=0.009*0.40=14.4cm2, which is 7.2cm2 per side. A 3HA16 reinforcement will be chosen by either side (6HA16 for the entire column), which will give As=12.03cm2. Nominal rigidity evaluation:
EI = K c .E cd . I c + K s .E s . I s . Is
: Inertia
ρ=
As 12,06 *10 −4 = = 0.00754 Ac 0.40 * 0.40
0.002 ≤ ρ = if
Is = 2 *
As 2
K s = 1 and
As < 0.01 Ac 2
2
12,06 * 10−4 0.40 h * − c = 2* * − 0.05 = 2,71.10− 5 m 4 2 2 2 Kc =
k1 * k2 1.12 * 0.069 = = 0.031 1 + ϕef 1 + 1.49
Therefore:
EI = 0.031 * 26230 * 2,133 * 10−3 + 1 * 200000 * 2,71 * 10−5 = 7.15MNm²
531
ADVANCE DESIGN VALIDATION GUIDE
Second order loads calculation:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed
M 0 Ed = 0.039 MNm N Ed = 0.300MN
(normal force acting at ULS).
β = 1.234 NB = π ² *
M Ed
EI 7.15 =π²* = 0.49 MN 2 l0 12²
1.234 = 0.039 * 1 + = 0.115MNm 0.49 − 1 0.300
Reinforcement calculation: Frames are calculated again from the interaction diagram:
µ=
0.115 M Ed = = 0.108 2 b * h * f cd 0.40 * 0.40² *16.67
v=
N Ed 0.300 = = 0.112 b * h * f cd 0.40 * 0.40 * 16.67
ω = 0.18 is obtained, which gives:
∑A
s
=
ω * b * h * f cd f yd
=
0.18 * 0.40² *16.67 = 11.04cm² 434.78
Therefore set up a section 5.52cm ² per side must be set; this will be the final column reinforcement Finite elements modeling ■ ■ ■
532
Linear element: S beam, 7 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area(cm2) (reference value: 11.04cm2)
Theoretical value (cm2) (reference value: 11.16 cm2 = 2 x 5.58 cm2)
5.57.2.3 Reference results Result name
Reference value 2
Az
5.57.3
Result description
5.58 cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Az
Az
5.58 cm²
0.0000 %
533
ADVANCE DESIGN VALIDATION GUIDE
5.58 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 32) Test ID: 5102 Test status: Passed
5.58.1
Description
Verifies a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature Bilinear stressstrain diagram (Class XC1). The column is made of concrete C25/30. Determination of the axial and bending efforts at ultimate limit state is performed. The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the reinforcement by considering a section symmetrically reinforced.
5.58.2
Background
Nominal curvature method is applied in this example. Verifies the adequacy of a square cross section column made from concrete C25/30. This test was evaluated by the French control office SOCOTEC.
5.58.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases are used: ■
Loadings from the structure: 147.15 kN (15 t) axial force ► 14.71 kNm (1.5 tm) bending moment applied to the top of the column ► The selfweight is neglected ►
■
Exploitation loadings (ψ 2 ► ► ► ► ► ► ►
534
= 0.3 ):
63.77 kN (6.5 t) axial force 6.87 kNm (0.7 tm) bending moment applied to the top of the column Concrete cover 5cm Spacing between the reinforcement bar legs: s = 30cm Concrete C25/30 Steel reinforcement S500B The column is considered isolated and braced
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■
Height: h = a = 0.40 m, Width: b = a = 0.40 m, Length: L = 6.00 m, Concrete cover: c = 5 cm
Materials properties Concrete class C25/30 and steel reinforcement S500B are used. The following characteristics are used in relation to theses materials: ■ ■
Reinforcement steel ductility: Class B Exposure class: X0
■
Characteristic compressive cylinder strength of concrete at 28 days:
■ ■
Mean value of concrete cylinder compressive strength: fcm = fck + 8 = 33 MPa Characteristic yield strength of reinforcement: f yk = 500 MPa
Design value of concrete compressive strength:
f cd =
Design value of yield strength of reinforcement:
f yd =
f ck
γc
f yk
γs
=
f ck = 25MPa
25 = 16.67 MPa 1.5
=
500 MPa = 434.78MPa 1.15
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free at the top.
535
ADVANCE DESIGN VALIDATION GUIDE
Loading The beam is subjected to the following load combinations: ■
Ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q NEd = 1.35 * 147.15kN + 1.5 * 63.77kN = 294.3kN = 0.2943MN MEd = 1.35 * 14.71 + 1.5 * 6.87 kNm = 30.16kNm = 0.03016MNm
■
Characteristic combination of actions defined by: CCQ = 1.0 x G + 1.0 x Q
■
e0 =
M Ed 0.03016MNm = = 0.10m N Ed 0.2943MN
5.58.2.2 Reference results in calculating the concrete column Geometric characteristics of the column The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * L = 12m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 *12m = = 104 0.40m a
Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 ) M EQP
M Ed
creep coefficient
serviceability first order moment under quasipermanent load combination ULS first order moment (including the geometric imperfections)
First order eccentricity evaluation:
e1 = e0 + ei
ei =
536
12m l0 = = 0.03m 400 400
ADVANCE DESIGN VALIDATION GUIDE
The first order moment provided by the quasipermanent loads:
e1 = e0 + ei =
M Eqp 0 N Eqp 0
+ ei =
14.71kNm + 0.30 * 6.87 kNm + 0.03m = 0.13m 147.15kN + 0.30 * 63.77 kN
N Eqp1 = 147.15kN + 0.30 * 63.77 kN = 166.281kN
M Eqp1 = N Eqp1 * e1 = 166.281kN * 0.13m = 21.616kNm = 0.0216MNm The first order ULS moment is defined latter in this example: MEd1 = 0.0383 MNm The creep coefficient
ϕ (∞,t0 )
is defined as follows:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) = β (t0 ) =
ϕ RH
16.8 16.8 = = 2.92 f cm 25 + 8
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0= 28 days concrete age).
RH 100 =1+ 0.1 * 3 h0 1−
50 1− 2 * Ac 2 * 0.4m * 0.4m 100 = 1.85 h0 = = = 0.2m = 200mm ⇒ ϕ RH = 1 + u 2 * (0.4m + 0.4m ) 0.1 * 3 200
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.85 * 2.92 * 0.488 = 2.64 The effective creep coefficient calculation:
ϕef = ϕ (∞, t0 ) *
M EQP1 M Ed 1
= 2.64 *
0.0216 MNm = 1.49 0.0383MNm According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
The necessity of buckling calculation (second order effects): For an isolated column, the slenderness limit verification is done using the next formula:
λlim =
20 * A * B * C n According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1)
Where:
n=
0.2943MN N Ed = = 0.11 Ac * f cd 0.40² m 2 * 16.67 MPa
A=
1 1 = (1 + 0.2 * ϕef ) 1 + 0.2 *1.46 = 0.77
B = 1 + 2 * ω = 1.1
because the reinforcement ratio in not yet known
537
ADVANCE DESIGN VALIDATION GUIDE
C = 1.7 − rm = 0.70
λlim =
because the ratio of the first order moment is unknown
20 * 0.77 * 1.1 * 0.7 = 35.75 0.11
λ = 104 > λlim = 35.75 Therefore, the second order effects must be considered. The second order effects  buckling calculation The stresses for the ULS load combination are: NEd = 1.35 * 147.15kN + 1.5 * 63.77kN = 294.3kN = 0.2943MN MEd = 1.35 * 14.71 + 1.5 * 6.87 kNm = 30.16kNm = 0.03016MNm Therefore, it must be determined: ■ The eccentricity of the first order ULS moment, due to the applied loadings ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:
e0 =
M Ed 0.03016MNm = = 0.10m 0.2943MN N Ed
Additional eccentricity:
ei =
l0 12m = = 0.03m 400 400
The first order eccentricity  loadings correction The forces correction, used for the combined flexural calculations, is done as following:
N Ed = 0.2943MN e1 = e0 + ei = 0.10m + 0.03m = 0.13m M Ed = e1 * N Ed = 0.13m * 0.2943MN = 0.0383MNm According to clause 6.1 (4) from EN 199211, for sections subjected to combined bending and compression efforts, the design bending moment value should be at least M = NEd * e0, where
20mm 20mm 20mm e0 = max h = max 400mm = max = 20mm 13.3mm 30 30 Corrected bending moment is bigger than this value, so clause 6.1 (4) is fulfilled.
538
ADVANCE DESIGN VALIDATION GUIDE
Reinforcement calculation in the first order situation To apply the nominal curvature method a starting section frame is needed. For this it will be used a concrete section considering only the first order effects. Advance Design iterates as many time as necessary. The needed reinforcement will be determined assuming a combined bending with compressive effort. All the results above were obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid of the tensioned steel.
h 0.40m M ua = M G 0 + N * (d − ) = 0.0383MNm + 0.2943MN * 0.35m − = 0.0824MNm 2 2 Verifying if the section is partially compressed:
h h 0.40m 0.40m * (1 − 0.4 * ) = 0.8 * * (1 − 0.4 * ) = 0.496 d d 0.35m 0.35m
µ BC = 0.8 *
µcu =
M ua 0.0824MNm = = 0.101 bw * d ² * f cd 0.40m * 0.35² m 2 * 16.67 MPa
µcu = 0.101 < 0.496 = µ BC
therefore the section is partially compressed.
Calculations of the steel reinforcement in pure bending
µ cu = 0.101
[
]
α u = 1.25 * 1 − (1 − 2 * 0.101) = 0.133 zc = d * (1 − 0.4 * α u ) = 0.35m * (1 − 0.4 * 0.133) = 0.331m As =
0.0824MNm M ua = = 5.73 * 10− 4 m 2 = 5.73cm² zc * f yd 0.331m * 434.78MPa
Calculations of steel reinforcement in combined bending with compressive effort For the combined bending:
A = A'−
N 0.2943MN = 5.73 * 10− 4 m 2 − = −1.04cm 2 f yd 434.78MPa
The minimum reinforcement percentage must be applied:
As , min =
0.10 * N Ed 0.10 * 0.2943MN = = 0.68 * 10 − 4 m 2 = 0.68cm² ≥ 0.002 * Ac = 3.2cm 2 f yd 434.78MPa
The proposed reinforcement area will be 4HA10 representing 3.14cm2.
539
ADVANCE DESIGN VALIDATION GUIDE
The second order effects calculation The second order effect will be determined by applying the nominal curvature method. Calculation of nominal curvature Considering a symmetrical reinforcement 4HA10 (3.14cm²), the curvature can be determined by the following formula:
1 1 = K r * Kϕ * r r0 According to Eurocode 2 – EN 199211 2004; Chapter 5.8.8.3 (1) With:
f yd
434.78MPa ε yd 1 Es = = = 200000MPa = 0.0138m −1 r0 0.45 * d 0.45 * d 0.45 * 0.35m
Kr
: is a correction factor depending on axial load,=>
n=
0.2943MN N Ed = = 0.110 Ac * f cd 0.40² m 2 * 16.67 MPa
ω=
As * f yd Ac * f cd
Kr =
nu − n ≤1 nu − nbal
3.14.10−4 m 2 * 434.78MPa = = 0.0512 0.40² m 2 * 16.67 MPa
nu = 1 + ω = 1 + 0.0512 = 1.0512 nbal = 0.4
Kr = Kϕ
1.0512 − 0.110 = 1.445 ≤ 1 ⇒ K r = 1 1.0512 − 0.40
: is a factor for taking account of creep=>
β = 0.35 +
Kϕ = 1 + β * ϕ ef ≥ 1
f ck λ 25 104 − = 0.35 + − = −0.218 200 150 200 150
Kϕ = 1 + β * ϕef = 1 − 0.218 * 1.49 = 0.68 ≥ 1 ⇒ Kϕ = 1 Therefore the curvature becomes:
1 1 = K r * Kϕ * = 0.0138m −1 r r0 Calculation moment:
M Ed = M 0 Ed + M 2 Where:
M 0 Ed
M2
540
: first order moment including the geometrical imperfections.
: second order moment
ADVANCE DESIGN VALIDATION GUIDE
The second order moment must be calculated from the curvature:
M 2 = N Ed * e2 1 l2 12² m 2 e2 = * 0 = 0.0138m −1 * = 0.248m r c 8 Note: c = 8 according to 5.8.8.2 (4) from EN 199211; because the moment is constant (no horizontal force at the top of column).
M 2 = N Ed * e2 = 0.2943MN * 0.248m = 0.073MNm M Ed = M 0 Ed + M 2 = 0.0383MN + 0.073MN = 0.1113MNm The reinforcement must be sized considering the demands of the second order effects:
N Ed = 0.2943MN M Ed = 0.1113MNm Reinforcement calculation corresponding to the second order: The input parameters in the diagram are:
µ=
0.1113MNm M Ed = = 0.104 2 b * h * f cd 0.40m * 0.40² m 2 * 16.67 MPa
v=
0.2943MN N Ed = = 0.110 b * h * f cd 0.40m * 0.40m *16.67 MPa
ω = 0.15 ; =>
∑ As =
ω * b * h * f cd f yd
=
0.15 * 0.40² m 2 * 16.67 MPa = 9.20cm² ; 434.78MPa
which means 4.60cm2 per
side. Buckling checking The verification will be made considering the reinforcement area found previously (9.20cm2)
541
ADVANCE DESIGN VALIDATION GUIDE
Curvature evaluation: The reinforcement has an influence only on the K r parameter:
1 = 0.0138m −1 r0 Kr =
nu − n ≤1 nu − nbal n = 0.110
ω=
As * f yd Ac * f cd
9.20 * 10−4 m 2 * 434.78MPa = = 0.15 0.40² m 2 * 16.67 MPa
nu = 1 + ω = 1 + 0.15 = 1.15
nbal = 0.4 Kr = Kϕ
1.15 − 0.110 = 1.387 ≤ 1 ⇒ K r = 1 1.15 − 0.40
 coefficient which takes account of the creep
Kϕ = 1 + β * ϕef ≥ 1
Kϕ = 1 + β * ϕef = 1 − 0.218 * 1.49 = 0.68 ≥ 1 ⇒ Kϕ = 1 The curvature becomes:
1 1 = K r * Kϕ * = 0.0138m −1 It r r0
was obtained the same curvature and therefore the same second order
moment, which validates the section reinforcement found.
542
ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element.
Theoretical reinforcement area (cm2) (reference value: 9.20 cm2 = 2 x 4.60 cm2)
5.58.2.3 Reference results Result name
Reference value 2
As
5.58.3
Result description
4.60 cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Az
Az
4.64 cm²
0.8696 %
543
ADVANCE DESIGN VALIDATION GUIDE
5.59 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to compression by nominal rigidity method Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 33) Test ID: 5109 Test status: Passed
5.59.1
Description
Verifies a square concrete column subjected to compression by nominal rigidity method Bilinear stressstrain diagram (Class XC1). The column is made of concrete C30/37. The verification of the axial force, applied on top, at ultimate limit state is performed. Nominal rigidity method. The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part, the translations along X and Y axis are also blocked. This test is based on the example from "Applications of Eurocode 2" (J. & JA Calgaro Cortade).
5.59.2
Background
The purpose of this test is to determine the second order effects by applying the method of nominal rigidity of a square cross section made from concrete C30/37, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.59.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases are used: Loadings from the structure (G): ► 1260 kN axial force (1.260 MN) axial force ► The selfweight is neglected The column is considered isolated and braced
ψ 2 = 0.3 2 2 The reinforcement is set to 4HA20 (2 x 6.28cm = 12.56cm )
544
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: Height: h = 0.30 m; Width: b = 0.30 m; Length: L = 4.74 m; ■ Concrete cover: c = 5 cm (for the concrete cover of the bottom / top external fiber measured along the local z axis, Advance Design is having the following notation: ebz = 5 cm; ehz = 5 cm; while for the concrete cover measured along the local y axis the following notations are used: eby = 5 cm; ehy = 5 cm ) Materials Properties Used materials: Concrete C30/37 Steel reinforcement S500B The following characteristics are used in relation to this / these material(s): Characteristic compressive cylinder strength of concrete at 28 days: fck = 30 MPa Value of concrete cylinder compressive strength: fcm = fck + 8(MPa) Characteristic yield strength of reinforcement: fyk = 500 MPa Design value of concrete compressive strength:
f cd =
Design value of yield strength of reinforcement:
f yd =
f ck
γc
f yk
γs
=
30MPa = 20 MPa 1.5
=
500 MPa = 434.78MPa 1.15
Boundary conditions The column is considered connected to the ground by a fixed connection and to the top part, the translations along X and Y axes are also blocked. Loading The beam is subjected to the following load combinations: ► ►
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G The characteristic combination of actions is: CCQ = 1.0 x G
545
ADVANCE DESIGN VALIDATION GUIDE
The ultimate limit state (ULS) combination is: NEd = 1.35 * 1.260 MN = 1.701 MN The characteristic combination of actions is: NEqp = 1.0 * 1.260 MN = 1.260 MN Initial eccentricity:
e0 =
Mu 0 = = 0m N u 1.7
5.59.2.2 Reference results in calculating the concrete column Load calculation: Additional eccentricity due to geometric imperfections: Buckling length: L0 = 0.7*4.74 = 3.32 m
L 332 ei = max 2cm; 0 = max 2cm; = max(2cm;0.83cm ) = 2cm = 0.02m 400 400 According to Eurocode 2 – EN 199211 2004; Chapter 5.2(7) The first order eccentricity:
e1 = e0 + ei = 0.02m
The necessity of buckling calculation (second order effect): Calculating the slenderness of the column:
λ=
L0 * 12 3.32 * 12 = = 38.34 0.3 h
For an isolated column, the slenderness limit check is done using the next formula:
λlim =
20 * A * B * C n
According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1) Where:
n=
1.7 N Ed = = 0.944 Ac * f cd 0.30² * 20
A = 0.7 if ϕ ef
is not known, if it is,
A=
1 1 + 0.2 * ϕ ef
B = 1.1 if ω (reinforcement ratio) is not known, if it is, B = 1 + 2 * ω = 1 + 2 * In this case B
C = 0.70
546
if
= 1+ 2*
12.56 * 10−4 * 434.78 = 1.27 0.3² * 20
rm is not known, if it is, C = 1.7 − rm
As * f yd Ac * f cd
ADVANCE DESIGN VALIDATION GUIDE
In this case:
20 * A * B * C 20 * 0.7 * 1.27 * 0.7 = = 12.81 n 0.944
λlim =
λ = 38.34 > λlim = 12.81 Therefore, the second order effects must be considered. Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕ ef = ϕ (∞, t 0 ).
M 0 EQP M 0 Ed
According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2) Where: ■
ϕ (∞,t0 )
final creep coefficient
■
M 0 EQP
is the first order bending moment in quasipermanent load combination (SLS)
■
M 0 Ed
■
t0 is the concrete age
is the first order bending moment in design load combination (ULS)
The ratio of the moment is in this case:
M 0 Eqp M 0 Ed
=
N eqp * e1 N ed *e1
=
N eqp N ed
=
1.26 = 0.74 1.7
According to annex B of EN 199211, the creep coefficient
ϕ (∞,t0 )
is defined as:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) With:
β ( f cm ) =
β (t0 ) =
ϕ RH
16.8 16.8 = = 2.73MPa f cm 30 + 8
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0 = 28 days concrete age).
RH 1− 100 * α * α = 1 + 2 1 0.1 * 3 h0
α 1 = α 2 = 1 if f cm ≤ 35MPa 35 α1 = f cm h0 =
0.7
and
35 α 2 = f cm
0.2
if
f cm > 35MPa
2 * Ac 2 * 300 * 300 = = 150mm u 2 * (300 + 300)
547
ADVANCE DESIGN VALIDATION GUIDE
Where: ■ ■ ■
RH relative humidity: RH=50% h0: is the notional size of the member (in mm) u: column section perimeter 0.7
f cm
ϕ RH
50 RH 1− 1− 100 * 0.944 * 0.984 = 1.86 100 * α * α = 1 + = 1 + 1 2 0.1 * 3 150 0.1 * 3 h0
= 0.944
and
35 α2 = 38
0.2
35 = 38MPa ≥ 35MPa ⇒ α 1 = 38
= 0.984
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.86 * 2.73 * 0.488 = 2.48
ϕef = ϕ (∞, t0 ) *
M 0 Eqp M 0 Ed
= 2.48 * 0.74 = 1.835
1 + ϕef = 1 + 1.835 = 2.835 Calculation of nominal rigidity: The nominal rigidity of a post or frame member, it is estimated from the following formula:
EI = K c * Ecd * I c + K s * Es * I s Where:
E cd
: is the design value of the modulus of elasticity of concrete
I c : is the moment of inertia of concrete cross section E s : is the design value of the modulus of elasticity of reinforcement I s : is the second moment of area of reinforcement, about the center of area of the concrete K c : is a factor for effects of cracking, creep etc. K s : is a factor for contribution of reinforcement
548
ADVANCE DESIGN VALIDATION GUIDE
According to Eurocode 2 – EN 199211 2004; Chapter 5.8.7.2 (1)
k1 * k2 1 + ϕe
Kc =
1 + ϕef = 2.835 f ck = 20
k1 =
n*λ 170
k2 = n=
30 = 1.22 20
N Ed 1.7 = = 0.944 Ac * f cd 0.3² * 20
λ = 38.34
k2 =
n * λ 0.944 * 38.34 = = 0.213 ⇒ k2 = 0.20 170 170
Kc =
Ecd =
k1 * k 2 1.22 × 0.20 = = 0.086 1 + ϕe 2.835
Ecm
γ CE
f +8 E cm = 22 * ck 10 Ecd = Ic =
Ecm
γ CE
=
0.3
30 + 8 = 22 * 10
0.3
= 32837 MPa
32837 = 27364MPa 1.2
b * h3 0.3 * 0.33 = = 0.000675m 4 12 12
Ks = 1 E s = 200GPa 2
2
A 12.56 *10 −4 0.25 − 0.05 d − d' −5 4 I s = 2 * theo * * = 1.256 *10 m = 2* 2 2 2 2 EI = 0.086 * 27364 * 0.000675 + 1 * 200000 * 1.256 * 10 −5 = 4.1MNm²
549
ADVANCE DESIGN VALIDATION GUIDE
Stresses correction: The total moment, including second order effects, is defined as an increased value of the first order moment:
M Ed
1.234 β = 0.034 * 1 + = 0.070 MNm = M 0 Ed * 1 + NB 3.67 − 1 −1 N Ed 1.7
M 0 Ed = 0.0231MNm (moment of first order (ULS) taking into account geometric imperfections) N Ed = 1.155MN (normal force acting at ULS)
β=
β=
π² c0
π² 8
NB =
and
c0 = 8
for a constant first order moment (no horizontal force at the top of column).
= 1.234
π ² * EI Lf ²
=
π ² * 4.10 3.32²
= 3.67 MN
The calculation made with flexural:
N Ed = 1.7 MN
and
M Ed = 0.070 MNm
2 Therefore, a 2 x 6.64cm 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 and it sets up 4HA20.
550
ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■ ■
Linear element: S beam, 6 nodes, 1 linear element. 2 Theoretical reinforcement area(cm )
Theoretical reinforcement area(cm2) (reference value: 13.08 cm2 = 2 x 6.64cm2)
5.59.2.3 Reference results Result name
Reference value 2
Az
5.59.3
Result description
6.64 cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Az
Az
6.53615 cm²
0.0295 %
551
ADVANCE DESIGN VALIDATION GUIDE
5.60 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 26) Test ID: 5072 Test status: Passed
5.60.1
Description
Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the crosssectional area of the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used.
5.60.2
Background
Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the crosssectional area of the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used. This test was evaluated by the French control office SOCOTEC.
5.60.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: The dead load will be represented by a linear load of 40kN/m
■
Exploitation loadings: The live load will be considered from one linear load of 25kN
■ ■ ■
Structural class: S1 Reinforcement steel ductility: Class B The reinforcement will be displayed like in the picture below:
The objective is to verify: ■ ■
552
The shear stresses results The crosssectional area of the shear reinforcement, Asw
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.70 m, Width: b = 0.35 m, Length: L = 5.75 m, Concrete cover: c=3.5cm Effective height: d=h(0.06*h+ebz)=0.623m; d’=ehz=0.035m
■ ■
Stirrup slope: α= 90° Strut slope: θ=45˚
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.30) restrained in translation along Y, Z and rotation along X Inner: None. ►
■
Loading: For a beam subjected to a point load, Pu, the shear stresses are defined by the formula below:
VEd =
Pu * l 2 According to EC2 the Pu point load is defined by the next formula: Pu = 1,35*Pg + 1,50*Pq=1.35*40kN+1.50*25kN=91.5kN
In this case :
VEd =
5.75 * 91.5 = 263KN 2
553
ADVANCE DESIGN VALIDATION GUIDE
5.60.2.2 Reference results in calculating the lever arm zc: The lever arm will be calculated from the design formula for pure bending:
Pu = 91.5kN / ml M Ed =
91.5 * 5.75² = 378.15kNm 8
M Ed 0.378 = = 0.167 bw * d ² * f cd 0.35 * 0.623² * 16.67
µcu =
(
)
(
)
α u = 1.25 * 1 − 1 − 2 * µcu = 1.25 * 1 − 1 − 2 * 0.167 = 0.230 zc = d * (1 − 0.4 * α u ) = 0.623 * (1 − 0.4 * 0.230) = 0.566m Calculation of reduced shear force When the shear force curve has no discontinuities, the EC2 allows to consider, as a reduction, the shear force to a horizontal axis In case of a member subjected to a distributed load, the equation of the shear force is:
V ( x) = Pu * x −
Pu * l 2
Therefore:
x = z * cot θ = 0.566 * cot 45° = 0.566m
VEd , red = 91.5 * 0.566 − 263 = −211kN Warning: Advance Design does not apply the reduction of shear force corresponding to the distributed loads (cutting edge at x = d). Calculation of maximum design shear resistance:
VRd, max = α cw * ν1 * fcd * z u * b w *
(cotα + cotθ ) 1 + cot 2 θ According to: EC2 Part 1,1 EN 1992112002 Chapter 6.2.3.(3)
Where:
α cw = 1
coefficient taking account of the state of the stress in the compression chord and
f v1 = 0,6 * 1 − ck 250 When the transverse frames are vertical, the above formula simplifies to: VRd,max =
v 1 * f cd * z u * b w tgθ + cot θ
In this case:
θ = 45°
and
α = 90°
v1 strength reduction factor for concrete cracked in shear
25 f = 0.54 v1 = 0,6 * 1 − ck = 0.6 * 1 − 250 250
554
ADVANCE DESIGN VALIDATION GUIDE
zu = 0.9 * d = 0.9 * 0.623 = 0.56m VRd , max =
0.54 * 16.67 * 0.566 * 0.35 = 0.891MN = 891kN 2
VEd = 236kN < VRd , max = 891kN Calculation of transversal reinforcement: Given the vertical transversal reinforcement (α = 90°), the following formula is used:
Asw VEd . * tgθ 0,211 * tg 45 = = = 8,67cm² / ml 500 s zu * f ywd 0,566 * 1,15 Finite elements modeling ■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. 2 Advance Design gives the following results for Atz (cm /ml)
5.60.2.3 Reference results Result name
Result description
Fz
Fz corresponding to the 101 combination (ULS) [kNm]
263 kN
2
8.67 cm2/ml
Theoretical reinforcement area [cm /ml]
At,z
5.60.3
Reference value
Calculated results
Result name
Result description
Value
Error
Fz
Fz
263.062 kN
0.0008 %
Atz
Atz
8.68636 cm²
0.1887 %
555
ADVANCE DESIGN VALIDATION GUIDE
5.61 EC2 / NF EN 199211/NA  France: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 30) Test ID: 5098 Test status: Passed
5.61.1
Description
Verifies a T cross section beam made from concrete C30/37 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with the crosssectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value. The beam model and the results were provided by Bouygues.
5.61.2
Background
Verifies a T cross section beam made from concrete C30/37 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with the crosssectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value. The beam model was provided by Bouygues. This test was evaluated by the French control office SOCOTEC.
5.61.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Concrete: C35/40 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
556
■
Loadings from the structure: The dead load will be represented by two linear loads of 23.50kN/m and 43.50kN/m
■
Exploitation loadings: The live load will be considered from one linear load of 25.00kN/m
■ ■ ■
Structural class: S3 Reinforcement steel ductility: Class B Exposure class: XC1
■
Characteristic compressive cylinder strength of concrete at 28 days:
ADVANCE DESIGN VALIDATION GUIDE
■ ■ ■ ■ ■
Partial factor for concrete: 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 objective is to verify: ■ ■ ■
The shear stresses results The longitudinal reinforcement corresponding to a 6.4cm concrete cover The transverse reinforcement corresponding to a 4.3cm concrete cover
Units Metric System Geometry Below are described the beam cross section characteristics: ■
Length: L = 8.10 m,
■ ■
Stirrup slope: α= 90° Strut slope: θ=22.00˚
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x = 0) restrained in translation along X, Y and Z, Support at end point (x = 10.00) restrained in translation along Y, Z and rotation along X Inner: None.
► ►
■
557
ADVANCE DESIGN VALIDATION GUIDE
Loading:
Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreduced shear, that will be used for further calculations.
5.61.2.2 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:
2 For a 4.3 cm concrete cover and dinit=0.521, the reference value will be 54.3 cm :
5.61.2.3 Reference results in calculating the transversal reinforcement
Asw VEd . * tgθ = s zu * f ywd Where:
zu = 0.9 * d = 0.9 * 0.521 = 0.469m cot θ = 2.5 ⇒ θ = 22° α = 90° Asw VEd . * tgθ 0.589 * tg 22° = = = 11.67cm² / ml 500 s zu * f ywd 0.469 * 1,15 The minimum reinforcement percentage calculation: 558
ADVANCE DESIGN VALIDATION GUIDE
0,08 * f ck 0,08 * 30 Asw ≥ ρ w, min * bw * sin α = * bw * sin α = * 0.80 * sin 90° = 7.01cm² / ml 500 s f yk Finite elements modeling ■ Linear element: S beam, ■ 9 nodes, ■ 1 linear element. 2 Advance Design gives the following results for Atz (cm /ml)
5.61.2.4 Reference results Result name
Result description
Reference value
My
My corresponding to the 101 combination (ULS) [kNm]
1193.28 kNm
Fz
Fz corresponding to the 101 combination (ULS) [kNm]
589.28 kN
Az(6.4cm cover)
Az longitudinal reinforcement corresp. to a 6.4cm cover [cm2/ml]
57.07 cm2/ml
Az(4.3cm cover)
Az longitudinal reinforcement corresp. to a 4.3cm cover [cm2/ml]
54.28 cm2/ml
At,z,1
At,z transversal reinforcement for the beam end [cm /ml]
11.68 cm2/ml
At,z,2
At,z transversal reinforcement for the middle of the beam [cm2/ml]
7.01 cm2/ml
5.61.3
2
Calculated results
Result name
Result description
Value
Error
My
My
1193.28 kN*m
0.0000 %
Fz
Fz
589.275 kN
0.0000 %
Az
Az 4.3cm cover
54.2801 cm²
0.0366 %
Atz
Atz beam end
11.6782 cm²
0.0703 %
Atz
Atz middle
7.01085 cm²
0.0121 %
559
ADVANCE DESIGN VALIDATION GUIDE
5.62 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column using the method based on nominal curvature Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 36) Test ID: 5125 Test status: Passed
5.62.1
Description
Verifying a rectangular concrete column using the method based on nominal curvature  Bilinear stressstrain diagram (Class XC1) Verifies the adequacy of a rectangular cross section column made from concrete C30/37. Method based on nominal curvature The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by an articulated connection (all the translations are blocked and all the rotations are permitted) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked. This example is provided by the “Calcul des Structures en beton” book, by JeanMarie Paille, edition Eyrolles.
5.62.2
Background
The calculation is based on nominal curvature method. Verifies the adequacy of a rectangular cross section made from concrete C30/37. The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.62.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
■
Loadings from the structure: ► NG = 0.40 MN axial force ► The selfweight is neglected Exploitation loadings: ► NQ = 0.30 MN axial force ►
■
560
= 0,3
Wind loads: ► Hw = 0.250 MN wind horizontal effect ► 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 coefficient ψ 2
The quasi permanent coefficients are: ψ 0
Concrete cover 5cm Concrete C30/37 Steel reinforcement S500B Relative humidity RH=50% Concrete age t0=30days 2 The reinforcement is set to 8HA16 (16.08cm )
= 1 and ψ 2 = 0
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the column cross section characteristics: ■ ■ ■ ■
Height: h = 0.35 m, Width: b = 0.60 m, Length: L = 5.00 m, Concrete cover: c = 5 cm
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a articulated connection (all the translations are blocked and all the rotations are permitted) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked. Loading The column is subjected to the following load combinations: ■
Load combinations:
N ED = 1.35 * N G + 1.5 * N Q = 1.35 * 0.4 + 1.5 * 0.3 = 0.99MN e0 =
Mu = 0cm Nu
l 500 ei = max 2cm; 0 = max 2cm; = 2cm 400 400 First order eccentricity: e1 = e 0 + e i = 2cm = 0.02m
561
ADVANCE DESIGN VALIDATION GUIDE
5.62.2.2 Reference results in calculating the concrete column Geometric characteristics of the column: The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = l = 5m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
2 3 * l0 2 3 * 5 = = 49.5 a 0.35
λ=
Effective creep coefficient calculation: The creep coefficient is determined by the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 ) M EQP The
creep coefficient
serviceability firs order moment under quasipermanent load combination
M EQP
value is calculated using the following combination:
M EQP = G + 0.3Q + 0 * W
M 0 Eqp = N qp * e1 = (0.4 + 0.3 * 0.3) * 0.02 = 0.0098MNm
M Ed
ULS first order moment (including the geometric imperfections)
M 0 ED
value is calculated using the following combination:
M 0 Eqp = N ed * e1 + 1.5 * HW *
M oED = 1.35G + 1.5 * Q + 1.5 * W
L 5 = 0.99 * 0.02 + 1.5 * 0.25 * = 0.48855MNm 4 4
The moment report becomes:
M 0 Eqp M 0 Ed
=
0.0098 = 0.02 0.48855
The creep coefficient
ϕ (∞,t0 )
is:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) = β (t0 ) =
562
16.8 16.8 = = 2.73MPa f cm 30 + 8
1 1 = = 0.482 0.20 0.1 + 300.20 0.1 + t0
(for t0= 30 days concrete age).
ADVANCE DESIGN VALIDATION GUIDE
ϕ RH
RH 1− 100 * α * α = 1 + 1 2 0.1 * 3 h0
RH = relative humidity; RH = 50% Where α1
h0 =
= α 2 = 1 if f cm
35 ≤ 35MPa if not α1 = f cm
and
35 α 2 = f cm
0.2
2 * Ac 2 * 0.35m * 0.60m = = 0.221m = 221mm u 2 * (0.35m + 0.60m )
f cm = 38MPa > 35MPa
35 α 2 = f cm
ϕ RH
0.7
0.2
35 = 38
35 α1 = f cm
therefore
0.7
35 = 38
0.7
= 0.944
and
0.2
= 0.984
50 RH 1− 1− 100 100 * α1 * α 2 = 1 + * 0.944 * 0.984 = 1.752 = 1+ 0.1 * 3 221 0.1 * 3 h0
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.752 * 2.73 * 0.482 = 2.30 The effective creep coefficient calculation:
ϕef = ϕ (∞, t0 ) *
M EQP M Ed
= 2.30 * 0.02 = 0.046 According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
1 + ϕef = 1 + 0.046 = 1.046 The necessity of buckling calculation (second order effect): The M2 moment is calculated from the curvature formula:
M 2 = N Ed * e2
1 l02 e2 = * r c l0 is the buckling length: l0 = l = 5m “c” is a factor depending on the curvature distribution. According to chapter 5.8.8.2 (4) from EN 199211, c = 10.
1 1 = K r * Kϕ * r0 r According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(1)
ε yd f yd 434.78 2.1739 1 and ε yd = = = = r0 (0.45 * d ) Es 200000 1000
563
ADVANCE DESIGN VALIDATION GUIDE
2.1739 1 1000 = = = 0.0161m −1 r0 (0.45 * d ) (0.45 * 0.30)
ε yd
Kr is the correction coefficient depending of the normal force:
nu − n ≤1 nu − nbal
Kr =
According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(3)
n=
0.99 N Ed = 0.236 = Ac * f cd 0.6 * 0.35 * 20
nu = 1 + ω
ω=
As f yd Ac f cd
=
16.08 * 10−4 * 434.78 = 0.166 0.6 * 0.35 * 20
nu = 1 + ω = 1. + 0.166 = 1.166 nbal = 0,4 Kr =
nu − n 1.166 − 0.236 = = 1.214 nu − nbal 1.166 − 0.4
Condition: K r
≤ 1 , therefore it will be considered: K r = 1
Kϕ creep coefficient: Kϕ = 1 + β * ϕef ≥ 1 According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(4)
β = 0,35 + λ=
f ck λ − 200 150
l0 12 5 12 = = 49.5 0.35 h
β = 0.35 +
f ck λ 30 49.5 − = 0.35 + − = 0.17 200 150 200 150
Kϕ = 1 + β * ϕef = 1 + 0.17 * 0.046 = 1.008 Therefore:
1 1 = K r * Kϕ * = 1 * 1.008 * 0.0161 = 0.0162 r r0 5² 1 l2 e2 = * 0 = 0.0162 * = 0.04057 m r c 10
M 2 = N Ed * e2 = 0.990 * 0.04057 = 0.04017 MNm M Ed = M 0 Ed + M 2 = 0.48855 + 0.04017 = 0.5287 MNm The frames must be sized considering the demands of the second order effects, as follows:
N Ed = 0.990 MN 564
ADVANCE DESIGN VALIDATION GUIDE
M Ed = 0.5287 MNm Reinforcement calculation according the second order effects: The calculation is considering the combined bending and axial solicitation and it is using the following values:
N Ed = 0.990 MN
M Ed = 0.5287 MNm The calculation resulted in 38.02cm2 tensioned reinforcement and a 20cm2 compressed reinforcement, meaning a total of 58.02cm2 reinforcement area. Buckling checking: The calculation is performed considering the reinforcement area found previously (58.02cm2): Curvature calculation: The reinforcement area has an influence only over the Kr parameter:
1 = 0.0161m −1 r0
nu − n ≤1 nu − nbal
Kr =
n = 0.112
ω=
As * f yd Ac * f cd
=
58,02 * 10−4 * 434.78 = 0.60 0.60 * 0.35 * 20
nu = 1 + ω = 1 + 0.60 = 1.60
nbal = 0,4
Kr =
1.60 − 0.112 = 1.24 ≤ 1 ⇒ K r = 1 1.60 − 0.40
Kϕ = 1
the creep coefficient
Kϕ = 1 + β .ϕ ef ≥ 1
Therefore the curvature becomes:
1 1 = K r * Kϕ * = 0.0161m −1 r r0 Considering this result, the same curvature is obtained, which means that the second order moment is the same. The reinforcement section is correctly chosen. Finite elements modeling ■ ■ ■
Linear element: S beam, 5 nodes, 1 linear element.
565
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area (cm2) (reference value: 38.02cm2)
Theoretical value (cm2) (reference value: 76.02 cm2)
5.62.2.3 Reference results Result name
566
Reference value 2
Ay
5.62.3
Result description
38.02cm2
Tensioned reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Ay
Ay
38.01 cm²
0.0263 %
ADVANCE DESIGN VALIDATION GUIDE
5.63 EC2 / NF EN 199211/NA  France: Verifying a square concrete column using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 37) Test ID: 5126 Test status: Passed
5.63.1
Description
Verifies the adequacy of a square concrete column made of concrete C25/30, using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1). The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked), while on the top part translations along X, Y axis and rotation along Z axis are blocked.
5.63.2
Background
Simplified Method Verifies the adequacy of a square cross section made from concrete C25/30. This test was evaluated by the French control office SOCOTEC.
5.63.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
■ ■ ■ ■ ■ ■
Loadings from the structure: ► 800kN axial force ► The selfweight is neglected Exploitation loadings: ► 800kN axial force Concrete cover 5cm Concrete C25/30 Steel reinforcement S500B Relative humidity RH=50% Buckling length L0=0.70*4=2.80m
567
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.40 m, Width: b = 0.400 m, Length: L = 4.00 m, Concrete cover: c=5cm
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked. Loading The beam is subjected to the following load combinations: ■
Load combinations:
N ED = 1.35 * 800 + 1.5 * 800 = 2280kN
568
ADVANCE DESIGN VALIDATION GUIDE
5.63.2.2 Reference results in calculating the concrete column Scope of the method:
L0 L L L L L * 12 2.8 * 12 = 0 = 0 = 0 = 0 = 0 = = 24.25 4 2 a i a 0.4 I a a 12 A 12 12 2 a
λ=
According to Eurocode 2 EN 199211 (2004) Chapter 5.8.3.2(1) The method of professional rules can be applied as:
λ < 120 20 < f ck < 50 MPa h > 0.15m Reinforcement calculation:
λ = 24.25 < 60 , therefore: α=
0.86 λ 1+ 62
2
=
0.86 24.25 1+ 62
2
= 0.746
According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283 Not knowing the values for ρ and δ , we can considered k h = 0.93
k s = 1.6 − 0.6 *
As =
1 f yd
f yk 500
= 1.6 − 0.6 *
500 =1 500
N ed 1 2.280 * * − 0.4 * 0.4 *16.67 = 14,24cm² − b * h * f cd = k * k * 434 . 78 0 . 93 * 1 * 0 . 746 α h s According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283
Finite elements modeling ■ ■ ■
Linear element: S beam, 5 nodes, 1 linear element.
569
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area(cm2) (reference value: 14.24cm2=4*3.56cm2)
5.63.2.3 Reference results Result name
570
Reference value 2
Ay
5.63.3
Result description
3.57cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Ay
Ay
3.28231 cm²
0.2337 %
ADVANCE DESIGN VALIDATION GUIDE
5.64 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to a small compression force and significant rotation moment to the top  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 40) Test ID: 5153 Test status: Passed
5.64.1
Description
Verifies a square cross section column made of concrete C30/37 subjected to a small compression force and significant rotation moment to the top  Bilinear stressstrain diagram is used (Class XC1). The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a symmetrically reinforced section. The column is considered connected to the ground by a fixed connection and free to the top part.
5.64.2
Background
Nominal rigidity method. Verifies the adequacy of a rectangular cross section made from concrete C30/37. The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.64.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: 15kN axial force ► 150kMm rotation moment applied to the column top ► The selfweight is neglected Exploitation loadings: ► 7kN axial force ► 100kNm rotation moment applied to the column top ►
■
■ ■ ■ ■ ■ ■ ■ ■
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 = 40cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced
571
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.50 m, Width: b = 0.50 m, Length: L = 5.80 m, Concrete cover: c = 5cm
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading The beam is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*15+150*7=30.75kN=0.03075MN MEd=1.35*150+1.50*100=352.50kNm=0.352MNm
■
572
e0 =
MEd 0.352 = = 11.45m NEd 0.03075
ADVANCE DESIGN VALIDATION GUIDE
5.64.2.2 Reference results in calculating the concrete column Geometric characteristics of the column: The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * l = 11.60m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 * 11.60 = = 80.37 a 0.50
Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 )
creep coefficient
M EQP
serviceability firs order moment under quasipermanent load combination
M Ed
ULS first order moment (including the geometric imperfections)
First order eccentricity evaluation:
e1 = e0 + ei ei = 0.03m The first order moment provided by the quasipermanent loads:
N Eqp1 = 15 + 0.30 * 7 = 17.10kN
M Eqp1 = N Eqp1 * e1 = 17.10 * 10.56 = 180.58kNm = 0.181MNm The first order ULS moment is defined latter in this example:
The creep coefficient
ϕ (∞,t0 )
is defined as follows:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) = β (t0 ) =
16.8 16.8 = = 2.72 30 + 8 f cm
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0= 28 days concrete age).
573
ADVANCE DESIGN VALIDATION GUIDE
RH 1− 100 * α * α = 1 + 1 2 0.1 * 3 h0
ϕ RH
f cm
35 α1 = f cm > 35MPa therefore:
h0 =
0.7
35 = 38
0.7
2 * Ac 2 * 500 * 500 = = 250mm ⇒ ϕ RH u 2 * (500 + 500 )
35 = 0.944 α 2 = f cm and
0.2
35 = 38
0.2
= 0.984
50 1− 100 * 0.944 * 0.984 = 1.72 = 1 + 0.1 * 3 250
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 The effective creep coefficient calculation:
ϕef = ϕ (∞, t0 ) *
M EQP M Ed
= 2.28 *
0.181 = 1.17 0.352 According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
The second order effects; The buckling calculation: For an isolated column, the slenderness limit check is done using the next formula:
λlim =
20 * A * B * C n According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1)
Where:
n=
N Ed 0.031 = = 0.0062 Ac * f cd 0.50² * 20
A=
1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.17 = 0.81
B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70
λlim =
because the reinforcement ratio in not yet known
because the ratio of the first order moment is not known
20 * 0.81 * 1.1 * 0.7 = 158.42 0.0062
λ = 80.37 < λlim = 158.42 Therefore, the second order effects can be neglected. Calculation of the eccentricities and solicitations corrected for ULS: The stresses for the ULS load combination are: NEd= 1.35*15 + 1.50*7= 30.75kN = 0,03075 MN MEd= 1.35*150 + 1.50*100= 352.5kN= 0.3525MNm
574
ADVANCE DESIGN VALIDATION GUIDE
Therefore, we must calculate: ■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:
e0 =
M Ed 0.3525 = = 11.46m N Ed 0.03075
Additional eccentricity:
ei =
l0 11.6 = = 0.03m 400 400
The first order eccentricity: stresses correction: The forces correction, used for the combined flexural calculations:
N Ed = 0.03075MN e1 = e0 + ei = 11.49m M Ed = e1 * N Ed = 11.49 * 0.03075 = 0.353MNm M = N Ed * e0 20mm 20mm 20mm = 20mm e0 = max h = max 500mm = max 16.7mm 30 30 According to Eurocode 2 – EN 199211 2004; Chapter 6.1.(4)
575
ADVANCE DESIGN VALIDATION GUIDE
Reinforcement calculation in the first order situation: The theoretical reinforcement will be determined by the following diagram
The input parameters of the diagram are:
µ=
0.353 M Ed = = 0.141 2 b * h * f cd 0.50 * 0.502 * 20
ν=
N Ed 0.03075 = = 0.00615 b * h * f cd 0.5 * 0.5 * 20
Therefore:
ω = 0.35 The reinforcement area will be:
∑ As =
ω * 0.502 * 20 434.78
= 40.25cm 2
The total area will be 40.25cm2. Finite elements modeling ■ ■ ■
576
Linear element: S beam, 7 nodes, 1 linear element.
which means 20.13cm2 per face.
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area(cm2) (reference value: 19.32cm2)
Theoretical value (cm2) (reference value: 38.64 cm2= 2 x 19.32cm2)
5.64.2.3 Reference results Result name
Reference value 2
Az
5.64.3
Result description
19.32 cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Az
Az
19.32 cm²
4.0000 %
577
ADVANCE DESIGN VALIDATION GUIDE
5.65 EC2 / NF EN 199211/NA  France: Verifying a square concrete column subjected to a significant compression force and small rotation moment to the top  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 41) Test ID: 5195 Test status: Passed
5.65.1
Description
Verifies a square cross section concrete column made of concrete C30/37 subjected to a significant compression force and a small rotation moment to the top  Bilinear stressstrain diagram is used (Class XC1). The purpose of this test is to determine the second order effects by applying the nominal rigidity method and then calculate the frames by considering a symmetrically reinforced section. The column is considered connected to the ground by a fixed connection and free to the top part.
5.65.2
Background
Nominal rigidity method. Verifies the adequacy of a rectangular cross section made from concrete C30/37. The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.65.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: 150kN axial force ► 15kMm rotation moment applied to the column top ► The selfweight is neglected Exploitation loadings: ► 100kN axial force ► 7kNm rotation moment applied to the column top ►
■
578
■
ψ 2 = 0,3
■ ■ ■ ■ ■ ■
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Concrete cover 3cm and 5cm Transversal reinforcement spacing a=40cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.50 m, Width: b = 0.50 m, Length: L = 5.80 m, Concrete cover: c = 5cm
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading The beam is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: NEd = 1.35*150+1.50*100 = 352.50kN = 0.035MN MEd = 1.35*15+1.50*7 = 30.75kNm = 0.03075MNm
■
e0 =
MEd 0.03705 = = 0.087m NEd 0.353
579
ADVANCE DESIGN VALIDATION GUIDE
5.65.2.2 Reference results in calculating the concrete column Geometric characteristics of the column: The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * l = 11.60m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 * 11.60 = = 80.37 a 0.50
Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 )
creep coefficient
M EQP
serviceability firs order moment under quasipermanent load combination
M Ed
ULS first order moment (including the geometric imperfections)
First order eccentricity evaluation:
e1 = e0 + ei ei = 0.03m The first order moment provided by the quasipermanent loads:
e1 = e0 + ei =
M Eqp 0 N Eqp 0
+ ei =
15 + 0.30 * 7 + 0.30 = 0.125m 150 + 0.30 * 100
N Eqp1 = 150 + 0.30 * 100 = 180kN M Eqp1 = N Eqp1 * e1 = 180 * 0.125 = 22.50kNm = 0.0225MNm The first order ULS moment is defined latter in this example:
M Ed 1 = 0.041MNm The creep coefficient
ϕ (∞,t0 )
is defined as follows:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) = β (t0 ) =
580
16.8 16.8 = 2.72 = 30 + 8 f cm
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0= 28 days concrete age).
ADVANCE DESIGN VALIDATION GUIDE
ϕ RH
RH 1− 100 * α * α = 1 + 1 2 0.1 * 3 h0
f cm
35 > 35MPa therefore: α1 = f cm
0.7
35 = 38
0.7
35 = 0.944 and α 2 = f cm
0.2
35 = 38
0.2
= 0.984
50 1− 2 * Ac 2 * 500 * 500 100 * 0.944 * 0.984 = 1.72 h0 = = = 250mm ⇒ ϕ RH = 1 + u 2 * (500 + 500 ) 0.1 * 3 250
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 The effective creep coefficient calculation:
ϕef = ϕ (∞, t0 ) *
M EQP M Ed
= 2.28 *
0.0225 = 1.25 0.041 According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
The second order effects; The buckling calculation: For an isolated column, the slenderness limit verification is done using the next formula:
20 * A * B * C n
λlim =
According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1) Where:
n=
N Ed 0.353 = = 0.071 Ac * f cd 0.50² * 20
A=
1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.25 = 0.80
B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70
λlim =
because the reinforcement ratio in not yet known
because the ratio of the first order moment is not known
20 * 0.80 * 1.1 * 0.7 = 46.24 0.071
λ = 80.37 > λlim = 46.24 Therefore, the second order effects must be taken into account. Calculation of the eccentricities and solicitations corrected for ULS: The stresses for the ULS load combination are: ■ 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: ■ ■
The eccentricity of the first order ULS moment, due to the stresses applied The additional eccentricity considered for the geometrical imperfections 581
ADVANCE DESIGN VALIDATION GUIDE
Initial eccentricity:
e0 =
M Ed 0.03075 = = 0.087 m 0.353 N Ed
Additional eccentricity:
ei =
l0 11.6 = = 0.03m 400 400
The first order eccentricity: stresses correction: The forces correction, used for the combined flexural calculations:
N Ed = 0.353MN e1 = e0 + ei = 0.117 m M Ed = e1 * N Ed = 0.117 * 0.353 = 0.041MNm M = N Ed * e0 20mm 20mm 20mm = 20mm e0 = max h = max 500mm = max mm 16 . 7 30 30 According to Eurocode 2 – EN 199211 2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation: To apply the nominal rigidity method, we need an initial reinforcement area to start from. For this, the concrete section 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 determined solicitations were calculated from the center of gravity of the concrete section alone. Those stresses must be reduced to the centroid of tensioned steel:
h 0.50 M ua = M G 0 + N * (d − ) = 0.041 + 0.353 * 0.45 − = 0.112MNm 2 2 Verification about the partially compressed section:
µ BC = 0,8 *
µcu =
582
0,50 0,50 h h ) = 0,494 * (1 − 0,4 * * (1 − 0,4 * ) = 0,8 * 0,45 0,45 d d
0.112 M ua = = 0.055 bw * d ² * f cd 0.50 * 0.45² * 20
ADVANCE DESIGN VALIDATION GUIDE
µcu = 0.055 < µ BC = 0.494
therefore the section is partially compressed
The calculation for the tensioned steel in pure bending:
µcu = 0.055
[
]
α u = 1,25 * 1 − (1 − 2 * 0,055) = 0,071 zc = d * (1 − 0,4 * α u ) = 0,45 * (1 − 0,4 * 0,071) = 0,437 m A =
M ua 0,112 = = 5,89cm² zc * f yd 0,437 * 434,78
The calculation for the compressed steel in bending: For the compound bending:
A = A−
0.353 N = 5.89 * 10− 4 − = −2.23cm 2 434.78 Fyd
The minimum column percentage reinforcement must be considered:
As , min =
0,10 * N Ed 0.10 * 0.353 = = 0.81cm² 434.78 Fyd
Therefore, a 5cm2 reinforcement area will be considered.
583
ADVANCE DESIGN VALIDATION GUIDE
5.65.2.3 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:
EI = K c * Ecd * I c + K s * Es * I s Where:
Ecd =
Ecm 1.2
f cm = f ck + 8Mpa = 38Mpa f Ecm = 22000 * cm 10
Ecd = Ic =
0.3
38 = 22000 * 10
0.3
= 32836.57 Mpa
Ecm 32837 = = 27364 Mpa 1.2 1.2
b * h3 0.50 4 = = 5,208.10− 3 m 4 12 12 inertia of the concrete section only
Es = 200000 Mpa Is
: Inertia
ρ=
As 5.10−4 = = 0.002 Ac 0.50 * 0.50
0.002 ≤ ρ = f ck = 20
k1 =
n=
As < 0.01 Ac 30 = 1.22 20
0.353 N Ed = = 0.071 Ac * f cd 0.50² * 20
k2 = n *
80.37 λ = 0.071 * = 0.0336 ≤ 0.20 170 170 2
2
A h 5 * 10−4 0.50 Is = 2 * s * − c = 2 * * − 0.05 = 2.10− 5 m 4 2 2 2 2
Ks = 1
and
Kc =
k1 * k2 1.22 * 0.0336 = = 0.018 1 + ϕef 1 + 1.25
Therefore:
EI = 0.018 * 27364 * 5,208 * 10−3 + 1 * 200000 * 2 * 10−5 = 6.56MNm²
584
ADVANCE DESIGN VALIDATION GUIDE
Stress correction: The total moment, including second order effects, is defined as a value plus the moment of the first order:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed
M 0 Ed = 0.041MNm
(moment of first order (ULS) taking into account geometric imperfections)
N Ed = 0.353MN (normal force acting at ULS). In addition:
β=
β=
π² c0
π² 8
and
c0 = 8
because the moment is constant (no horizontal force at the top of post).
= 1.234
NB = π ² *
6.56 EI =π²* = 0.48MN 2 11.60² l0
It was therefore a moment of 2nd order which is:
M Ed
1.234 = 0.041 * 1 + = 0.182MNm 0.48 − 1 0.353
There is thus a second order moment of 0.182MNm
585
ADVANCE DESIGN VALIDATION GUIDE
Calculation of the flexural combined reinforcement The theoretical reinforcement will be determined by the following diagram:
M Ed = 0.182MNm N Ed = 0.353MN
The input parameters of the diagram are:
µ=
0.182 M Ed = = 0.073 2 b * h * f cd 0.50 * 0.502 * 20
ν=
0.353 N Ed = = 0.071 b * h * f cd 0.5 * 0.5 * 20
Therefore:
ω = 0.11 The reinforcement area will be:
∑ As =
ω * b * h * f cd 0.11 * 0.502 * 20 = = 12.65cm 2 434.78 f yd
This means a total of 12.65cm2 The initial calculations must be repeated by increasing the section; a 6cm2 reinforcement section will be considered. Additional iteration: One more iteration by considering an initial section of 6.5cm ²
586
ADVANCE DESIGN VALIDATION GUIDE
Estimation of the nominal rigidity:
EI = K c * Ecd * I c + K s * Es * I s Where:
Ecd =
Ecm 1.2
f cm = f ck + 8Mpa = 38Mpa Ecm
f = 22000 * cm 10
Ecd = Ic =
0.3
38 = 22000 * 10
0.3
= 32836.57 Mpa
Ecm 32837 = = 27364 Mpa 1.2 1.2
b * h3 0.50 4 = = 5,208.10− 3 m 4 12 12
considering only the concrete section only
Es = 200000 Mpa Is
: Inertia
As 6.5 * 10−4 = = 0.0026 ρ= Ac 0.50 * 0.50 0.002 ≤ ρ = k1 =
n=
f ck = 20
As < 0.01 Ac 30 = 1.22 20
N Ed 0.353 = = 0.071 Ac * f cd 0.50² * 20
k2 = n *
λ 80.37 = 0.071 * = 0.0336 ≤ 0.20 170 170 2
2
A h 6.5 * 10−4 0.50 Is = 2 * s * − c = 2 * − 0.05 = 2.6 × 10− 5 m 4 2 2 2 2
K s = 1 and
Kc =
k1 * k2 1.22 * 0.0336 = = 0.018 1 + ϕef 1 + 1.25
Therefore:
EI = 0.018 * 27364 * 5,208 *10−3 + 1 * 200000 * 2.6 *10−5 = 7.78MNm²
587
ADVANCE DESIGN VALIDATION GUIDE
Stress correction: The total moment, including second order effects, is defined as a value plus the moment of the first order:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed
M 0 Ed = 0.041MNm
(moment of first order (ULS) taking into account geometric imperfections)
N Ed = 0.353MN (normal force acting at ULS). In addition:
β=
β=
π² c0
π² 8
and
c0 = 8
because the moment is constant (no horizontal force at the top of post).
= 1.234
NB = π ² *
7.78 EI =π²* = 0.57 MN 2 11.60² l0
It was therefore a moment of 2nd order which is:
M Ed
1.234 = 0.041 * 1 + = 0.123MNm 0.57 − 1 0.353
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:
M Ed = 0.123MNm N Ed = 0.353MN
588
ADVANCE DESIGN VALIDATION GUIDE
The input parameters of the diagram are:
µ=
M Ed 0.123 = = 0.049 2 b * h * f cd 0.50 * 0.502 * 20
ν=
N Ed 0.353 = = 0.071 b * h * f cd 0.5 * 0.5 * 20
Therefore:
ω = 0.05 The reinforcement area will be:
∑ As =
ω * b * h * f cd 0.05 * 0.502 * 20 = = 5.75cm 2 f yd 434.78
This means a total of 5.75cm2 Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element.
589
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area(cm2) (reference value: 5.75cm2=2*2.88cm2)
Theoretical value (cm2) (reference value: 6.02 cm2)
5.65.2.4 Reference results Result name
590
Reference value 2
Az
5.65.3
Result description
2.88 cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Az
Az
3.01 cm²
4.6957 %
ADVANCE DESIGN VALIDATION GUIDE
5.66 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column subjected to compression to top – Based on nominal rigidity method  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 35) Test ID: 5123 Test status: Passed
5.66.1
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 stressstrain diagram (Class XC1). Based on nominal rigidity method The purpose of this test is to determine the second order effects by applying the nominal rigidity method, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection and free to the top part.
5.66.2
Background
The purpose of this test is to determine the second order effects by applying the nominal stiffness method on a rectangular cross section made of concrete C30/37, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.66.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: 0.45 MN axial force ► 0.10 MNm rotation moment applied to the column top ► The selfweight is neglected Exploitation loadings: ► 0.50 MN axial force ► 0.06 MNm rotation moment applied to the column top ►
■
■
ψ 2 = 0,3
■
The column is considered isolated and braced
591
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.45 m, Width: b = 0.70 m, Length: L = 4.50 m, Concrete cover: c = 5 cm (for the concrete cover of the bottom / top external fiber measured along the local z axis, Advance Design is having the following notation: ebz = 5 cm; ehz = 5 cm; while for the concrete cover measured along the local y axis the following notations are used: eby = 5 cm; ehy = 5 cm )
Materials Properties Used materials: Concrete C30/37 Steel reinforcement S500B The following characteristics are used in relation to this / these material(s): Characteristic compressive cylinder strength of concrete at 28 days: fck = 30 MPa Value of concrete cylinder compressive strength: fcm = fck + 8(MPa) = 38MPa Characteristic yield strength of reinforcement: fyk = 500 MPa Design value of concrete compressive strength:
f cd =
Design value of yield strength of reinforcement:
f yd =
f ck
γc
f yk
γs
=
30MPa = 20 MPa 1.5
=
500 MPa = 434.78MPa 1.15
Boundary conditions The column is considered connected to the ground by a fixed connection and free to the top part. Loading The beam is subjected to the following load combinations: ► ►
592
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q The characteristic combination of actions is: CCQ = 1.0 x G + 0.3 x Q
ADVANCE DESIGN VALIDATION GUIDE
The ultimate limit state (ULS) combination is: NEd = 1.35 * 0.45 MN + 1.5 * 0.5 MN = 1.3575 MN ► MEd = 1.35 * 0.1 MNm + 1.5 * 0.06 MNm = 0.225 MNm The characteristic combination of actions is: ►
► ►
NEqp = 1.0 * 0.45 MN + 0.30 * 0.5 MN = 0.6 MN MEqp = 1.0 * 0.1 MNm + 0.30 * 0.06 MNm = 0.118 MNm
Initial eccentricity:
e0 =
Mu 0.225 = = 0.166m N u 1.3575
5.66.2.2 Reference results in calculating the concrete column Geometric characteristics of the column: The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * l = 2 * 4.50 = 9m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 * 9 = = 69.28 0.45 a
Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed
According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2) Where: Where: ■
ϕ (∞,t0 )
final creep coefficient
■
M 0 EQP
is the first order bending moment in quasipermanent load combination (SLS)
■
M 0 Ed
■
t0 is the concrete age
is the first order bending moment in design load combination (ULS)
M 0 Eqp = 1.0 * 0.1 MNm + 0.30 * 0.06 MNm = 0.118 MNm
M 0 Ed = 1.35 * 0.1 MN + 1.5 * 0.06 MNm = 0.225 MNm
593
ADVANCE DESIGN VALIDATION GUIDE
The moment report becomes:
M 0 Eqp M 0 Ed
=
0.118 = 0.524 0.225
The creep coefficient
ϕ (∞,t0 )
is defined as follows:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) With:
β ( f cm ) = β (t0 ) =
ϕ RH
16.8 16.8 = = 2.73MPa f cm 30 + 8
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0 = 28 days concrete age).
RH 1− 100 * α * α = 1 + 1 2 0.1 * 3 h0
RH = relative humidity; RH = 50%
α 1 = α 2 = 1 if f cm ≤ 35MPa 35 α1 = f cm h0 =
0.7
and
35 α 2 = f cm
0.2
if
f cm > 35MPa
2 * Ac 2 * 450 * 700 = = 274mm u 2 * (450 + 700)
Where: ■ ■ ■
RH relative humidity: RH = 50% h0: is the notional size of the member (in mm) u: column section perimeter
35 f cm = 38MPa ≥ 35MPa ⇒ α 1 = 38
594
0.7
= 0.944
and
35 38
α2 =
0.2
= 0.984
ADVANCE DESIGN VALIDATION GUIDE
ϕ RH
RH 50 1− 1− 100 * α * α = 1 + 100 * 0.944 * 0.984 = 1.70 = 1 + 1 2 0.1 * 3 h0 0.1 * 3 274
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.70 * 2.73 * 0.488 = 2.26 The effective creep coefficient calculation:
ϕ ef = ϕ (∞, t 0 ) *
M EQP M Ed
= 2.26 * 0.524 = 1.18 , according to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
1 + ϕef = 1 + 1.18 = 2.18 The necessity of buckling calculation (second order effect): For an isolated column, the slenderness limit check is done using the next formula:
λlim =
20 * A * B * C n
, according to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1)
Where:
n=
1.3575 N Ed = = 0.215 Ac * f cd 0.45 * 0.70 * 20
A=
1 1 = (1 + 0.2 * ϕ ef ) 1 + 0.2 *1.18 = 0.81
B = 1.1 if ω (reinforcement ratio) is not known, if it is, B = 1 + 2 * ω = 1 + 2 * In this case B
= 1.1 , because the reinforcement ratio in not yet known.
C = 0.70
rm is not known, if it is, C = 1.7 − rm
if
In this case C
λlim =
As * f yd Ac * f cd
= 0.70 , because the ratio between the first order moments is not known.
20 * 0.81 * 1.1 * 0.7 0.215
= 26.90
λ = 69.28 > λlim = 26.90 Therefore, the second order effects must be considered.
5.66.2.3 The eccentricity calculation and the corrected loads on ULS: Initial eccentricity:
e0 =
0.225 = 0.166m 1.3575
Additional eccentricity:
ei =
l0 9 = = 0.0225m 400 400
595
ADVANCE DESIGN VALIDATION GUIDE
First order eccentricity stresses correction:
N Ed = 1.3575MN e1 = e0 + ei = 0.166m + 0.0225m = 0.1885m = 18.85cm M Ed = N Ed * e1 = 1.3575MN * 0.1885m = 0.256 MNm Reinforcement calculation in the first order situation: To play the nominal rigidity method, a starting section frame is needed. For this, a concrete section considering only the first order effects will be used. Advance Design iterates as many time as necessary. The needed frames will be determined assuming a compound bending with compressive stress. All the results below were obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid of the tensioned steel.
h 0.45 M ua = M G 0 + N * (d − ) = 0.256 + 1.3575 * 0.40 − = 0.494MNm 2 2 Verification if the section is partially compressed:
µ BC = 0.8 *
µ cu =
h h 0.45 0.45 * (1 − 0.4 * ) = 0.8 * * (1 − 0.4 * ) = 0.495 d d 0.40 0.40
M ua 0.494 = = 0.220 bw * d ² * f cd 0.70 * 0.40² * 20
µ cu = 0.220 < 0.495 = µ BC , therefore the section is partially compressed.
596
ADVANCE DESIGN VALIDATION GUIDE
Calculations of steel reinforcement in pure bending:
µcu = 0.220
[
]
α u = 1.25 * 1 − (1 − 2 * 0.220) = 0.315 z c = d * (1 − 0.4 * α u ) = 0.40 * (1 − 0.4 * 0.315) = 0.350m A =
M ua 0.494 = = 32.46cm² z c * f yd 0.350 * 434.78
Calculations of steel reinforcement in combined bending: For the combined bending:
A = A'−
N 1.3575 = 32.46 *10 − 4 − = 1.24cm 2 434.78 f yd
The minimum reinforcement percentage:
As ,min =
0.10 * N Ed 0.10 *1.3575 = = 3.12cm² < 0.002 * Ac = 6.3cm 2 f yd 434.78
2 Therefore, the reinforcement will be 8HA10 representing a 6.28 cm section.
The second order effects calculation: The second order effect will be determined by applying the method of nominal rigidity: Calculation of nominal rigidity: It is estimated nominal rigidity of a post or frame member from the following formula:
EI = K c * Ecd * I c + K s * Es * I s Where:
E cd
: is the design value of the modulus of elasticity of concrete
I c : is the moment of inertia of concrete cross section E s : is the design value of the modulus of elasticity of reinforcement ( E s = 200000 MPa ) I s : is the second moment of area of reinforcement, about the center of area of the concrete K c : is a factor for effects of cracking, creep etc. K s : is a factor for contribution of reinforcement
597
ADVANCE DESIGN VALIDATION GUIDE
With:
Ecd =
Ecm 1.2
f cm = f ck + 8MPa = 38MPa f Ecm = 22000 * cm 10
Ecd = Ic =
0.3
38 = 22000 * 10
0.3
= 32837 MPa
Ecm 32837 = = 27364 MPa 1.2 1.2
b * h 3 0.70 * 0.45 3 = = 5.316 * 10 −3 m 4 12 12
(concrete only inertia)
Es = 200000 MPa
ρ=
As 6.28 * 10 −4 = = 0.002 Ac 0.70 * 0.45
0.002 ≤ ρ = k1 =
n=
f ck = 20
As < 0.01 Ac 30 = 1.22 MPa 20
N Ed 1.3575 = = 0.215 Ac . f cd 0.45 * 0.70 * 20
k2 = n *
69.28 λ = 0.215 * = 0.088 ≤ 0.20 170 170 2
2
As h 6.28.10 −4 0.45 Is = 2* * − c = 2* * − 0.05 = 1.92.10 −5 m 4 2 2 2 2
K s = 1 and K c =
k1 * k2 1.22 * 0.088 = = 0.049 1 + ϕef 1 + 1.18
Therefore:
EI = 0.049 * 27364 * 5.316 * 10−3 + 1 * 200000 * 1.92 * 10−5 = 10.97 MNm² Stresses correction: The total moment, including second order effects, is defined as a value plus the time of the first order:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed
M 0 Ed = 0.256 MNm
(moment of first order (ULS)) taking into account geometric imperfections
N Ed = 1.3575MN (normal force acting at ULS). 598
ADVANCE DESIGN VALIDATION GUIDE
β=
β=
π² c0
π² 8
and
c0 = 8
the moment is constant (no horizontal force at the top of post).
= 1.234
NB = π ² *
EI 10.82 =π²* = 1.32MN 2 l0 9²
The second order moment is:
M Ed 2
1.234 = 0.256 * 1 + = −11.18MNm 1.32 − 1 1.3575
An additional iteration must be made by increasing the ratio of reinforcement.
599
ADVANCE DESIGN VALIDATION GUIDE
Additional iteration: The iteration is made considering 8HA12 or As = 9.05 cm2 Therefore
ρ=
9.05 * 10 −4 = 0.0029 0.70 * 0.45
Followings are obtained:
Therefore:
N Ed 2 = 1.3575MN M Ed 2 = 2.48MNm
600
ADVANCE DESIGN VALIDATION GUIDE
Given the mentioned reports, there must be another reiteration. The reinforcement section must be increased to 8HA20 or As = 25.13 cm2 and ρ
= 0.008 .
The followings results are obtained:
Therefore:
N Ed 2 = 1.3575MN
M Ed 2 = 0.563MNm 2 Using the “Concrete Tools EC2”, combined bending analytical calculation, a 28.80 cm value is found. be another iteration.
⇒ There must
601
ADVANCE DESIGN VALIDATION GUIDE
The new considered reinforcement section is As = 30 cm2 and ρ
= 0.0095 :
Therefore:
N Ed 2 = 1.3575MN M Ed 2 = 0.499 MNm 2 Using the “Concrete Tools EC2”, combined bending analytical calculation, a 22.22 cm value is found reinforcement of the column is correct.
We therefore retain a theoretical section of 30 cm².
602
⇒ the
ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■
Linear element: S beam, 6 nodes, 1 linear element.
Theoretical reinforcement area(cm2) (reference value: 30 cm2 = 2 x 15 cm2)
5.66.2.4 Reference results Result name
Reference value 2
Az
5.66.3
Result description
15 cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Az
Az
16 cm²
0.0313 %
603
ADVANCE DESIGN VALIDATION GUIDE
5.67 EC2 / NF EN 199211/NA  France: Verifying a circular concrete column using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 39) Test ID: 5146 Test status: Passed
5.67.1
Description
Verifies the adequacy of a concrete (C25/30) column with circular cross section using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1). Simplified Method The column is considered connected to the ground by an articulated connection (all the translations are blocked). To the top part the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.
5.67.2
Background
Determines the longitudinal reinforcement of a circular cross section made from concrete C25/30, using the simplified method (with professional recommendations). This test was evaluated by the French control office SOCOTEC.
5.67.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure (G): 5000 kN axial force ► The selfweight is neglected Buckling length L0 = 5.00 m ►
■
Units Metric System Geometry Below are described the beam cross section characteristics: ■
604
Height: H = 5.00 m,
ADVANCE DESIGN VALIDATION GUIDE
■ ■
Radius: r = 0.40 m, Concrete cover: c = 5 cm (for the concrete cover of the bottom / top external fiber measured along the local z axis, Advance Design is having the following notation: ebz = 5 cm; ehz = 5 cm; while for the concrete cover measured along the local y axis the following notations are used: eby = 5 cm; ehy = 5 cm )
Materials properties Concrete class C25/30 and steel reinforcement S500B are used. The following characteristics are used in relation to these materials: ■ ■ ■
Reinforcement steel ductility: Class B Exposure class: XC1 Relative humidity RH=50%
■
Characteristic compressive cylinder strength of concrete at 28 days:
■
Characteristic yield strength of reinforcement: f yk = 500 MPa
Design value of concrete compressive strength:
f cd =
Design value of yield strength of reinforcement:
f yd =
f ck
γc
f ck = 25MPa
=
25 = 16.67 MPa 1.5
=
500 MPa = 434.78MPa 1.15
f yk
γs
Boundary conditions The column is considered connected to the ground by a hinged connection (all the translations are blocked) while at the top part, the translations along X and Y axis are blocked and the rotation along Z axis is also blocked. Loading The beam is subjected to the following load combinations: ► The ultimate limit state (ULS) combination is: Cmax = 1.35 x G The ultimate limit state (ULS) combination is: ►
NEd = 1.35 * 5000 kN = 6750 kN = 6.750 MN
605
ADVANCE DESIGN VALIDATION GUIDE
5.67.2.2 Reference results in calculating the concrete column Scope of the method:
According to Eurocode 2 EN 199211 (2004): Chapter 5.8.3.2(1) L0: buckling length L0 = L = 5 m According to Eurocode 2 EN 199211 (2004): Chapter 5.8.3.2(2)
i=
I , radius of giration of uncraked concrete section A
Where:
I=
I=
π *r4 4
, second moment of inertia for circular cross sections
π * (0.4m) 4 = 0.02011m 4 4
A = π * r 2 , cross section area A = π * (0.4m) 2 = 0.503m 2 Therefore:
i=
λ=
π *r4
I = A
r 2 r 0.40m = = = 0.20m 4 2 2
4 = π *r2
L0 L0 2 * L0 2 * 5.00m = = = = 25 r 0.40m r i 2
The method of professional rules can be applied If the following conditions are met:
■
λ < 120 20 < f ck < 50MPa
■
h > 0.15m
■
Reinforcement calculation:
λ = 25 ≤ 60 , therefore: α=
0.84 λ 1+ 52
2
=
0.84 25 1+ 52
2
= 0.68
k h = (0.70 + 0.5 * D) * (1 − 8 * ρ * δ )
section
Therefore:
606
for D < 0.60 m, if not
k h = 1.00 , where D is the diameter of the cross
ADVANCE DESIGN VALIDATION GUIDE
k h = 1.00
k s = 1.6 − 0.65 * f yd =
f yk 1.15
=
f yk 500
for f yk > 500 MPa and
λ > 30 ,if not k s = 1.00
500 = 434.78MPa 1.15
Therefore:
k s = 1.00
As =
N ed 1 1 6.750 * − π * r 2 * f cd = * − π * 0.4 2 *16.67 = 35.58cm² f yd k h * k s * α 434 . 78 1 . 00 * 1 . 00 * 0 . 68
Finite elements modeling ■ ■ ■
Linear element: S beam, 6 nodes, 1 linear element. 2
Theoretical reinforcement area(cm ) (reference value: 35.58 cm2 = 2 x 17.79 cm2)
5.67.2.3 Reference results Result name
Result description
Reference value
Ay
Reinforcement area [cm2]
17.79 cm2
5.67.3
Calculated results
Result name
Result description
Value
Error
Az
Az
17.4283 cm²
0.0000 %
607
ADVANCE DESIGN VALIDATION GUIDE
5.68 EC2/NF EN 199211/NA  France: Verifying a square concrete column subjected to a small rotation moment and significant compression force to the top with Nominal Curvature MethodBilinear stressstrain diagram(Class XC1)(SOCOTEC FranceTest 43) Test ID: 5211 Test status: Passed
5.68.1
Description
Verifies the adequacy of a square cross section column made of concrete C30/37 subjected to a small rotation moment and significant compression force to the top with Nominal Curvature Method  Bilinear stressstrain diagram (Class XC1). The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed. The purpose of this test is to determine the second order effects by applying the nominal curvature method, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection and free to the top part.
5.68.2
Background
The calculation is based on the nominal curvature 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 curvature method, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.68.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: 150 kN axial force ► 15 kMm rotation moment applied to the column top (about y axis) ► the selfweight is neglected Exploitation loadings: ► 100 kN axial force ► 7 kNm rotation moment applied to the column top (about y axis) ►
■
■ ■ ■ ■ ■ ■ ■
608
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Concrete cover 3 cm (on local y axis) and 5cm (on local z axis) Transversal reinforcement spacing a = 40cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.50 m, Width: b = 0.50 m, Length: L = 5.80 m, Concrete cover: c = 5 cm
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading The column is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: NEd = 1.35 * 150 + 1.50 * 100 = 352.50kN = 0.353MN MEd = 1.35 * 15 + 1.50 * 7 = 30.75kNm = 0.03075MNm
■
609
ADVANCE DESIGN VALIDATION GUIDE
5.68.2.2 Reference results in calculating the concrete column Geometric characteristics of the column: The column has a fixed connection on the bottom end and it is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * l = 11.60m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 * 11.60 = = 80.37 a 0.50
Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 )
creep coefficient
M EQP
serviceability firs order moment under quasipermanent load combination
M Ed
ULS first order moment (including the geometric imperfections)
First order eccentricity evaluation:
e1 = e0 + ei
ei =
l0 11.60 = = 0.029m 400 400
The first order moment provided by the quasipermanent loads:
e1 = e0 + ei =
M Eqp 0 N Eqp 0
+ ei =
15 + 0.30 * 7 + 0.029 = 0.124m 150 + 0.30 * 100
N Eqp1 = 150 + 0.30 * 100 = 180kN M Eqp1 = N Eqp1 * e1 = 180 * 0.124 = 22.32kNm = 0.02232 MNm The first order ULS moment is defined latter in this example (see chapter “The first order eccentricity: stresses correction”):
M Ed 1 = 0.041MNm The creep coefficient
ϕ (∞,t0 )
is defined as follows:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) =
610
16.8 16.8 = = 2.72 30 + 8 f cm
ADVANCE DESIGN VALIDATION GUIDE
β (t0 ) =
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
ϕ RH
RH 1− 100 * α * α = 1 + 1 2 0.1 * 3 h0
f cm
35 > 35MPa therefore: α1 = f cm
0.7
(for t0= 28 days concrete age).
35 = 38
0.7
35 = 0.944 and α 2 = f cm
0.2
35 = 38
0.2
= 0.984
50 1− 2 * Ac 2 * 500 * 500 100 * 0.944 * 0.984 = 1.72 h0 = = = 250mm ⇒ ϕ RH = 1 + u 2 * (500 + 500 ) 0.1 * 3 250
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 The effective creep coefficient calculation:
ϕ ef = ϕ (∞, t 0 )*
M EQP M Ed
= 2.28 *
0.02232 = 1.24 0.041
The second order effects; The buckling calculation For an isolated column, the slenderness limit check is done using the next formula:
λlim =
20 * A * B * C n According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1)
Where:
n=
0.353 N Ed = = 0.071 Ac * f cd 0.50² * 20
A=
1 1 = (1 + 0.2 * ϕef ) 1 + 0.2 *1.25 = 0.80
B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70
λlim =
because the reinforcement ratio in not yet known
because the ratio of the first order moment is not known
20 * 0.80 * 1.1 * 0.7 = 46.24 0.071
λ = 80.37 > λlim = 46.24
611
ADVANCE DESIGN VALIDATION GUIDE
Therefore, the second order effects cannot be neglected. Calculation of the eccentricities and solicitations corrected for ULS The stresses for the ULS load combination are: NEd= 1.35*150 + 1.50*100= 352.5kN = 0,3525 MN MEd= 1.35*15 + 1.50*7= 352.5kN= 0.03075MNm Therefore it must be determined: ■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:
e0 =
M Ed 0.03075 = = 0.087 m 0.353 N Ed
Additional eccentricity:
ei =
l0 11.6 = = 0.029m 400 400
The first order eccentricity: stresses correction For reinforced concrete section subjected to a combination of bending moment and compression, EN 199211 chapter §6.1 (4) recommends to consider the design value of bending moment at least M
= N Ed * e0 , where:
20mm 20mm 20mm = 20mm e0 = max h = max 500mm = max 16.7mm 30 30 In this case, the corrected forces used for the combined flexural calculations are:
N Ed = 0.353MN e1 = e0 + ei = 0.087 + 0.029 = 0.116m M Ed = e1 * N Ed = 0.116 * 0.353 = 0.041MNm Reinforcement calculation in the first order effects To apply the nominal curvature method, we need an initial reinforcement area to start from. For this, the concrete section will be sized considering only the first order effect. 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 combined bending with compressive stress. The determined solicitations were calculated from the center of gravity of the concrete section alone. Those stresses must be reduced to the centroid of tensioned steel:
612
ADVANCE DESIGN VALIDATION GUIDE
h 0.50 M ua = M G 0 + N * (d − ) = 0.041 + 0.353 * 0.45 − = 0.112MNm 2 2 Verification about the partially compressed section:
h d
h d
µ BC = 0.8 * * (1 − 0.4 * ) = 0.8 *
µcu =
0.50 0.50 * (1 − 0.4 * ) = 0.494 0.45 0.45
0.112 M ua = = 0.055 bw * d ² * f cd 0.50 * 0.45² * 20
µcu = 0.055 < µ BC = 0.494
therefore the section is partially compressed
The calculation for the tensioned steel in pure bending:
µcu = 0.055
[
]
α u = 1.25 * 1 − (1 − 2 * 0.055) = 0.071 z c = d * (1 − 0.4 * α u ) = 0.45 * (1 − 0.4 * 0.071) = 0.437 m A =
M ua 0.112 = = 5.89cm² z c * f yd 0.437 * 434.78
The calculation for the compressed steel in bending: For combined bending:
The minimum column percentage reinforcement must be considered:
Therefore a 5cm2 reinforcement area will be considered
613
ADVANCE DESIGN VALIDATION GUIDE
5.68.2.3 Nominal curvature method (second order effects) The curvature calculation: Considering a reinforcement of 5cm² (considered symmetric), one can determine the curvature from the following formula:
1 1 = K r * Kϕ * r r0 According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(1)
f yd
434.78 ε yd Es 1 = = = 200000 = 0.0107 m −1 r0 (0.45 * d ) (0.45 * d ) 0.45 * 0.45 Kr is the correction coefficient depending of the normal force:
nu − n ≤1 nu − nbal
Kr =
According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(3)
n=
ω=
0.353 N Ed = = 0.0706 Ac * f cd 0.50² * 20
As * f yd Ac * f cd
5 * 10−4 * 434.78 = = 0.0435 0.50² * 20
nu = 1 + ω = 1 + 0.0435 = 1.0435 nbal = 0,4
Kr =
1.0435 − 0.0706 = 1.51 1.0435 − 0.40
Condition: K r
≤ 1 , therefore we consider: K r = 1
Kϕ creep coefficient: K ϕ = 1 + β * ϕ ef ≥ 1 According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(4)
, Therefore:
1 1 = K r * Kϕ * = 1 * 1 * 0.0107 = 0.0107m −1 r r0 Calculation moment: The calculation of bending moment is defined by the following formula:
M Ed = M 0 Ed + M 2
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ADVANCE DESIGN VALIDATION GUIDE
Where:
M 0 Ed
M2
is moment of the first order including geometric imperfections.
is nominal moment of second order.
The 2nd order moment is calculated from the curvature:
M 2 = N Ed * e2 e2 =
1 l02 11.60² * = 0.0107 * = 0.1799m r c 8
Note: c = 8 according to chapter 5.8.8.2 (4) from EN 199211; because the moment is constant (no horizontal force at the top of column).
M 2 = N Ed * e2 = 0.353 * 0.1799 = 0.0635MNm M Ed = M 0 Ed + M 2 = 0.041 + 0.0635 = 0.1045MNm The reinforcement must be sized considering the demands of the second degree effects, as follows: NEd = 0.353 MN MEd = 0.1045 MNm Reinforcement calculation according the second order effects: The interaction diagram will be used. The input parameters in the diagram are:
µ=
M Ed 0.1045 = = 0.0418 2 b * h * f cd 0.5 * 0.5 2 * 20
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ADVANCE DESIGN VALIDATION GUIDE
According to the diagram, it will be obtained minimal percentage reinforcement:
As ,min =
0.10 * N Ed 0.10 * 0.353 = = 0.81cm² ≥ 0.002 * Ac = 5cm 2 f yd 434.78
Finite elements modeling ■ ■ ■
616
Linear element: S beam, 7 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area (cm2) (reference value: Au = 5cm2)
Theoretical value (cm2) (reference value: 5 cm2)
5.68.2.4 Reference results Result name
Reference value 2
Amin
5.68.3
Result description
5 cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Amin
Amin
5 cm²
0.0000 %
617
ADVANCE DESIGN VALIDATION GUIDE
5.69 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to eccentric loading  Bilinear stressstrain diagram (Class X0) (evaluated by SOCOTEC France ref. Test 44) Test ID: 5213 Test status: Passed
5.69.1
Description
Verifies a rectangular cross section beam made of concrete C30/37 subjected to eccentric loading  Bilinear stressstrain diagram (Class X0). The verification of the bending stresses at ultimate limit state is performed. Simple Bending Design for Ultimate Limit State During this test, the determination of stresses is made along with the determination of the longitudinal and transversal reinforcement.  Support at start point (x = 0) fixed connection  Support at end point (x = 5.00) fixed connection
5.69.2
Background
Simple Bending Design for Ultimate Limit State Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist eccentric loading. During this test, the calculation of stresses is made along with the calculation of the longitudinal and transversal reinforcement. This test was evaluated by the French control office SOCOTEC.
5.69.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
618
■
Loadings from the structure: G = 12 kN/m The dead load is neglected
■ ■
Exploitation loadings (category A): Q = 3kN/m, The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■ ■ ■ ■
Height: h = 0.30 m, Width: b = 0.18 m, Length: L = 5.00 m, 2 Section area: A = 0.054 m , Concrete cover: c=5cm 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 The boundary conditions are described below: ■
Outer: Support at start point (x=0) fixed connection, Support at end point (x = 5.00) fixed connection Inner: None.
► ►
■
619
ADVANCE DESIGN VALIDATION GUIDE
Reference results in calculating the concrete beam The moment is defined by the following formulas:
TA = P * e *
3 3 b = 0,5 * P * = P 5 10 a+b
TB = − P * e *
a 1 2 = −0,5 * P * = − P a+b 5 5
Note: the diagram of bending moment is the same as the shear force multiplied by the eccentricity.
The ULS load calculation:
Pu = 1.35 *12 + 1.5 * 3 = 20.70kN Before the point of application of torque, the torque is:
TEd =
3 3 * Pu = * 20,7 = 6.21KNm 10 10
After the point of application of torque, the torque is:
TEd = −
Pu − 20.7 = = −4.14 KNm 5 5
Result ADVANCE Design 2012  Moment: (in kNm) Reference value: 6.21 kNm and 4.14kNm
5.69.2.2 Calculation in pure bending Bending moment at ULS:
Pu = 1,35 * 12 + 1,5 * 3 = 20,7 KN
620
ADVANCE DESIGN VALIDATION GUIDE
M Ed =
6 * Pu 6 * 20,7 = = 24,84 KNm = 0,0248MNm 5 5
Simple bending design:
0,0248 = 0,110 0,18 * 0.25² * 20
µcu =
(
)
α u = 1,25 * 1 − 1 − 2 * 0,110 = 0,146 zc = 0.25 * (1 − 0,4 * 0,146) = 0,235m Au =
0,0248 = 2,43 * 10− 4 m² = 2,43cm² 0,235 * 434,78
5.69.2.3 Torsion reinforcement Torsion longitudinal reinforcement  Before application of the moment
ΣAl =
TEd * uk * cot θ 2 * Ak * f yd
According to: EC2 Part 1,1 EN 1992112004 Chapter 6.3.2(3)
uk = 2 * ((b − tef ) + (h − tef ))
Ak = (b − tef ) * (h − tef ) with:
A 0.18 * 0.30 tef = max 2 * c; = max 2 * 0.035; = max(0.07;0.05625) = 0.07m u 2 * (0.18 + 0.3)
uk = 2 * ((18 − 7) + (30 − 7)) = 68cm = 0.68m Ak = (18 − 7) * (30 − 7) = 253cm² = 0.0253m² ΣAl =
0.0062 * 0.68 = 1.92cm² 2 * 0.0253 * 434.78
Torsion longitudinal reinforcement  After application of the moment
ΣAl =
0.00414 * 0.68 = 1.28cm² 2 * 0.0253 * 434.78
Torsion transversal reinforcement  Before application of the moment
AswT TEd 0.0062 = = = 2.82cm² / ml sT 2 * Ak * f yd * cot θ 2 * 0.0253 * 434.78 Torsion transversal reinforcement  After application of the moment
AswT TEd 0.00414 = = = 1.88cm² / ml sT 2 * Ak * f yd * cot θ 2 * 0.0253 * 434.78
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ADVANCE DESIGN VALIDATION GUIDE
Concrete verification: Calculation of the maximum allowable stress under torsional moment:
TRd , max = 2 * v * α cw * f cd * Ak * t ef ,i * sin θ * cos θ According to: EC2 Part 1,1 EN 1992112004 Chapter 6.3.2(4)
f v = 0,6 * 1 − ck = 0.528 250 TRd , max = 2 * 0.528 * 20 * 0.0253 * 0.07 * 0.70 * 0.70 = 0.018MN Calculation of the maximum allowable stress under shear:
VRd , max = bw * z * v * f cd *
cot θ + cot α 1 + cot ²θ
According to: EC2 Part 1,1 EN 1992112004 Chapter 6.2.3(3)
z = zc = 0.235m
(from the design in simple bending)
Vertical reinforcement:
cot α = 1
30 v = 0,6 * 1 − = 0.528 250 VRd , max = bw * zc * v * f cd *
1 cot θ + cot α = 0.18 * 0.235 * 0.528 * 20 * = 0.223MN 2 1 + cot ²θ
Because of the combined shear/moment effect, it must be calculated:
TEd
TRd , max
+
VEd ≤ 1,0 VRd , max
According to: EC2 Part 1,1 EN 1992112004 Chapter 6.3.2(4)
0.0062 0.0124 + = 0.400 ≤ 1,0 0.018 0.223 Finite elements modeling ■ ■ ■
622
Linear element: S beam, 6 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
ULS load combinations(kNm) Longitudinal reinforcement (Al=1.92cm2 and Al=1.28cm2)
Transversal reinforcement (2.82cm2/ml and 1.88cm2/ml)
5.69.2.4 Reference results Result name
Result description
Reference value
Al,1
Longitudinal reinforcement for the first part [cm ]
1.92 cm2
Al,2
Longitudinal reinforcement for the second part [cm2]
1.28 cm2
Ator,y,1
Transversal reinforcement for the first part [cm2/ml]
2.82 cm2/ml
Ator,y,2
2
1.88 cm2/ml
5.69.3
2
Transversal reinforcement for the second part [cm /ml]
Calculated results
Result name
Result description
Value
Error
Al
Al,1
1.91945 cm²
0.0286 %
Al
Al,2
1.27964 cm²
0.0281 %
Ator,y / face
Ator,y,1
2.82273 cm²
0.0968 %
Ator,y / face
Ator,y,2
1.88182 cm²
0.0968 %
623
ADVANCE DESIGN VALIDATION GUIDE
5.70 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam supporting a balcony  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 45) Test ID: 5225 Test status: Passed
5.70.1
Description
Verifies the adequacy of a rectangular cross section beam made of concrete C25/30 supporting a balcony  Bilinear stressstrain diagram (Class XC1). Simple Bending Design for Ultimate Limit State Verifies the column resistance to rotation moment along its length. During this test, the determination of stresses is made along with the determination of the longitudinal and transversal reinforcement.
5.70.2
Background
Simple Bending Design for Ultimate Limit State Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist rotation moment along its length. During this test, the calculation of stresses is performed, along with the calculation of the longitudinal and transversal reinforcement. This test was evaluated by the French control office SOCOTEC.
5.70.2.1 Model description ■ ■ ■ ■
Reference: Guide de validation Eurocode 2 EN 1992112002; Analysis type: static linear (plane problem); Element type: linear. The geometric dimension of the beam are:
The following load cases and load combination are used: ■ ■ ■ ■ ■ ■ ■ ■ ■
624
Concrete type: C25/30 Reinforcement type: S500B Exposure class: XC1 Balcony load: 1kN/m2 The weight of the beam will be considered in calculation Concrete density: 25kN/m3 The beam is considered fixed at both ends Concrete cover: 40mm Beam length: 4.00m
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.75 m, Width: b = 0.25 m, Length: L = 4.00 m, 2 Section area: A = 0.1875 m , Concrete cover: c=4cm
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) fixed connection, Support at end point (x = 4.00) fixed connection Inner: None.
► ►
■
5.70.2.2 Reference results in calculating the concrete beam 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 (selfweight and distributed load) is determined by summing the resulting vertical loads along the eaves and 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 halfwidth). The results are displayed in the table below:
625
ADVANCE DESIGN VALIDATION GUIDE
Load calculation: From previously calculated results, the following stresses can be determined:
VEd
■
Shear:
■
Bending moment:
■
Torque:
M Ed
TEd
Shear and bending moment: One can determine the load at ULS taken over by the beam:
Pu = 1,35 * 13.44 + 1,5 * 2 = 21.14 KN / m For a beam fixed on both ends, the following values will be obtained: Maximum shear (ULS):
VEd =
Pu * l 21,14 * 4 = 42,3KN = 0,042 MN = 2 2
Bending moment at the supports:
M Ed =
Pu * l ² 21,14 * 4² = = 28,2 KNm = 0,028MNm 12 12
Maximum Moment at middle of span:
M Ed =
Pu * l ² 21,14 * 4² = = 14,1KNm = 0,014 MNm 24 24
Torsion moment: For a beam subjected to a torque constant:
mtu = 1,35 * 8,59 + 1,5 * 2,25 = 14,97 KNm / m = 0,015MNm / m TEd = mtu *
626
4 l = 0,015 * = 0,03MNm 2 2
ADVANCE DESIGN VALIDATION GUIDE
5.70.2.3 Bending rebar Span reinforcement (at bottom fiber)
µcu =
0,014 = 0,007 0,25 * 0.70² * 16.67
(
)
α u = 1,25 * 1 − 1 − 2 * 0,007 = 0,0088 zc = 0.70 * (1 − 0,4 * 0,0088) = 0,697 m Au =
0,014 = 4.62 * 10− 4 m² = 0.46cm² 0,697 * 434,78
Minimum reinforcement percentage verification:
As , min
Cracking matrix required (calculation hypothesis):
As , min
f ct , eff * bw * d 0.26 * = Max f yk 0.0013 * b * d w
f ct , eff = f ctm = 2.56 Mpa
f ct , eff 2.56 * bw * d = 0.26 * * 0.25 * 0.70 = 2.27cm² 0.26 * = Max = 2.27cm² f yk 500 0.0013 * bw * d = 0.0013 * 0.25 * 0.70 = 2.27cm²
Therefore, it retains 2.27 cm ². Reinforcement on supports (at top fiber)
µcu =
0,028 = 0,014 0,25 * 0.70² * 16.67
(
)
α u = 1,25 * 1 − 1 − 2 * 0,014 = 0,018 zc = 0.70 * (1 − 0,4 * 0,018) = 0,695m Au =
0,028 = 9.27 * 10− 4 m² = 0.93cm² 0,695 * 434,78
It also retains 2.27 cm ² (minimum percentage). Shear reinforcement
VEd = 0,042MN The transmission to the support is not direct; it is considered a connecting rod inclined by 45˚, therefore
cot θ = 1
Concrete rod verification:
VRd , max = bw * z * v * f cd *
cot θ + cot α 1 + cot ²θ According to: EC2 Part 1,1 EN 1992112004 Chapter 6.2.3(4)
z = zc = 0.695m (from the design in simple bending of support)
627
ADVANCE DESIGN VALIDATION GUIDE
Vertical frames:
25 v = 0,6 * 1 − = 0.54 250 VRd , max = bw * zc * v * f cd *
VRd , max =
cot θ + cot α bw * zc * v * f cd 0.25 * 0.695 * 0.54 * 16.67 = = = 0.78MN 1 + cot ²θ tgθ + cot θ 2
v1 * f cd * zu * bw tgθ + cot θ According to: EC2 Part 1,1 EN 1992112004 Chapter 6.2.3(3)
VEd = 0.042MN < VRd , max = 0.78MN Calculation of transverse reinforcement:
Asw VEd . * tgθ 0.042 = = = 1.39cm² / ml s zu * f yd 0.695 * 434.78
(over shear)
From the minimum reinforcement percentage:
Asw ≥ ρ w,min .bw . sin α s According to: EC2 Part 1,1 EN 1992112004 Chapter 9.2.2(5) With:
ρ w, min =
0,08 * f ck 0,08 * 25 = 0.0008 = 500 f yk
Asw ≥ 0.0008 * 0.25 = 2cm² / ml s Therefore:
Asw ≥ 2cm² / ml s Torsion calculation: Torsion moment was calculated before:
TEd = 0,03MNm
Torsional shear stress:
τ t ,i =
TEd 2 * tef ,i * Ak According to: EC2 Part 1,1 EN 1992112004 Chapter 6.3.2(1)
tef ,i
2 * c = 8cm A ( 25 * 75) = max = = 9.375cm = 9.375cm u 2(25 + 75)
Ak = (25 − 9.375) * (75 − 9.375) = 1025cm² = 0.1025m² 628
ADVANCE DESIGN VALIDATION GUIDE
TEd 0.03 = = 1.56Mpa 2 * t ef ,i * A k 2 * 0.09375 * 0.1025
τ t,i =
Concrete verification: Calculate the maximum allowable stress in the rods:
TRd ,max = 2 * v * α cw * f cd * Ak * tef ,i * sin θ * cos θ According to: EC2 Part 1,1 EN 1992112004 Chapter 6.3.2(4)
f v = 0,6 * 1 − ck = 0.54 250 TRd , max = 2 * 0.54 * 16.67 * 0.1025 * 0.09375 * 0.70 * 0.70 = 0.085MN Because of the combined share/moment effect, we must calculate:
TEd
TRd , max
+
VEd ≤ 1,0 VRd , max According to: EC2 Part 1,1 EN 1992112004 Chapter 6.3.2(4)
0.03 0.042 + = 0.399 ≤ 1,0 0.085 0.78 Torsion longitudinal reinforcement
ΣAl =
TEd * uk * cot θ 2 * Ak * f yd According to: EC2 Part 1,1 EN 1992112004 Chapter 6.3.2(3)
uk = 2 * [(b − tef ) + (h − tef )] = 2 * [(25 − 9.375) + (75 − 9.375)] = 162.5cm = 1.625m
ΣAl =
0.03 * 1.625 = 5.47cm² 2 * 0.1025 * 434.78
Torsion transversal reinforcement
0.03 AswT TEd = = = 3.36cm² / ml 2 * Ak * f yd * cot θ 2 * 0.1025 * 434.78 sT Therefore
AswT ≥ 3,36cm² / ml for each face. sT
Finite elements modeling ■ ■ ■
Linear element: S beam, 11 nodes, 1 linear element.
629
ADVANCE DESIGN VALIDATION GUIDE
ULS load combinations(kNm) Torsional moment (Ted=29.96kNm)
Longitudinal reinforcement (5.46cm2)
2 Transversal reinforcement (3.36cm /ml)
5.70.2.4 Reference results Result name
Result description
Mx
Torsional moment [kNm]
29.96 cm2
Al
Longitudinal reinforcement [cm ]
5.46 cm2
Ator,y
Transversal reinforcement [cm2/ml]
3.36 cm2/ml
5.70.3
630
Reference value 2
Calculated results
Result name
Result description
Value
Error
Mx
Mx
29.9565 kN*m
0.1450 %
Al
Al
5.4595 cm²
0.1920 %
Ator,y / face
Ator,y/face
3.35969 cm²
0.0092 %
ADVANCE DESIGN VALIDATION GUIDE
5.71 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column subjected to compression on the top – Method based on nominal stiffness  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 34) Test ID: 5114 Test status: Passed
5.71.1
Description
Verifies the adequacy of a rectangular cross section column made of concrete C30/37. The verification of the axial stresses applied on top, at ultimate limit state is performed. Method based on nominal stiffness  The purpose of this test is to determine the second order effects by applying the method 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.
5.71.2
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, and then calculate the frames by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.71.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: 300kN axial force ► The selfweight is neglected Exploitation loadings: ► 500kN axial force ►
■ ■
ψ 2 = 0,3
■ ■ ■ ■ ■ ■ ■
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Concrete cover 5cm Transversal reinforcement spacing a=30cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced
631
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.40 m, Width: b = 0.60 m, Length: L = 4.00 m, Concrete cover: c = 5 cm along the long section edge and 3cm along the short section edge
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading The beam is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*0.30+1.5*0.50=1.155MN NQP=1.35*0.30+0.30*0.50=0.450MN
■
632
ADVANCE DESIGN VALIDATION GUIDE
5.71.2.2 Reference results in calculating the concrete column Geometric characteristics of the column: The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * l = 2 * 4 = 8m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 * 8 = = 69.28 a 0.40
Effective creep coefficient calculation: The creep coefficient is calculated using the next formula:
ϕef = ϕ (∞, t0 ).
M EQP M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 )
creep coefficient
M EQP
serviceability first order moment under quasipermanent load combination
M Ed
ULS first order moment (including the geometric imperfections)
The ratio between moments becomes:
M 0 Eqp M 0 Ed
N eqp * e1
=
N ed *e1
The creep coefficient
=
N eqp N ed
=
ϕ (∞,t0 )
0.450 = 0.39 1.155
is defined as:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) = β (t0 ) =
ϕ RH
16.8 16.8 = = 2.73MPa f cm 30 + 8
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0= 28 days concrete age).
RH 1− 100 * α * α = 1 + 1 2 0.1 * 3 h0
RH = relative humidity; RH=50% Where α1
h0 =
= α 2 = 1 if f cm
35 ≤ 35MPa if not α1 = f cm
0.7
and
35 α 2 = f cm
0.2
2 * Ac 2 * 400 * 600 = = 240mm u 2 * (400 + 600 ) 633
ADVANCE DESIGN VALIDATION GUIDE
f cm = 38MPa > 35MPa , Therefore:
35 α1 = f cm
ϕ RH
0.7
35 = 38
0.7
35 = 0.944 and α 2 = f cm
0.2
35 = 38
0.2
= 0.984
50 RH 1− 1− 100 * α * α = 1 + 100 * 0.944 * 0.984 = 1.73 = 1 + 1 2 0.1 * 3 240 0.1 * 3 h0
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.73 * 2.73 * 0.488 = 2.30 The effective creep coefficient calculation:
ϕef = ϕ (∞, t0 ) *
M EQP M Ed
= 2.30 * 0.39 = 0.90 According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
The necessity of buckling calculation (second order effect): For an isolated column, the slenderness limit check is done using the next formula:
λlim =
20 * A * B * C n
According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1) Where:
n=
1.155 N Ed = = 0.241 Ac * f cd 0.40 * 0.60 * 20
A=
1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 * 0.90 = 0.85
B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70
λlim =
because the reinforcement ratio in not yet known
because the ratio of the first order moment is not known
20 * 0.85 * 1.1 * 0.7 = 26.66 0.112
λ = 69.28 > λlim = 26.66 Therefore, the second order effects must be considered.
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ADVANCE DESIGN VALIDATION GUIDE
5.71.2.3 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:
ei =
l0 8 = = 0.02m 400 400
First order eccentricity stresses correction:
20mm 20mm 20mm e0 = max h = max 400mm = max = 20mm 13.3mm 30 30 According to Eurocode 2 – EN 199211 2004; Chapter 6.1.(4) The corrected solicitations which are taken into account when calculating the column under combined bending effort and 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 concrete section considering only the first order effects. Advance Design iterates as many time as necessary. The needed frames will be determined assuming a compound bending with compressive stress. All the results above were obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid of the tensioned steel.
First order moment in the centroid of the tensioned reinforcement is:
h 0.40 M ua = M G 0 + N * (d − ) = 0.0231 + 1.155 * 0.35 − = 0.196MNm 2 2 Verification if the section is partially compressed:
µ BC = 0,8 *
µcu =
h h 0,40 0,40 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,496 d d 0,35 0,35
0.196 M ua = = 0.133 bw * d ² * f cd 0.60 * 0.35² * 20
µcu = 0.133 < 0.496 = µ BC
therefore the section is partially compressed.
635
ADVANCE DESIGN VALIDATION GUIDE
Calculations of steel reinforcement in pure bending:
µcu = 0.133
[
]
α u = 1,25 * 1 − (1 − 2 * 0,133) = 0,179 zc = d * (1 − 0,4 * α u ) = 0,35 * (1 − 0,4 * 0,179) = 0,325m A =
M ua 0,196 = = 13.87cm² zc * f yd 0,325 * 434,78
Calculations of steel reinforcement in combined bending: For the combined bending:
A = A'−
N 1.155 = 13.87 * 10− 4 − = −12.69cm 2 f yd 434.78
The minimum reinforcement percentage:
As , min =
0,10 * N Ed 0.10 * 1.155 = = 2.66cm² ≥ 0.002 * Ac = 4.80cm 2 f yd 434.78
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:
EI = K c * Ecd * I c + K s * Es * I s With:
Ecd =
Ecm 1.2
f cm = f ck + 8MPa = 38MPa f Ecm = 22000 * cm 10
Ecd = Ic =
0.3
38 = 22000 * 10
b * h3 0.60 * 0.403 = = 3,2.10− 3 m4 12 12
: Inertia of the steel rebars
ρ=
636
= 32837 MPa
Ecm 32837 = = 27364 MPa 1.2 1.2
Es = 200000 MPa Is
0.3
As 6,09.10 −4 = = 0.00254 Ac 0.60 * 0.40
(concrete only inertia)
ADVANCE DESIGN VALIDATION GUIDE
0.002 ≤ ρ =
30 f ck = = 1.22 Mpa 20 20
k1 =
n=
As < 0.01 Ac
N Ed 1.155 = = 0.241 Ac . f cd 0.40 * 0.60 * 20
k2 = n *
λ 69.28 = 0.241* = 0.098 ≤ 0.20 170 170 2
2
A h 6,09.10 −4 0.40 * Is = 2 * s * − c = 2 * − 0.05 = 1,37.10 −5 m4 2 2 2 2
K s = 1 and K c =
k1 * k2 1.22 * 0.098 = = 0.063 1 + ϕef 1 + 0.90
Therefore:
EI = 0.063 * 27364 * 3,2 * 10 −3 + 1 * 200000 * 1,37 * 10 −5 = 8.2566 MNm² Stresses correction: The total moment, including second order effects, is defined as a value plus the time of the first order:
M Ed
β = M 0 Ed * 1 + NB − 1 N Ed
M 0 Ed = 0.0231MNm (moment of first order (ULS) taking into account geometric imperfections) (normal force acting at ULS).
β=
β=
π² c0
π² 8
and
c0 = 8
the moment is constant (no horizontal force at the top of post).
= 1.234
NB = π ² *
EI 8.2566 =π²* = 1.27327 MN 2 l0 8²
The second order efforts are:
N Ed 2 = 1.155MN M Ed 2 = 0.3015MN .m The reinforcement calculations considering the second order effect: The reinforcement calculations for the combined flexural, under the second order effect, are done using the Graitec EC2 tools. The obtained section is null, therefore the minimum reinforcement area defined above is sufficient for the defined efforts.
637
ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■
Linear element: S beam, 5 nodes, 1 linear element.
Theoretical reinforcement area (cm2) and minimum reinforcement area (cm2) (reference value: 2.66cm2 and 4.8cm2)
Theoretical value (cm2) (reference value: 5.32 cm2)
5.71.2.4 Reference results Result name
2.66 cm2
Reinforcement area [cm ] 2
Amin
638
Reference value 2
Ay
5.71.3
Result description
4.80 cm2
Minimum reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Ay
Ay
2.66 cm²
0.0000 %
Amin
Amin
4.8 cm²
0.0000 %
ADVANCE DESIGN VALIDATION GUIDE
5.72 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete column using the simplified method – Professional rules  Bilinear stressstrain diagram (Class XC1) (evaluated by SOCOTEC France  ref. Test 38) Test ID: 5127 Test status: Passed
5.72.1
Description
Verifies a rectangular cross section concrete C25/30 column using the simplified method –Professional rules Bilinear stressstrain diagram (Class XC1). The column is considered connected to the ground by an articulated connection (all the translations are blocked). At the top part, the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.
5.72.2
Background
Simplified Method Verifies the adequacy of a rectangular cross section made from concrete C25/30. This test was evaluated by the French control office SOCOTEC.
5.72.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
■ ■ ■ ■ ■
Loadings from the structure: ► 800kN axial force ► The selfweight is neglected Concrete cover 5cm Concrete C25/30 Steel reinforcement S500B Relative humidity RH=50% Buckling length L0=6.50m
639
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.30 m, Width: b = 0.50 m, Length: L = 6.50 m, Concrete cover: c=5cm
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a articulated connection (all the translations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked. Loading The beam is subjected to the following load combinations: ■
Load combinations:
N ED = 1.35 * 800 = 1080kN 5.72.2.2 Reference results in calculating the concrete column Scope of the method:
λ=
L0 L L L L L * 12 6.5 * 12 = 0 = 0 = 0 = 0 = 0 = = 75.06 a i 0.3 a I a4 a2 12 A 12 12 2 a According to Eurocode 2 EN 199211 (2004) Chapter 5.8.3.2(1)
The method of professional rules can be applied as:
λ < 120 20 < f ck < 50 MPa h > 0.15m Reinforcement calculation:
60 < λ = 75.06 ≤ 120 1.3
32 α = λ
1.3
32 = 75.06
= 0.33 According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283
k h = (0.75 + 0.5 * h) * (1 − 6 * ρ * δ ) = 0.93 k s = 1.6 − 0.6 *
As =
640
1 f yd
f yk 500
= 1.6 − 0.6 *
500 =1 500
N ed 1 1.08 − b * h * f cd = − 0.3 * 0.5 *16.67 = 23.43cm² * * α k * k * 434 . 78 0 . 93 * 1 * 0 . 33 h s
ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■
Linear element: S beam, 8 nodes, 1 linear element.
Theoretical reinforcement area(cm2) (reference value: 23.43cm2=4*5.86cm2)
5.72.2.3 Reference results Result name
Reference value 2
Ay
5.72.3
Result description
5.85cm2
Reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Ay
Ay
6.91808 cm²
0.8908 %
641
ADVANCE DESIGN VALIDATION GUIDE
5.73 EC2/NF EN 199211/NAFrance: Verifying a square concrete column subjected to a significant rotation moment and small compression force to the top with Nominal Curvature MethodBilinear stressstrain diagram(Class XC1)(SOCOTEC FranceTest 42) Test ID: 5205 Test status: Passed
5.73.1
Description
Verifies a square cross section column made of concrete C30/37 subjected to a significant rotation moment and small compression force to the top with Nominal Curvature Method  Bilinear stressstrain diagram (Class XC1). The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed. The purpose of this test is to determine the second order effects by applying the nominal curvature method, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection and free to the top part.
5.73.2
Background
Verifies the adequacy of a square cross section made from concrete C30/37. The purpose of this test is to determine the second order effects by applying the nominal curvature method, and then calculate the member by considering a section symmetrically reinforced. This test was evaluated by the French control office SOCOTEC.
5.73.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure (G): 15 kN axial force ► 150 kMm rotation moment applied to the column top ► The selfweight is neglected ►
■
Exploitation loadings (Q) from B exploitation category (ψ 2
7 kN axial force ► 100 kNm bending moment applied to the column top The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Distance between the reinforcement legs: a = 40 cm = 0.40 m The column is considered isolated and braced ►
■ ■ ■
642
= 0.3 ):
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■
Height and width are equal (square shape): b = h = 0.50 m, Length: l = 5.80 m, Concrete cover: c = 5 cm (for the concrete cover of the bottom / top external fiber measured along the local z axis, Advance Design is having the following notation: ebz = 5 cm; ehz = 5 cm; while for the concrete cover measured along the local y axis the following notations are used: eby = 3 cm; ehy = 3 cm)
Materials properties The following characteristics are used in relation to concrete class C30/37 and steel reinforcement S500B materials: ■ ■ ■
Reinforcement steel ductility: Class B Exposure class: XC1 Concrete density: 25kN/m3
■
Characteristic compressive cylinder strength of concrete at 28 days:
■
Characteristic yield strength of reinforcement: f yk = 500 MPa
■
Design value of concrete compressive strength:
f cd =
■
Design value of yield strength of reinforcement:
f yd =
■
Design value of modulus of elasticity of reinforcing steel: Es = 200000 MPa
f ck
γc
f yk
γs
f ck = 30MPa
=
30 = 20MPa 1.5
=
500 MPa = 434.78MPa 1.15
Boundary conditions The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading The beam is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: ► ►
■
e0 =
NEd = 1.35 * 15kN + 1.50 * 7kN = 30.75 kN = 0.03075 MN MEd = 1.35 * 150kNm + 1.50 * 100kNm = 352.50 kNm = 0.3525 MNm
M Ed 0.352 MNm = = 11.45m N Ed 0.03075MN 643
ADVANCE DESIGN VALIDATION GUIDE
5.73.2.2 Reference results in calculating the concrete column Geometric characteristics of the column The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
l0 = 2 * l = 11.60m According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:
λ=
2 3 * l0 2 3 *11.60m = = 80.37 h 0.50m
Effective creep coefficient calculation The creep coefficient is calculated using the next formula:
ϕ ef = ϕ (∞, t0 ).
M Eqp M Ed According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
Where:
ϕ (∞,t0 ) M Eqp
M Ed
creep coefficient
serviceability first order moment under quasipermanent load combination ULS first order moment (including the geometric imperfections)
First order eccentricity evaluation:
e1 = e0 + ei ei = 0.029m Eccentricity e1 is determined considering the first order moment provided by the quasipermanent loads:
e1 = e0 + ei =
M Eqp 0 N Eqp 0
+ ei =
150kNm + 0.30 *100kNm + 0.029m ≅ 10.56m 15kN + 0.30 * 7 kN
N Eqp1 = 15kN + 0.30 * 7 kN = 17.10kN M Eqp1 = N Eqp1 * e1 = 17.10kN *10.56m = 180.58kNm = 0.181MNm The first order ULS moment was defined at the beginning, in the “Loading” chapter:
M Ed = 0.3525MNm The creep coefficient
ϕ (∞,t0 )
is defined as follows:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )
β ( f cm ) = β (t0 ) =
644
16.8 16.8 = = 2.72 30 + 8 f cm
1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t0
(for t0= 28 days concrete age).
ADVANCE DESIGN VALIDATION GUIDE
ϕ RH
RH 1− 100 * α * α = 1 + 1 2 0.1 * 3 h0
f cm
35 > 35MPa therefore: α1 = f cm
0.7
35 = 38
0.7
35 = 0.944 and α 2 = f cm
0.2
35 = 38
0.2
= 0.984
h0 is the notional size of the member (in mm):
h0 =
2 * Ac 2 * 0.5m * 0.5m = = 0.250m = 250mm ⇒ ϕ RH u 2 * (0.5m + 0.5m )
50 1− 100 * 0.984 = 1.72 * 0 . 944 = 1 + 0.1* 3 250
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.72 * 2.72 * 0.488 = 2.28 After a numerical replacement of each term, effective creep coefficient becomes:
ϕ ef = ϕ (∞, t0 ) *
M EQP M Ed
= 2.28 *
0.181MNm = 1.17 0.352 MNm According to Eurocode 2 – EN 199211 2004; Chapter 5.8.4(2)
The second order effects; The buckling calculation For an isolated column, the slenderness limit check is done using the next formula:
λlim =
20 * A * B * C n According to Eurocode 2 – EN 199211 2004; Chapter 5.8.3.1(1)
Where:
n=
0.031 N Ed = = 0.0062 Ac * f cd 0.50² * 20
A=
1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.17 = 0.81
B = 1 + 2 * ω = 1.1 C = 1.7 − rm = 0.70
λlim =
because the reinforcement ratio in not yet known because the ratio of the first order moment is not known
20 * 0.81 * 1.1 * 0.7 = 158.42 0.0062
λ = 80.37 < λlim = 158.42 Therefore, the second order effects can be neglected.
645
ADVANCE DESIGN VALIDATION GUIDE
Calculation of the eccentricities and solicitations corrected for ULS The solicitations for the ULS load combination are: ■ NEd = 0.03075 MN ■ MEd = 0.3525 MNm Therefore it must be determined: ■ The eccentricity of the first order at ULS, due to the applied stresses ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:
e0 =
M Ed 0.3525MNm = = 11.46m 0.03075MN N Ed
Additional eccentricity:
ei =
l0 11.6m = = 0.029m 400 400
The corrected forces, which are used for the combined flexural calculations, are the following:
N Ed = 0.03075MN e1 = e0 + ei = 11.46m + 0.029m ≅ 11.49m M Ed = e1 * N Ed = 11.49m * 0.03075MN = 0.353MNm According to clause 6.1 (4) from EN 199211, for sections subjected to combined bending and compression efforts, the design bending moment value should be at least M = NEd * e0, where
h 500 mm e0 = max 20 mm; = max 20 mm; = max( 20 mm; 16.7 mm ) = 20 mm 30 30 Corrected bending moment is bigger than this value, so clause 6.1 (4) is fulfilled. Reinforcement calculation in the first order situation The theoretical reinforcement will be determined by using the following diagram:
646
ADVANCE DESIGN VALIDATION GUIDE
The input parameters of the diagram are:
µ=
M Ed 0.353MNm = = 0.141 2 b * h * f cd 0.50m * 0.50 2 m 2 * 20 MPa
ν=
N Ed 0.03075MN = = 0.00615 b * h * f cd 0.5m * 0.5m * 20 MPa
Therefore:
ω = 0.35 The total reinforcement area will be:
∑ As =
ω * 0.50 2 m 2 * 20MPa = 40.25cm 2 434.78MPa
which means 20.125cm2 per face.
The nominal curvature method (second order effect) In the following, the steps necessary to calculate the curvature are shown. Considering a reinforcement of 40.25cm² (considered symmetric), one can determine the curvature from the following formula:
1 1 = K r * Kϕ * r0 r According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(1)
f yd
434.78MPa Es 1 = = 200000 MPa = 0.0107 m −1 = r0 (0.45 * d ) (0.45 * d ) 0.45 * 0.45m
ε yd
Kr is the correction coefficient depending of the normal force:
Kr =
nu − n ≤1 nu − nbal
According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(3)
N Ed 0.03075MN = 0.00615 = Ac * f cd 0.50² m 2 * 20 MPa
n=
ω=
As * f yd Ac * f cd
=
40.25 *10 −4 m 2 * 434.78MPa = 0.350 0.50² m 2 * 20 MPa
nu = 1 + ω = 1 + 0.350 = 1.350 nbal = 0.4
Kr =
1.350 − 0.00615 = 1.41 1.350 − 0.40
Condition: K r
Kϕ
≤ 1 , therefore it will be considered: K r = 1
creep coefficient:
Kϕ = 1 + β * ϕ ef ≥ 1
According to Eurocode 2 – EN 199211(2004): Chapter 5.8.8.3(4)
647
ADVANCE DESIGN VALIDATION GUIDE
therefore
Calculation of design bending moment The calculation of design bending moment is estimated from the formula:
M Ed = M 0 Ed + M 2 Where:
M 0 Ed
M2
is moment of the first order including geometric imperfections;
is nominal moment of second order.
The 2nd order moment is given by the curvature:
M 2 = N Ed * e2 1 l2 11.60² e2 = * 0 = 0.0107 * = 0.1799m r c 8 Note: c = 8 according to 5.8.8.2 (4) from EN 199211; because the moment is constant (no horizontal force at the top of column).
M 2 = N Ed * e2 = 0.03075MN * 0.1799m = 0.00553MNm M Ed = M 0 Ed + M 2 = 0.353MNm + 0.00553MNm = 0.35853MNm The reinforcement must be sized considering the demands of the second order effects, as follows:
N Ed = 0.03075MN M Ed = 0.35853MNm Reinforcement calculation according to the second order effects The interaction diagram will be used. The input parameters in the diagram are:
µ=
648
0.35853 M Ed = 0.143 = 2 b * h * f cd 0.5 * 0.52 * 20
ADVANCE DESIGN VALIDATION GUIDE
From the diagram ω
∑ As =
= 0.35 will be obtained, which gives:
ω * b * h * f cd 0.35 * 0.50² m 2 * 20 MPa = = 0.004025m 2 = 40.25cm² f yd 434.78MPa
Meaning a 20.125 cm2 per side (which confirms the initial section)
649
ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element.
Theoretical longitudinal reinforcement area (cm2) (reference value: 40.25 cm2 = 2 * 20.125 cm2)
5.73.2.3 Reference results Result name
Result description
Reference value
Az
Longitudinal reinforcement area [cm2]
20.125 cm2
5.73.3
650
Calculated results
Result name
Result description
Value
Error
Az
Az
19.32 cm²
4.0000 %
ADVANCE DESIGN VALIDATION GUIDE
5.74 EC2 / NF EN 199211/NA  France: Verifying a square concrete beam subjected to a normal force of traction  Bilinear stressstrain diagram (Class X0) (evaluated by SOCOTEC France ref. Test 46 II) Test ID: 5231 Test status: Passed
5.74.1
Description
Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction  Bilinear stressstrain diagram (Class X0). Tie sizing Bilinear stressstrain 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 The boundary conditions are described below:  Support at start point (x=0) fixed connection  Support at end point (x = 5.00) translation along the Z axis is blocked
5.74.2
Background
Determines the reinforcement of a rectangular cross section made from concrete C25/30, subjected to a normal force of traction, analyzed with both bilinear stressstrain and inclined stressstrain laws. This test was evaluated by the French control office SOCOTEC.
5.74.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: 233.33 kN axial force ► The selfweight is neglected Exploitation loadings: ► 56.67 kN axial force ►
■
651
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.15 m, Width: b = 0.15 m, Length: L = 5.00 m, 2 Section area: A = 0.00225 m , Concrete cover: c = 3 cm (for the concrete cover of the bottom / top external fiber measured along the local z axis, Advance Design is having the following notation: ebz = 3 cm; ehz = 3 cm; while for the concrete cover measured along the local y axis the following notations are used: eby = 3 cm; ehy = 3 cm )
Materials Properties C20/25 Reinforcement S400, Class: B Characteristic compressive cylinder strength of concrete at 28 days: fck = 20 MPa Value of concrete cylinder compressive strength: fcm = fck + 8(MPa) = 28MPa Characteristic yield strength of reinforcement: fyk = 400 MPa Mean value of axial tensile strength of concrete:
2 3
f ctm = 0.30 * ( f ck ) = 2.21MPa
Design value of concrete compressive strength:
f cd =
Design value of yield strength of reinforcement:
f yd =
f ck
γc
f yk
γs
=
20 MPa = 13.33MPa 1 .5
=
400 MPa = 347.82 MPa 1.15
Boundary conditions The boundary conditions are described below: ■
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.
► ►
■
Loading The beam is subjected to the following load combinations: ►
652
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q
ADVANCE DESIGN VALIDATION GUIDE
The ultimate limit state (ULS) combination is: NEd = 1.35 * 233.3 kN + 1.5 * 56.67 kN = 400 kN = 0.400 MN
5.74.2.2 Reference results in calculating the concrete beam There will be two successive calculations, considering a bilinear stressstrain diagram constitutive law and then an inclined stressstrain diagram constitutive law. Calculations according a bilinear stressstrain diagram
As ,U ≥
N Ed
σ s ,U
=
0.400 = 11.50 * 10 − 4 m² = 11.50cm² 400 1.15
It will be used a 4HA20
⇒
A = 12.57cm2
Calculations according a inclined stressstrain diagram
S 400 B ⇒ σ s ,U = 373MPa As ,U ≥
N Ed
σ s ,U
=
0.400 = 10.72 * 10 − 4 m² = 10.72cm² 373
It can be seen that the gain is not negligible (about 7%). Checking the condition of nonfragility:
As ≥ Ac *
f ctm f yk
Ac = 0.15m * 0.15m = 0.0225m² As ≥ Ac *
f ctm 2.21 = 0.0225 * = 1.24cm² f yk 400
Finite elements modeling ■ ■ ■
Linear element: S beam, 6 nodes, 1 linear element.
ULS load combinations (kNm) In case of using the bilinear stressstrain diagram, the reinforcement will result: (Az = 11.50 cm2 = 2*5.75 cm2)
653
ADVANCE DESIGN VALIDATION GUIDE
In case of using the inclined stressstrain diagram, the reinforcement will result: (Az = 10.74 cm2 = 2*5.36 cm2)
5.74.2.3 Reference results Result name
Result description
Reference value
Az,1
Longitudinal reinforcement obtained using the bilinear stressstrain diagram [cm2] Longitudinal reinforcement obtained using the inclined stressstrain diagram [cm2]
5.75 cm2
Az,2
5.74.3
654
5.36 cm2
Calculated results
Result name
Result description
Value
Error
Az
Azi
5.75001 cm²
0.0000 %
ADVANCE DESIGN VALIDATION GUIDE
5.75 NTC 2008  Italy: Longitudinal reinforcement on a simply supported beam Test ID: 6278 Test status: Passed
5.75.1
Description
Verifies the beam''s longitudinal reinforcement area on the support with linear load applied along the linear element
5.75.2
Background
5.75.2.1 Model description ■ ■
Analysis type: static linear (plane problem); Element type: linear.
Units Metric Geometry ■ ■
Length: L = 13.50 m, Area: A = 3000 cm²
Materials properties
655
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■ ■
Hinge at end x = 4.50, 9.00, 13.50 m Restrained in translation on x,y axes and restrained in rotation on x axis x = 0.00 m.
Loading Uniformly distributed load of G1k = 33.00 kN/m Uniformly distributed load of G2k = 25.74 kN/m Uniformly distributed load of Q1k = 13.20 kN/m
5.75.2.2 Reference results Calculation method used to obtain the reference solution
5.75.2.3 Theoretical results Result name Longitudinal reinforcement
5.75.3
656
Result description
Reference value
Verification of the longitudinal reinforcement area on the support
25.23 cm2
Calculated results
Result name
Result description
Value
Error
Az
Theoretical reinforcement on the support
0 cm²
3.9635 %
ADVANCE DESIGN VALIDATION GUIDE
5.76 EC2 / CSN EN 199211  Czech Republic: Main reinforcement design for simple supported beam Test ID: 6261 Test status: Passed
5.76.1
Description
CSN EN199211: Required tension reinforcement design for simple supported rectangular beam
5.76.2
Background
Design of main reinforcement for simple supported beam Verifies values of minimal and theoretical reinforcement areas for rectangular beam made of concrete C20/25, reinforcement B500B. Design is considered for Ultimate Limit State with uniform distribution of stresses in compressed concrete and unlimited reshaping of tensioned reinforcement. Self weight is taken into account. Bilinear stressstrain diagram for reinforcement.
5.76.2.1 Model description ■ Reference: Manually calculated example ■ Analysis type: static linear (plane problem) ■ Element type: linear Load cases and load combinations: ■ ■ ■
Dead load: g = 10 kN/m, γG = 1.35, Live load: q = 15 kN/m, γQ = 1.5 CULS = 1.35*(g+g´) + 1.5*q
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■
Height: h = 450 mm, Width: b = 250 mm, Length: L = 6000 mm, 3 2 Section area: A = 112.5*10 mm , Reinforcement center of gravity position: d´ = 38 mm, Effective height: d = 412 mm
657
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Concrete: C20/25 ■ Weight: 2500 kg/m3 ■ γc = 1.15 Reinforcement: B500B ■ ■ ■
fyk = 500 MPa γs = 1.15 Es = 200 GPa
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (Z=0), hinged connection, Support at end point (Z = 6.0) hinged connection restrained to rotate around Z Inner: none
► ►
■
Loading The beam is subjected to following load combination: ■
Ultimate Limit State: qEd = 1.35*(g + g´) + 1.5*q = 1.35*(10 + 0,25*0,45*2500*9.81) + 1.5*15 = 39.72 kN/m
5.76.2.2 Reinforcement design Material characteristics ■
Concrete C20/25: ►
►
fcd = αcc ∗
fyk γc
20
= 1.0 ∗ 1.5 = 13.3 MPa
According to: CSN EN199211 Chapter 3.1.6(1)P, Formula (3.15)
fctm = 2.2 MPa According to: CSN EN199211 Chapter 3.1.3, Table 3.1
■
Reinforcement B500B: fyk = 500 MPa
► ►
fyd =
fyk γs
500
= 1.15 = 435 MPa
According to: CSN EN199211 Chapter 3.2.7(2), Picture 3.8
Design moment for Ultimate Limit State
Reinforcement design ■
As , req
1 1 ∗ 𝑞 ∗ 𝐿2 = ∗ 39.72 ∗ 62 = 178.74 𝑘𝑁𝑚 8 𝐸𝑑 8
Theoretical reinforcement
b ∗ d ∗ f cd 2 ∗ M Ed = ∗ 1 − 1 − f yd b ∗ d 2 ∗ f cd ■
𝑀𝐸𝑑 =
0.25 ∗ 0.412 ∗ 13.3 ∗ 10 6 = 435 ∗ 10 6
2 ∗ 178.74 ∗ 103 ∗ 1− 1− 0.25 ∗ 0.412 2 ∗ 13.3 ∗ 103
Minimal reinforcement 𝑓𝑐𝑡𝑚 2.2 0.26 ∗ ∗ 𝑏𝑓 ∗ 𝑑 = 0.26 ∗ ∗ 0.25 ∗ 0.412 = 119 ∗ 10−6 𝑚2 𝑓𝑦𝑘 500 𝐴𝑆𝑚𝑖𝑛 = 𝑚𝑎𝑥 � 0.0013 ∗ 𝑏𝑓 ∗ 𝑑 = 0.0013 ∗ 0.25 ∗ 0.412 = 134 ∗ 10−6 𝑚2
658
2 = 12.42cm
ADVANCE DESIGN VALIDATION GUIDE
ASmin = 1.34 cm2 According to: CSN EN199211 Chapter 9.2.1.1(1), Formula (9.1N) Finite elements modeling ■ ■ ■
Linear element: beam, 7 nodes, 1 linear element.
FEM Results Bending moment for Ultimate Limit State (reference value 178.74 kNm)
Concrete design results Theoretical reinforcement area (reference value 12.42 cm2)
2 Minimal reinforcement area (reference value 1.34 cm )
5.76.2.3 Reference results Result name
Result description
My
Bending moment for Ultimate Limit State [kNm]
Reference value 178.74 kNm
2
Az
Theoretical reinforcement area [cm ]
12.42 cm2
Amin
Minimal reinforcement area [cm2]
1.34 cm2
5.76.3
Calculated results
Result name
Result description
Value
Error
My
Design moment
178.756 kN*m
0.0090 %
Az
RC required
12.4214 cm²
0.0113 %
Amin
RC minimal
1.339 cm²
0.0746 %
659
ADVANCE DESIGN VALIDATION GUIDE
5.77 EC2 / NF EN199211  France: Verifying a T concrete section  Inclined stressstrain diagram (Class XC3) Test ID: 6274 Test status: Passed
5.77.1
Description
The purpose of this test is to calculate the longitudinal reinforcement of a TBeam with inclinated stressstrain diagram with a XC3 class concrete.
5.77.2
Background
Verifies the reinforcement area on a T concrete section (C35/45)..
5.77.2.1 Model description ■ Reference: Calcul des structures en béton – JeanMarie Paillé; ■ Analysis type: static linear (plane workspace); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: G = 53 kN/m + self weight
■ ■
Exploitation loadings (ψ0=0.8, ψ1=0.8, ψ2=0.5): Q = 80 kN/m The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q
Units Metric System Geometry Below are described the beam cross section characteristics:
660
■ ■ ■ ■
Height: h = 1.25 m, Width: b = 0.55 m, Length: L = 14.00 m, Concrete cover: cnom=4 cm
■
Reinforcement S500, Class: B, ss=500 MPa – Inclinated stressstrain diagram
ADVANCE DESIGN VALIDATION GUIDE
■ ■
Fck=355 MPa The load combinations will produce the following efforts: MEd= 214.5 kN.m
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) hinged connection, Support at end point (x=14) hinged connection. Inner: None.
► ►
■
5.77.2.2 Reference results in calculating the concrete column Moment values MEd =pEd x L²/8 Combination type
Value (kN.m)
ULS combination
5 255
SLS CQ combination
2 695
Bottom reinforcement calculation σ= 454 MPa,
As =
M Ed (1 − 0.4α )d × σ
=
5.255 = 110 cm² (1 − 0.4 × 0.122)×1.10 × 454
Finite elements modeling ■ ■ ■
Linear element: S beam, 29 nodes, 1 linear element.
5.77.2.3 Reference results Result name
Result description
Reference value
As
Longitudinal reinforcement obtained using the inclined stressstrain diagram [cm2]
110 cm2
5.77.3
Calculated results
Result name
Result description
Value
Error
My
ULS Bending moment
5261.45 kN*m
0.1227 %
My
SLS CQ Bending moment
2699.59 kN*m
0.1703 %
Az
Longitudinal bottom reinforcement
107.285 cm²
2.4682 %
661
ADVANCE DESIGN VALIDATION GUIDE
5.78 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to Pivot B efforts – Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 4 II) Test ID: 5893 Test status: Passed
5.78.1
Description
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement. The purpose of this test is to verify the software results for Pivot B efforts. For these tests, the constitutive law for reinforcement steel, on the inclined stressstrain diagram is applied. The objective is to verify the longitudinal reinforcement corresponding to Class B reinforcement steel ductility.
5.78.2
Background
This test was evaluated by the French control office SOCOTEC. This test performs the verification of the α 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:
xu ε cu 2 ε cu 2 .d =α = ⇒ x u = α .d = d ε ud + ε cu 2 ε ud + ε cu 2 The limit for α depends of the ductility class: ■
For a Class A steel: ε ud = 22.50 °
■
For a Class B steel: ε ud = 45 °
■
For a Class C steel: ε ud = 67.50 °
°°
°°
⇒ α u = 0.1346
⇒ α u = 0.072
°°
⇒ α u = 0.049
The purpose of this test is to verify the software results for Pivot A efforts. For these tests, it will be used the constitutive law for reinforcement steel, on the inclined stressstrain diagram.
S 500 A ⇒ σ su = 432,71 + 952,38.ε su ≤ 454 MPa S 500 B ⇒ σ su = 432,71 + 727,27.ε su ≤ 466 MPa S 500C ⇒ σ su = 432,71 + 895,52.ε su ≤ 493MPa The Pivot efforts types are described below: ■ ■ ■
662
Pivot A: Simple traction and simple bending or combined Pivot B: Simple or combined bending Pivot C: Combined bending with compression and simple compression
ADVANCE DESIGN VALIDATION GUIDE
5.78.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■
Loadings from the structure: G = 25 kN/m+ dead load, Exploitation loadings (category A): Q = 50kN/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 ■ There will be considered a Class B reinforcement steel ■ The calculation will be made considering inclined stressstrain diagram The objective is to verify: ■ ■ ■
The stresses results The longitudinal reinforcement corresponding to Class B reinforcement steel ductility The minimum reinforcement percentage
663
ADVANCE DESIGN VALIDATION GUIDE
Simply supported beam
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■
Height: h = 0.90 m, Width: b = 0.50 m, Length: L = 5.80 m, 2 Section area: A = 0.45 m , Concrete cover: c = 4.00 cm Effective height: d = h  (0.6 * h + ebz)=0.806 m; d’ = ebz = 0.040 m
Materials properties Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■
Exposure class XC1 3 Concrete density: 25kN/m There will be considered a Class B reinforcement steel ductility The calculation will be made considering inclined stressstrain diagram Cracking calculation required
■
Concrete C25/30: fcd =
fck 25 = = 16,67MPa γc 1,5
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■
2/3 = 0.30 * 25 2 / 3 = 2.56MPa fctm = 0.30 * fck
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1 ■
Steel S500 : f yd =
f yk γs
=
500 = 434,78MPa 1,15 According to: EC2 Part 1,1 EN 1992112002 Chapter 3.2
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.8) restrained in translation along Y and Z, and restrained rotation along X. Inner: None. ►
■
664
ADVANCE DESIGN VALIDATION GUIDE
Loading The beam is subjected to the following load combinations: 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
■
Load calculations:
M Ed =
123.94 * 5.80² = 521.16kNm 8
M Ecq =
86.25 * 5.80² = 362.68kNm 8
5.78.2.2 Reference results in calculating the concrete beam reduced moment limit For S500B reinforcement steel, we have µ lu = 0.372 (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. The calculation of the reinforcement is detailed below: ■ ■
Effective height: d=h(0.6*h+ebz)=0.806 m Calculation of reduced moment:
µ cu = ■
M Ed 0.52116 * 103 Nm = = 0,096 bw * d ² * f cd 0,50m * 0.806² m 2 * 16,67 MPa
The α value:
A design in simple bending is performed and it will be considered a design stress of concrete equal to
f cd .
zc = d * (1 − 0,4 * α cu ) = 0.806 * (1 − 0.4 * 0.127) = 0.765m ■
Tensioned reinforcement steel elongation :
■
Reinforcement steel stresses :
ε su =
1 − αu
αu
* ε cu 2 =
1 − 0,127 * 3.5 = 24.1 ° °° 0,127
σ su = 432,71 + 727,27 * ε su ≤ 466MPa σ su = 432,71 + 727,27 * 0,0241 = 450.25Mpa ≤ 466MPa Au =
M Ed 0.52116 = = 15.13cm² z c * f yd 0.765 * 450.25
Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element.
665
ADVANCE DESIGN VALIDATION GUIDE
ULS and SLS load combinations (kNm) Simply supported beam subjected to bending
ULS (reference value: 521.16kNm)
SLS (reference value: 362.68kNm)
Theoretical reinforcement area (cm2) 2 For Class B reinforcement steel ductility (reference value: A=15.13cm )
5.78.2.3 Reference results Result name
Result description
Reference value
My,ULS
My corresponding to 101 combination (ULS) [kNm]
521.16 kNm
My,SLS
My corresponding to 102 combination (SLS) [kNm]
362.68 kNm
2
Az
5.78.3
666
15.13 cm2
Theoretical reinforcement area (Class B) [cm ]
Calculated results
Result name
Result description
Value
Error
My
My corresponding to 101 combination (ULS)
521.125 kN*m
0.0000 %
My
My corresponding to 102 combination (SLS)
362.657 kN*m
0.0000 %
Az
Theoretical reinforcement area (Class B)
15.1105 cm²
0.0000 %
ADVANCE DESIGN VALIDATION GUIDE
5.79 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to tension load  Bilinear stressstrain diagram (Class XD2) (evaluated by SOCOTEC France  ref. Test 47I) Test ID: 5964 Test status: Passed
5.79.1
Description
Verifies a rectangular cross section beam made from concrete C25/30 to resist simple tension. Verification of the bending stresses at ultimate limit state and serviceability limit state is performed. Verification is done according to EN 199211 French annex.
5.79.2
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 this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
5.79.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■
Loadings from the structure: G = 135 kN/m (including dead load), Exploitation loadings (category A): Q = 150kN/m,
■ ■ ■
Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Quasipermanent combination of actions: CQP = 1.0 x G + 0.3 x Q
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.20 m, Width: b = 0.20 m, Length: L = 5.00 m, 2 Section area: A = 0.04 m , Concrete cover: c=5.00cm
667
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Rectangular solid concrete C25/30 and S500, class A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■
Exposure class XD2 Concrete density: 25kN/m3 Stressstrain law for reinforcement: Bilinear stressstrain diagram The concrete age t0=28 days Humidity RH=50%
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) , fixed connection, ► Support at end point (x = 5.80) restrained in translation along Z. Inner: None. ►
■
Loading The beam is subjected to the following load combinations: Load combinations: ■
Ultimate Limit State:
NEd = 1.35 * G + 1.5 * Q = 1.35 * 135 + 1.5 * 150 = 407 kN ■
Characteristic combination of actions:
Nser,cq = G + Q = 135 + 150 = 285 KN ■
Quasipermanent combination of actions:
Nser,qp = G + 0.3 * Q = 135 + 0.3 * 150 = 180 KN
5.79.2.2 Reference results in calculating the longitudinal reinforcement and the crack width Calculating for ultimate limit state:
As ,U ≥
N Ed
σ s ,U
=
0.407 = 9.37 * 10− 4 m² = 9.37cm² 500 1.15
Calculating for serviceability limit state:
As , ser ≥
N ser
σS
σ s = 0,8 * f yk = 0,8 * 500 = 400Mpa As ,ser ≥
668
0.285 = 7.13 *10 −4 m² = 7.13cm² 400
ADVANCE DESIGN VALIDATION GUIDE
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:
σs =
N ser 0.285 = 226.80 MPa ≤ σ s = 400 MPa ≤ σ s => As ,ser 0.001257
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:
sr , max = k3 * c +
0.425 * k1 * k2 * φ
ρ p , eff
According to: EC2 Part 1,1 EN 1992112004 Chapter 7.3.4(3) The effective depth “d” must be estimated by considering the real reinforcement of the beam:
d = 20 − 5 − 0.6 −
2 = 13.4cm 2
2.5 * (h − d ) 2.5 * (0.20 − 0.134) = 0.165 Ac , eff = b * min = 0.20 * min = 0.020 h 0.20 = 0 . 100 2 2 According to: EC2 Part 1,1 EN 1992112004 Chapter 7.3.4(2); Figure: 7.1
Ac ,eff = 0.040m²
ρ p , eff =
this value was multiplied by two because the section is fully stretched
As 12.57 * 10−4 = = 0.031 Ac , eff 0.040 According to: EC2 Part 1,1 EN 1992112004 Chapter 7.3.4(2)
φeq =
n1 * φ12 + n2 * φ22 = 20mm n1 * φ1 + n2 * φ2 According to: EC2 Part 1,1 EN 1992112004 Chapter 7.3.4(2)
sr , max = k3 * c +
0.425 * k1 * k2 * φ
ρ p , eff
Concrete cover of 5cm and HA6 reinforcement, therefore c=5+0.6=5.6cm
k 1 = 0.8 HA bars k 2 = 1 pure tension sizing 25 k 3 = 3.4 * c
2/3
sr , max = k3 * c +
25 = 3.4 * 56
2/3
= 1.99
0.425 * k1 * k2 * φ
ρ p , eff
= 1.99 * 56 +
0.425 * 0.8 * 1 * 20 = 331mm 0.031 669
ADVANCE DESIGN VALIDATION GUIDE
Calculation of average deformations
αe =
Es 200000 = = 6.35 31476 Ecm According to: EC2 Part 1,1 EN 1992112004 Chapter 7.3.4(2)
Steel stresses in SLSqp:
σs =
N ser −qp As ,ser
=
0.180 = 143.2Mpa 0.001257
σs − k t * ε sm − ε cm =
fct,eff
ρp,eff
* (1 + α e * ρp,eff ) Es
= 5.17 * 10− 4 ≥ 0,6 *
=
143 − 0.4 *
2.56 * (1 + 6.35 * 0.031) 0.031 = 200000
σs = 4.30 * 10− 4 Es
According to: EC2 Part 1,1 EN 1992112004 Chapter 7.3.4(2) Calculation of crack widths:
wk = sr , max * (ε sm − ε cm ) = 331 * (5.17 * 10−4 ) = 0.171mm According to: EC2 Part 1,1 EN 1992112004 Chapter 7.3.4(1) For an exposure class XD2, applying the French national annex, it retained a maximum opening of 0.20mm crack. This criterion is satisfied. Checking the condition of nonfragility: In case one control for cracking is required:
As ≥ Ac .
f ctm f yk
Because h< 50 MPa ⇒ η λ = 0,8 f ck = 25MPa < 50 MPa ⇒ η =1 f ck
f cu = η ⋅ α cc f cu = 1 ⋅1
γc
25 = 16,7 MPa 1,5
f ck ≤ 50 MPa ⇒ f ctm = 0,3[ f ck ] 3 2
f ctm = 0,3[25] 3 = 2,56 MPa 2
σ c = k1 ⋅ f ck σ c = 0,6 ⋅ 25 = 15MPa ■
f yd = f yd =
Steel:
f yk
γs 500 = 435MPa 1,15
σ s = k 3 ⋅ f yk σ s = 0,8 ⋅ 500 = 400MPa Boundary conditions ■ ■
Start point (x=0) : Hinged, End point (x=5) : Hinged.
5.81.2.2 Reference results Resistance ultimate limit state ■
Actions at m1:
ϖ = specific weight of reinforced concrete ϖ = 25 kN/m3 g = g1 + ϖ.bw .h g = 15,40 + 25.0,18.0,60 = 18,10 kN/m pu = 1, 35.g + 1, 5.q p u = 1,35.18,10 + 1,5.9 p u = 37,94 kN/m ■
Maximum bending moment:
M Ed = Pu
l eff2 8
677
ADVANCE DESIGN VALIDATION GUIDE
6,85 2 8 = 222,53mkN
M Ed = 37,94 M Ed
Serviceability limit state ■
Actions at m1:
pser = g + q pser = 18,10 + 9 pser = 27,10 kN/m ■
Maximum bending moment:
leff2
M ser = Pser M ser M ser
8
6,85 2 = 27,10 8 = 158,95mkN
Coefficient
γ
γ =
M Ed M ser
γ =
222,53 = 1,40 158,95
Calculation of tensioned reinforcements at ultimate limit state in the case of the deformation diagram
µ cu =
M Ed bw ⋅ d 2 ⋅ f cu
0,22253 = 0,245 0,18 ⋅ 0,55 2 ⋅ 16,7 1 ⇒ α u = 1 − 1 − 2 ⋅µ cu
µ cu =
] 1 [1 − 1 − 2 ⋅ 0,245 ] = 0,357 = 0,8 λ
αu
[
Ductility class and S 500 A steel and
f ck ⇒ µ AB f ck = 25MPa
α = 0,1346 ⇒ AB µ AB = 0,1019
678
σ −ε
steel inclined
ADVANCE DESIGN VALIDATION GUIDE
⇒ µ cu >< µ AB ⇒ Pivot ⇒ µ cu = 0,245 > 0,1019 = µ AB ⇒ Pivot B ⇒ ε c = ε cu 2 = ε cu 3 =
ε s1 = ε c
3,5 1000
1 − αu
αu
3,5 1 − 0,357 6,30 ⋅ = 1000 0,357 1000 435 2,175 = = 5 2 ⋅ 10 1000
ε s1 = ε yd
ε s1 >< ε yd ε s1 =
deformation of the σ − ε steel diagram 6,30 2,175 > = ε yd ⇒ 1000 1000
σ s1 = 432,71 + 952,38 ⋅ ε s1
454 (MPa)
σ s1 = 432,71 + 952,38 ⋅ 6,30 ⋅ 10 −3 = 439MPa λ ⇒ z c = d 1 − α u 2 0,8 0,357 = 0,472m z c = 0,551 − 2
As1,u =
M Ed z c ⋅ σ s1
As1,u =
0,22253 10 4 = 10,74cm 2 0,472 ⋅ 439
5.81.2.3 Theoretical results Result name
Result description
Reference value
Az
Bottom longitudinal reinforcement
10.74 cm²
My
ULS maximum bending moment
222.53 kN.m
5.81.3
Calculated results
Result name
Result description
Value
Error
My
ULS Bending moment
222.501 kN*m
0.0130 %
My
SLS Bending moment
158.95 kN*m
0.0000 %
Az
Bottom reinforcement area
10.7894 cm²
0.4600 %
679
ADVANCE DESIGN VALIDATION GUIDE
5.82 A23.304  Canada: Verifying the longitudinal reinforcement value on a oneway reinforced concrete slab Test ID: 6264 Test status: Passed
5.82.1
Description
This test is about the calculation of the bottom longitudinal reinforcement values for a oneway slab loaded with a distributed load (dead loads + live loads)
5.82.2
Background
This test checks the calculation of the bottom longitudinal reinforcement for a one way reinforced concrete slab simply supported. This slab is loaded by distributed loads (dead loads and live loads).
5.82.2.1 Model description ■ ■ ■
Reference: “Reinforced Concrete Design, a practical approach” – A23.304; Analysis type: static linear (gris workspace); Element type: planar.
The following load cases and load combination are used: ■
Distributed planar loads: DL: 6 kN/m² (including the self weight) ► LL: 5 kN/m² The ultimate limit state combination (ULS) is: 1.25 x DL + 1.5 x LL ►
■
Dead Loads
Live Loads
Units Metric System Geometry Below are described the concrete slab characteristics: ■ ■ ■ ■ ■ ■
680
Thickness: e = 300 mm, Width: b = 1000mm, Length: L = 6 m, Concrete cover: c=30 mm => means an effective slab depth equal to 270mm Concrete material: f’c= 25 Mpa (ϕc= 0.65) Steel reinforcement material: fy= 400Mpa (ϕs= 0.85)
ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions ■
The slab is considered as simply supported.
Finite elements modeling ■ ■
Planar element: shell, Mesh size: 0.50m
5.82.2.2 Reference results in calculating the concrete slab Calculation of the factored bending moment The factored load ( w f ) is determined according to Cl.4.1.3.2 of NBC 2005:
w f = 1.25 DL + 1.50 LL = 1.25 × 6 + 1.50 × 5 = 15 KN / m² This is a simply supported slab and the maximum bending moment at the midspan can be calculated from the following equation:
Mf =
w f .l ² 8
=
15 × 6² = 67.50 KN .m / m 8
With this equation, we consider the oneway slab as a rectangular beam of 1m width. Calculation of the required area of tensioned reinforcement We calculate the required area of tensioned reinforcement using the direct procedure, considering that
3,85.M r As = 0,0015. f c' .b. d − d ² − f c' .b
Mr = M f
3,85 × 67,50.106 As = 0,0015 × 25 × 1000 270 − 270² − 25 × 1000
= 750mm² / m
681
ADVANCE DESIGN VALIDATION GUIDE
Reinforcement calculated value The values calculated by Advance Design are the following ones:
5.82.2.3 Reference results Result name
Result description
Reference value
Ax,b
Longitudinal bottom reinforcement in the slab (mm²/m)
750 mm2/m
5.82.3
682
Calculated results
Result name
Result description
Value
Error
Axb
Longitudinal reinforcement value in the middle span
758.163 mm²
1.0884 %
ADVANCE DESIGN VALIDATION GUIDE
5.83 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement  Bilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 2) Test ID: 4970 Test status: Passed
5.83.1
Description
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
5.83.2
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 this test, the calculation of stresses will be made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. This test was evaluated by the French control office SOCOTEC.
5.83.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. 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 Quasipermanent combination of actions: CQP = 1.0 x G + 0.3 x Q
Simply supported beam
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■
Height: h = 0.60 m, Width: b = 0.25 m, Length: L = 5.80 m,
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■ ■ ■
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
Materials properties Rectangular solid concrete C25/30 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■
Exposure class XD1 3 Concrete density: 25kN/m Stressstrain law for reinforcement: Bilinear stressstrain diagram Cracking calculation required
■
Concrete C25/30: fcd =
fck 25 = = 16,67MPa γ c 1,5
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■
2/3 fctm = 0.30 * fck = 0.30 * 25 2 / 3 = 2.56MPa
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1 ■
Steel S500 : f yd =
f yk γs
=
500 = 434,78MPa 1,15 According to: EC2 Part 1,1 EN 1992112002 Chapter 3.2
■
f +8 E cm = 22000 * ck 10
0. 3
25 + 8 = 22000 * 10
0. 3
= 31476MPa
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1 Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x = 0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.80) restrained in translation along Y, Z and restrained in rotation along X. Inner: None. ►
■
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 Quasipermanent combination of actions: CQP = 1.0 x G + 0.3 x Q=15+3.75+0.3*20=24.75kN/ml
■
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Load calculations:
ADVANCE DESIGN VALIDATION GUIDE
M Ed =
55,31 * 5,80² = 232,59kN .m 8
M Ecq =
38,75 * 5,80² = 162,94kN .m 8
M Eqp =
24,75 * 5,80² = 104,07kN .m 8
5.83.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity:
(∞, t 0 ) =
RH
* β( fcm ) * β( t 0 ) According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.2
16.8
β( fcm ) =
fcm
16.8
=
25 + 8
= 2.925 According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.4
β( t 0 ) =
1
=
0.1 + t 00.20
1 0.1 + 28 0.20
= 0.488 at t0=28 days According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.5
The value of the φRH coefficient depends of the concrete quality:
RH
RH 1− 100 * α1 * α 2 = 1 + 0.1 * 3 h 0 According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.3a
α1 = α 2 = 1 if
fCM = 35MPa
35 If not: α1 = fcm
0.7
35 and α 2 = fcm
0.2
According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.8c In this case: fcm = fck + 8Mpa = 33Mpa ⇒ α1 = α 2 = 1
2 * Ac 2 * 250 * 600 = = 176.47mm ⇒ h0 = u 2 * ( 250 + 600)
50 100 = 1+ = 1.89 0.1 * 3 176.47 1−
RH
According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.6
(∞, t 0 ) =
RH
* β( fcm ) * β( t 0 ) = 1.89 * 2.925 * 0.488 = 2.70
The coefficient of equivalence is determined by the following formula:
Es E cm
αe =
1 + (∞, t 0 ) *
MEqp MEcar
Defined earlier: MEcar =
38,75 * 5,80² = 162,94kN.m 8
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MEqp =
24,75 * 5,80² = 104,07kN.m 8
This gives:
αe =
200000 = 17.32 31476 104.07 1 + 2.70 * 162.94
5.83.2.3 Reference results in calculating the concrete beam reduced moment limit Due to exposure class XD1, we will verify the section having noncompressed steel reinforcement, determining the reduced moment limit, using the formula defined by Jeans Roux in his book “Practice of EC2”: This value can be determined by the next formula if μ luc = K(α e ) *
f ck < 50MPa valid for a constitutive law to horizontal plateau:
fck ( 4.69 − 1.70 * γ ) * fck + (159.90 − 76.20 * γ )
(
K (α e ) = 10 −4 * a + b * α e + c * α e 2
)
The values of the coefficients "a", "b" and "c" are defined in the following table: Diagram for inclined tier
Diagram for horizontal plateau
a
75,3 * f ck − 189,8
71,2 * f ck + 108
b
− 5,6 * f ck + 874,5
− 5,2 * f ck + 847,4
c
0,04 * f ck − 13
0,03 * f ck − 12,5
This gives us: ■
a = 71,2 * 25 + 108 = 1888
■
b = −5,2 * 25 + 847,4 = 717.4
■
c = 0,03 * 25 − 12,5 = −11.75
■
K (α e ) = 10−4 (1888 + 717.4 * 17.27 − 11.75 * 17.27² ) = 1.079
Then:
232,58 = 1,43 162,94
■
γ=
■
µluc = K (α e ).
f ck = 0.2504 ( 4.69 − 1.70 * γ ) * f ck + (159.90 − 76.20 * γ )
Reference reinforcement calculation at SLU: The calculation of the reinforcement is detailed below: ■ ■
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Effective height: d=0.9*h=0.53m Calculation of reduced moment:
ADVANCE DESIGN VALIDATION GUIDE
µcu =
■
M Ed 0,233 = = 0,207 bw * d ² * f cd 0,25 * 0,519² * 16,67
Calculation of the lever arm zc:
zc = d * (1 − 0,4 * α u ) = 0,519 * (1 − 0,4 * 0,294) = 0,458m ■
Calculation of the reinforcement area:
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). ■
α1 =
α e * σc α e * σc + σ s
=
17.32 * 15 = 0,394 17.32 * 15 + 400
M ser , cq = 163kNm < M rb ⇒
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 γ. Then we calculate the reinforcement resulted from the SLS efforts (assuming the maximum constrain is reached on steel and concrete): ■
Neutral axes position :
■
Lever arm :
■
α1 = 0,394
α 0.394 zc = d * 1 − 1 = 0.519 * 1 − = 0.451m 3 3 M Ecq 0.163 Reinforcement section : As1, ser = = = 9.03cm² zc * σ s 0.451 * 400
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: ■
fct,eff * bw * d 0.26 * A s,min = Max f yk 0.0013 * b * d w
According to: EC2 Part 1,1 EN 1992112002 Chapter 9.2.1.1(1); Note 2 Where: fct,eff = fctm = 2.56MPa from cracking conditions
Therefore: ►
2.56 * 0.25 * 0.519 = 1.73 * 10− 4 m² 0.26 * A s,min = max = 1.73cm² 500 0.0013 * 0.25 * 0.519 = 1.70 * 10− 4 m²
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ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element.
ULS and SLS load combinations(kNm) Simply supported beam subjected to bending ULS (reference value: 232.59kNm)
SLS (reference value: 162.94kNm)
Theoretical reinforcement area(cm2) (reference value: 11.68cm2)
Minimum reinforcement area(cm2) (reference value: 1.73cm2)
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ADVANCE DESIGN VALIDATION GUIDE
5.83.2.4 Reference results Result name
Result description
Reference value
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
2
Az
2
Minimum reinforcement area [cm ]
Amin
5.83.3
11.68 cm2
Theoretical reinforcement area [cm ]
1.73 cm2
Calculated results
Result name
Result description
Value
Error
My
My USL
232.578 kN*m
0.0052 %
My
My SLS
162.936 kN*m
0.0025 %
Az
Az
11.6779 cm²
0.0180 %
Amin
Amin
1.73058 cm²
0.0335 %
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ADVANCE DESIGN VALIDATION GUIDE
5.84 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcement Bilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 6) Test ID: 4979 Test status: Passed
5.84.1
Description
The purpose of this test is to verify the software results for the My resulted stresses for the USL load combination and for the results of the theoretical reinforcement area Az.
5.84.2
Background
This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model. This test was evaluated by the French control office SOCOTEC.
5.84.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■ ■ ■
Loadings from the structure: G = 45 kN/m Exploitation loadings (category A): Q = 37.4kN/m, The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q
■ ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stressstrain diagram The objective is to verify: ■
The theoretical reinforcement area results
Simply supported beam
Units Metric System
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ADVANCE DESIGN VALIDATION GUIDE
Geometry Beam cross section characteristics:
■ ■ ■
Beam length: 8m Concrete cover: c=3.50 cm Effective height: d=h(0.6*h+ebz)=0.435 m; d’=ebz=0.035m
Materials properties Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■
Exposure class XC1 3 Concrete density: 25kN/m Reinforcement steel ductility: Class B The calculation is made considering bilinear stressstrain diagram
■
Concrete C25/30:
f cd =
f ck
γc
=
25 = 10.67 MPa 1,5
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■
f ctm = 0.30 * f ck2 / 3 = 0.30 * 25 2 / 3 = 2.56 MPa According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1
■
Steel S500B :
f yd =
f yk
γs
=
500 = 434.78MPa 1,15 According to: EC2 Part 1,1 EN 1992112002 Chapter 3.2
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, Support at end point (x = 8) restrained in translation along Y and Z, and restrained rotation along X. Inner: None.
► ►
■
Loading The beam is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*45+1.5*37.4=116.85kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=45+34.7=79.7kN/ml
■
Load calculations:
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ADVANCE DESIGN VALIDATION GUIDE
M Ed =
116.85 * 8² = 934.8kNm 8
5.84.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:
hf 0.15 M btu = beff * h f * d − * f cd == 1.00 * 0.15 * 0.435 − * 16.67 = 9kNm 2 2 Comparing Mbtu with MEd:
M btu = 900kNm < M Ed = 934.8kNm Therefore, the concrete section is not entirely compressed; Therefore, the calculations considering the T section are required.
5.84.2.3 Reference reinforcement calculation: Theoretical section 2: The moment corresponding to this section is:
hf M Ed 2 = (beff − bw ) * h f * f cd * d − 2 = 720kNm
0.15 = (1 − 0.20) * 0.15 *16.67 * 0.435 − = 2
According to this value, the steel section is:
A2 =
M Ed 2 0.720 = = 46.00cm² 0.15 hf d − * f yd 0.435 − * 347.8 2 2
Theoretical section 1: The theoretical section 1 corresponds to a calculation for a rectangular shape beam section.
M Ed 1 = M Ed − M Ed 2 = 935 − 720 = 215kNm
µ cu =
M Ed 1 0.215 = 0.341 = bw * d ² * Fcd 0.20 * 0.435² *16.67
For a S500B reinforcement and for a XC1 exposure class, there will be a: μ cu > μ lu = 0.372 , 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:
[
]
[
]
α u = 1.25 * 1 − (1 − 2 * µ cu ) = 1.25 * 1 − (1 − 2 * 0.341) = 0.544 z c1 = d * (1 − 0.4 * α u ) = 0.435 * (1 − 0.40 * 0.544) = 0.340m
A1 =
M Ed 1 0.215 = = 14.52cm² z c1 * f yd 0.340 * 347.78
Theoretical section 1: In conclusion, the entire reinforcement steel area is A=A1+A2=46+14.52=60.52cm2
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Finite elements modeling ■ ■ ■
Linear element: S beam, 9 nodes, 1 linear element.
ULS load combinations(kNm) Simply supported beam subjected to bending
Theoretical reinforcement area(cm2) For Class B reinforcement steel ductility (reference value: A=60.52cm2)
5.84.2.4 Reference results Result name
Result description
Reference value
Az (Class B)
Theoretical reinforcement area [cm2]
60.52 cm2
5.84.3
Calculated results
Result name
Result description
Value
Error
Az
Az
60.5167 cm²
0.0055 %
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ADVANCE DESIGN VALIDATION GUIDE
5.85 EC2 / NF EN 199211/NA  France: Verifying a T concrete section, without compressed reinforcement Bilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 7) Test ID: 4980 Test status: Passed
5.85.1
Description
The purpose of this test is to determine the theoretical reinforcement area Az for a T concrete beam subjected to bending efforts at ultimate limit state (ULS). The test confirms the absence of the compressed reinforcement for this model.
5.85.2
Background
Verifies the theoretical reinforcement area for a T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model. This test was evaluated by the French control office SOCOTEC.
5.85.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■ ■ ■
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 ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q
■ ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stressstrain diagram The objective is to test: ■
The theoretical reinforcement area
Simply supported beam
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ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Beam cross section characteristics:
■ ■ ■
Beam length: 8m Concrete cover: c=3.50 cm Effective height: d=h(0.6*h+ebz)=0.482 m; d’=ebz=0.035m
Materials properties Rectangular solid concrete C25/30 and S400B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■
Exposure class XC1 3 Concrete density: 25kN/m Reinforcement steel ductility: Class B The calculation is made considering bilinear stressstrain diagram
■
Concrete C25/30:
f cd =
f ck
γc
=
25 = 10.67 MPa 1,5
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■
f ctm = 0.30 * f ck2 / 3 = 0.30 * 25 2 / 3 = 2.56 MPa According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1
■
Steel S400B :
f yd =
f yk
γs
=
400 = 347.83MPa 1,15 According to: EC2 Part 1,1 EN 1992112002 Chapter 3.2
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x = 0) restrained in translation along X, Y and Z, Support at end point (x = 8) restrained in translation along Y and Z, and restrained rotation along X. Inner: None.
► ►
■
Loading The beam is subjected to the following load combinations: ■
Load combinations:
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ADVANCE DESIGN VALIDATION GUIDE
The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*0+1.5*18.2=27.3kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=0+18.2=18.2kN/ml ■
Load calculations:
M Ed =
27.3 * 8² = 218.40kNm 8
M Ecq =
18.20 * 8² = 145.60kNm 8
5.85.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:
hf M btu = beff * h f * d − 2
0.12 * f cd == 1.10 * 0.12 * 0.482 − *16.67 = 928kNm 2
Comparing Mbtu with MEd:
M Ed = 218.40kNm < M btu = 928kNm Therefore, the concrete section is not entirely compressed; This requires a calculation considering a rectangular section of b=110cm and d=48.2cm. Reference longitudinal reinforcement calculation:
µ cu =
M Ed 1 0.218 = = 0.051 bw * d ² * Fcd 1.10 * 0.482² *16.67
[
]
[
]
α u = 1.25 * 1 − (1 − 2 * µ cu ) = 1.25 * 1 − (1 − 2 * 0.051) = 0.066 z c = d * (1 − 0.4 * α u ) = 0.482 * (1 − 0.40 * 0.066) = 0.469m
A=
M Ed 0.218 = 13.38cm² = z c * f yd 0.469 * 347.83
Finite elements modeling ■ ■ ■
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Linear element: S beam, 9 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
ULS load combinations(kNm) Simply supported beam subjected to bending
Theoretical reinforcement area(cm2) For Class B reinforcement steel ductility (reference value: A=13.38cm2)
5.85.2.3 Reference results Result name Az (Class B)
5.85.3
Result description
Reference value 2
13.38 cm2
Theoretical reinforcement area [cm ]
Calculated results
Result name
Result description
Value
Error
Az
Theoretical reinforcement area
13.3793 cm²
0.0052 %
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ADVANCE DESIGN VALIDATION GUIDE
5.86 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam without compressed reinforcement – Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 8) Test ID: 4981 Test status: Passed
5.86.1
Description
Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. The purpose of the test is to verify the results using a constitutive law for reinforcement steel, on the inclined stressstrain diagram. The objective is to verify:  The stresses results  The longitudinal reinforcement corresponding to Class A reinforcement steel ductility  The minimum reinforcement percentage
5.86.2
Background
Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. The purpose of this test is to verify the software results for using a constitutive law for reinforcement steel, on the inclined stressstrain diagram. This test was evaluated by the French control office SOCOTEC.
5.86.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■
Loadings from the structure: G = 25 kN/m (including the dead load) Exploitation loadings (category A): Q = 30kN/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 ■ Quasipermanent combination of actions: CQP = 1.0 x G + 0.6 x Q ■ Reinforcement steel: Class A ■ The calculation is performed considering inclined stressstrain diagram ■ Cracking calculation required The objective is to verify: ■ ■ ■
698
The stresses results The longitudinal reinforcement corresponding to doth Class A reinforcement steel ductility The minimum reinforcement percentage
ADVANCE DESIGN VALIDATION GUIDE
Simply supported beam
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■ ■ ■
Height: h = 0.65 m, Width: b = 0.28 m, Length: L = 6.40 m, 2 Section area: A = 0.182 m , Concrete cover: c=4.50 cm Effective height: d=h(0.6*h+ebz)=0.806 m; d’=ebz=0.045m
Materials properties Rectangular solid concrete C30/37 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■
Exposure class XD3 3 Concrete density: 25kN/m Reinforcement steel ductility: Class A The calculation is performed considering inclined stressstrain diagram Cracking calculation required
■
Concrete C25/30:
f cd =
f ck
γc
=
30 = 20 MPa 1,5
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■
f ctm = 0.30 * f ck2 / 3 = 0.30 * 30 2 / 3 = 2.90 MPa According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1
■
f +8 E cm = 22000 * ck 10
■
Steel S500 : f yd =
f yk γs
0.3
=
30 + 8 = 22000 * 10
0.3
= 32837MPa
500 = 434,78MPa 1,15 According to: EC2 Part 1,1 EN 1992112002 Chapter 3.2
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ADVANCE DESIGN VALIDATION GUIDE
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 6.40) restrained in translation along Y and Z, and restrained rotation along X. Inner: None. ►
■
Loading The beam is subjected to the following load combinations: ■
Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*(25)+1.5*30=78.75kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=25+30=55kN/ml Quasipermanent combination of actions: CQP = 1.0 x G + 0.6 x Q=25+0.6*30=43kNm
■
Load calculations:
M Ed =
78.75 * 6.40² = 403.20kNm 8
M Ecq =
55 * 6.40² = 281.60kNm 8
M Eqp =
43 * 6.40² = 220.16kNm 8
5.86.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity:
ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.2
According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.4
at t0=28 days According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.5
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ADVANCE DESIGN VALIDATION GUIDE
The value of the φRH coefficient depends of the concrete quality:
RH
RH 1− 100 * α1 * α 2 = 1 + 0. 1 * 3 h 0 According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.3a
α1 = α 2 = 1 if fCM = 35MPa 35 If not: α1 = fcm
0.7
35 and α 2 = fcm
0.2
According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.8c In this case, fcm = fck + 8Mpa = 38Mpa 0.7
35 α1 = fcm
35 α 2 = fcm
0.2
35 = 38
0.7
35 = 38
0.2
= 0.944
= 0.984
2 * Ac 2 * 280 * 650 h0 = = = 195.7mm ⇒ u 2 * ( 280 + 650)
RH
50 1− 100 = 1+ 0.1 * 3 195.7
* 0.984 = 1.78
According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.6
(∞, t 0 ) =
RH
* β( fcm ) * β( t 0 ) = 1.78 * 2.73 * 0.488 = 2.37
The coefficient of equivalence is determined using the following formula:
αe =
Es E cm
= MEqp
1 + (∞, t 0 ) *
MEcar
200000 = 17.40 32837 220.16 1 + 2.37 * 281.60
5.86.2.3 Reference results in calculating the concrete beam reduced moment limit Due to exposure class XD3, we will verify the section having noncompressed steel reinforcement, determining the reduced moment limit, using the formula defined by Jeans Roux in his book “Practice of EC2”: This value can be determined by the next formula if inclined stressstrain diagram:
µ luc = K (α e ) *
f ck < 50 MPa and
valid for a constitutive law considering
f ck (4.62 − 1.66 * γ ) * f ck + (165.69 − 79.62 * γ )
(
K (α e ) = 10−4 * a + b * α e + c * α e
2
)
The values of the coefficients "a", "b" and "c" are defined in the following table: Diagram for inclined stressstrain diagram
Diagram for bilinear stressstrain diagram
a
75,3 * f ck − 189,8
71,2 * f ck + 108
b
− 5,6 * f ck + 874,5
− 5,2 * f ck + 847,4
c
0,04 * f ck − 13
0,03 * f ck − 12,5
701
ADVANCE DESIGN VALIDATION GUIDE
This gives us:
a = 75.3 * 25 + 189.8 = 2069.2 b = −5,2 * 30 + 874.5 = 706.5
c = 0,04 * 30 − 13 = −11.8
K (α e ) = 10 −4 (2069.2 + 706.5 * 17.60 − 11.80 * 17.60² ) = 1.087 Then:
γ=
437.76 = 1,425 307.2
µ luc = K (α e ).
f ck = 0.272 ( 4.62 − 1.66 * γ ) * f ck + (165.69 − 79.62 * γ )
Calculation of reduced moment:
µ cu =
M Ed 0.403 *10 3 Nm = = 0,225 bw * d ² * f cd 0,50m * 0.566² m 2 * 20MPa
µ cu = 0.225 < µ luc = 0.272
therefore, there is no compressed reinforcement
Reference reinforcement calculation at SLS: The calculation of the reinforcement is detailed below: ■ ■
Effective height: d = 0.566m Calculation of reduced moment:
µ cu = 0.225
■
Calculation of the lever arm zc:
z c = d * (1 − 0,4 *α u ) = 0,566 * (1 − 0,4 * 0,323) = 0,493m ■
Calculation of the reinforcement area:
Au = ■
0,233 M Ed = = 11,68 * 10− 4 m² = 11,68cm² zc * f yd 0,458 * 434,78
Calculation of the stresses from the tensioned reinforcement: Reinforcement steel elongation:
ε su =
1−αu
αu
* ε cu 2 =
1 − 0,323 * 3.5 = 7.35 ‰ 0,323
Tensioned reinforcement efforts:
σ su = 432,71 + 952,38 * ε su = 432,71 + 952,38 * 0.00735 = 439.71MPa ≤ 454MPa
702
ADVANCE DESIGN VALIDATION GUIDE
■
Reinforcement section calculation:
Au =
M Ed 0.403 = = 18.60 *10 −4 m² = 18.60cm² z c * f yd 0.493 * 439.71
Reference solution for minimal reinforcement area The minimum percentage for a rectangular beam in pure bending is: fct,eff * bw * d 0.26 * A s,min = Max f yk 0.0013 * b * d w
According to: EC2 Part 1,1 EN 1992112002 Chapter 9.2.1.1(1); Note 2 Where: fct, eff = fctm = 2.90MPa from cracking conditions
Therefore:
As ,min
2.90 * 0.28 * 0.566 = 2.39 * 10 −4 m² 0.26 * = 2.39cm² = max 500 0.0013 * 0.28 * 0.566 = 2.06 * 10 −4 m²
Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element.
ULS and SLS load combinations (kNm) Simply supported beam subjected to bending
ULS (reference value: 403.20kNm)
SLS (reference value: 281.60kNm)
703
ADVANCE DESIGN VALIDATION GUIDE
Theoretical reinforcement area(cm2) For Class A reinforcement steel ductility (reference value: A=18.60cm2)
Minimum reinforcement area(cm2) (reference value: 2.39cm2)
5.86.2.4 Reference results Result name
Result description
My,ULS
My corresponding to the 101 combination (ULS) [kNm]
403.20 kNm
My,SLS
My corresponding to the 102 combination (SLS) [kNm]
281.60 kNm
Az (Class A)
Theoretical reinforcement area [cm2]
18.60 cm2
2
Minimum reinforcement area [cm ]
Amin
5.86.3
704
Reference value
2.39 cm2
Calculated results
Result name
Result description
Value
Error
My
My USL
403.2 kN*m
0.0000 %
My
My SLS
281.6 kN*m
0.0000 %
Az
Az
18.6006 cm²
0.0000 %
Amin
Amin
2.38697 cm²
0.0000 %
ADVANCE DESIGN VALIDATION GUIDE
5.87 EC2 / NF EN 199211/NA  France: Verifying a square concrete beam subjected to a normal force of traction  Inclined stressstrain diagram (Class X0) (evaluated by SOCOTEC France ref. Test 46 I) Test ID: 5232 Test status: Passed
5.87.1
Description
Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction  Inclined stressstrain diagram (Class X0). Determines the necessary reinforcement of a reinforced concrete tie, subjected to a normal force of traction. Ultimate limit state load combinations are used. The tie member is fixed at one end, and blocked in translation along Z axis at the other end.
5.87.2
Background
Tie sizing Bilinear stressstrain diagram / Inclined stressstrain diagram Verifies the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction. This test was evaluated by the French control office SOCOTEC.
5.87.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane workspace); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: Fx,G = 233.33 kN The dead load is neglected
■ ■
Exploitation loadings (category A): Fx,Q = 56.67kN The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.15 m, Width: b = 0.15 m, Length: L = 5.00 m, 2 Section area: A = 0.0225 m , Concrete cover: c=3cm 705
ADVANCE DESIGN VALIDATION GUIDE
■
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 The boundary conditions are described below: ■
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.
► ►
■
5.87.2.2 Reference results in calculating the concrete beam There will be two successive calculations, considering a bilinear stressstrain diagram constitutive law and then a inclined stressstrain diagram constitutive law. Calculations according a bilinear stressstrain diagram
As ,U ≥
N Ed
σ s ,U
=
0.400 = 11,50 * 10 − 4 m² = 11,50cm² 400 1.15
It will be used a 4HA20=A=12.57cm
2
Calculations according a inclined stressstrain diagram
S 400 − ClassB ⇒ σ s ,U = 373MPa As ,U ≥
N Ed
σ s ,U
=
0.400 = 10,72 * 10 − 4 m² = 10,70cm² 373
It can be seen that the gain is not negligible (about 7%). Checking the condition of nonfragility:
As ≥ Ac *
f ctm f yk
f ctm = 0.30 * f ck2 / 3 = 0.30 * 202 / 3 = 2,21MPa
Ac = 0.15 * 0.15 = 0.0225m² As ≥ Ac *
f ctm 2.21 = 0.0225 * = 1.24cm² f yk 400
Finite elements modeling ■ ■ ■
706
Linear element: S beam, 6 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
ULS load combinations (kNm) In case of using the bilinear stressstrain diagram, the reinforcement will result: (Az=11.50cm2=2*5.75 cm2)
In case of using the inclined stressstrain diagram, the reinforcement will result: (Az=10.70cm2=2*5.35 cm2)
5.87.2.3 Reference results Result name
Result description
Reference value
Az,1
Longitudinal reinforcement obtained using the bilinear stressstrain diagram [cm2] Longitudinal reinforcement obtained using the inclined stressstrain diagram [cm2]
5.75 cm2
Az,2
5.87.3
5.37 cm2
Calculated results
Result name
Result description
Value
Error
Az
As,U  longitudinal reinforcement area when inclined stressstrain diagram is used
5.36526 cm²
0.0981 %
707
ADVANCE DESIGN VALIDATION GUIDE
5.88 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to uniformly distributed load, with compressed reinforcement Bilinear stressstrain diagram (evaluated by SOCOTEC France  ref. Test 3) Test ID: 4976 Test status: Passed
5.88.1
Description
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. The objective is to verify:  The stresses results  The longitudinal reinforcement  The verification of the minimum reinforcement percentage
5.88.2
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 this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. This test was evaluated by the French control office SOCOTEC.
5.88.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combinations are used: ■ ■
Loadings from the structure: G = 70 kN/m, Exploitation loadings (category A): Q = 80kN/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 ■ Quasipermanent combination of actions: CQP = 1.0 x G + 0.3 x Q The objective is to verify: ■ ■ ■
708
The stresses results The longitudinal reinforcement The verification of the minimum reinforcement percentage
ADVANCE DESIGN VALIDATION GUIDE
Simply supported beam
Units Metric System Geometry Beam cross section characteristics: ■ ■ ■ ■ ■ ■
Height: h = 1.25 m, Width: b = 0.65 m, Length: L = 14 m, 2 Section area: A = 0.8125 m , Concrete cover: c = 4.50 cm Effective height: d = h(0.6*h+ebz) = 1.130 m; d’ = ebz = 0.045m
Materials properties Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■
Exposure class XD1 Concrete density: 25kN/m3 Stressstrain law for reinforcement: Bilinear stressstrain diagram Cracking calculation required
■
Concrete C25/30: fcd =
fck 25 = = 16,67MPa γ c 1,5
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■
2/3 = 0.30 * 25 2 / 3 = 2.56MPa fctm = 0.30 * fck
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1 ■
Steel S500 : f yd =
f yk γs
=
500 = 434,78MPa 1,15 According to: EC2 Part 1,1 EN 1992112002 Chapter 3.2
■
f +8 E cm = 22000 * ck 10
0.3
25 + 8 = 22000 * 10
0.3
= 31476MPa
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1 Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 14) restrained in translation along Z. Inner: None. ►
■
709
ADVANCE DESIGN VALIDATION GUIDE
Loading The beam is subjected to the following load combinations: ■ Load combinations: The ultimate limit state (ULS) combination is: 3 Cmax = 1.35 x G + 1.5 x Q=1.35*70+1.5*80=214.5*10 N/ml
Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=70+80=150*103 N/ml Quasipermanent combination of actions: CQP = 1.0 x G + 0.3 x Q=70+0.3*80=94*103 N/ml ■
Load calculations: MEd =
214.5 * 14² = 5.255 * 10 6 Nm 8
MEcq = 150
150 * 14² = 3.675 * 10 6 Nm 8
94 * 14² = 2.303 * 10 6 Nm 8
MEqp =
5.88.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity: ■
(∞, t 0 ) =
RH
* β( fcm ) * β( t 0 ) According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.2
■
β( fcm ) =
16.8 fcm
16.8
=
25 + 8
= 2.925 According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.4
■
β( t 0 ) =
1 0.1 + t 00.20
=
1 0.1 + 28 0.20
= 0.488 at t0=28 days According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.5
The value of the φRH coefficient depends of the concrete quality:
■
RH
RH 1− 100 * α * α = 1 + 1 0. 1 * 3 h 2 0 According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.3a
■
α1 = α 2 = 1
■
35 If not: α1 = fcm
if
fcm ≤ 35MPa 0.7
35 and α 2 = fcm
0.2
According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.8c
f cm = f ck + 8Mpa = 33Mpa ⇒ α1 = α 2 = 1
■
In our case we have:
■
2 * Ac 2 * 650 * 1250 h0 = = = 427.63mm ⇒ 2 * (650 + 1250) u
50 100 = 1+ = 1.66 0.1 * 3 427.63 1−
RH
According to: EC2 Part 1,1 EN 1992112002; Annex B; Chapter B.1(1); B.6 ■
710
(∞, t 0 ) =
RH
* β( fcm ) * β( t 0 ) = 1.66 * 2.925 * 0.488 = 2.375
ADVANCE DESIGN VALIDATION GUIDE
The coefficient of equivalence is determined by the following formula: ■
Es E cm
αe =
1 + (∞, t 0 ) *
MEqp MEcar
Defined earlier:
M Ecar = 3675 * 103 Nm
M Eqp = 2303 * 103 Nm This gives:
αe =
200000 31476 1 + 2.70 *
= 15.82
2303 * 10 3 3975 * 10 3
5.88.2.3 Reference results in calculating the concrete beam reduced moment limit For the calculation of steel ULS, we consider the moment reduced limit of
µ lu = 0.372
for a steel grade 500Mpa.
Therefore, make sure to enable the option "limit µ (0.372 / S 500) " 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 Calculation of reduced moment:
M Ed 5255.25 * 103 Nm µcu = = = 0,383 bw * d ² * f cd 0,65m * 1.125² m 2 * 16,67 MPa
µcu = 0.383 < µlu = 0.372
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
:
M Ed 1 = µlu * bw * d ² * f cd = 0.372 * 0.65 * 1.125² * 16.67 = 5.10 * 106 Nm
[
]
[
]
■
The α value: α lu = 1.25 * 1 − (1 − 2 * μ lu ) = 1.25 * 1 − (1 − 2 * 0.372) = 0.618
■
Calculation of the lever arm zc: z lu = d * (1 − 0.4 * α lu ) = 1.130 * (1 − 0.4 * 0.618) = 0.851m
■
Calculation of the reinforcement area:
A1 =
M Ed 1 5.10 * 10−6 Nm = = 137.35cm² zlu . f yd 0.851m * 434.78MPa
711
ADVANCE DESIGN VALIDATION GUIDE
Compressed steel reinforcement reduction (Section As2): Reduction coefficient:
ε sc =
3,5 3.5 (0.618 *1.130 − 0.045) = 0.00327 (α lu * d − d ' ) = 1000 * α lu * d 1000 * 0.618 * 1.130
ε sc = 0.00327 > ε yd = 0.00217 ⇒ σ sc = f yd = 434.78MPa Compressed reinforcement calculation:
As 2 =
M Ed − M Ed 1 5.255 − 5.15 = = 2.32cm² (d − d ' )σ sc (1.130 − 0.045) × 434.78
The steel reinforcement condition:
A2 = As 2 .
σ sc f yd
= 2.32cm²
Total area to be implemented: In the lower part: As1=A1+A2=142.42 cm2 In the top part: As2=2.32 cm2 Reference reinforcement calculation at SLS: The concrete beam design at SLS will be made considering a limitation of the characteristic compressive cylinder strength of concrete at 28 days, at 0.6*fck The assumptions are: 150 * 14² = 3.675 * 10 6 Nm 8
■
The SLS moment: MEcq = 150
■
The equivalence coefficient: α e = 15.82
■ The stress on the concrete will be limited at 0.6*fck=15Mpa and the stress on steel at 0.8*fyk, or 400MPa Calculation of the resistance moment MRd for detecting the presence of the compressed reinforcement:
x1 = Fc =
αe * σc αe * σc + σs
.d =
15.82 * 15 × 1.125 = 0.419m 15.82 * 15 + 400
1 1 * b w * x1 * σ c = * 0.65 * 0.419 × 15 = 2.04 * 10 6 N 2 2
zc = d −
x1 0.419 = 1.125 − = 0.985m 3 3
Mrb = Fc * z c = 2.04 * 10 6 * 0.985 = 2.01 * 10 6 Nm Therefore the compressed reinforced established earlier was correct.
712
ADVANCE DESIGN VALIDATION GUIDE
Theoretical section 1 (tensioned reinforcement only)
M 1 = M rb = 2.01 * 106 Nm
α1 =
x1 0.419 = = 0.372 d 1.125
zc = d − A1 =
x1 0.419 = 1.125 − = 0.985m 3 3
M1 2.01 = = 51.01cm² zc × σ s 0.985 × 400
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
M 2 = M ser , cq − M rb = (3.675 − 2.01) * 106 = 1.665 * 106 Nm Compressed reinforcement stresses:
σ sc = α e * σ c *
α1 * d − d ' α1 * d
σ sc = 15.82 *15 *
0.372 * 1.125 − 0.045 = 211.78MPa 0.372 * 1.125
Compressed reinforcement area:
A' =
M2 1.645 = = 71.92cm² (d − d ' ) * σ sc (1.125 − 0.045) * 211.78
Complementary tensioned reinforcement area:
As = A' *
σ sc 211.78 = 71.92 * = 38.08cm² 400 σs
Section area: Tensioned reinforcement: 51.01+38.08=89.09 cm2 Compressed reinforcement: 71.92 cm2 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 of 2 tensioned reinforcement (after ULS: Au=141.44cm )
713
ADVANCE DESIGN VALIDATION GUIDE
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 calculating the SLS: Au=141.44cm2. For this particular value, it will be calculated the resistance obtained for tensioned reinforcement:
σs =
89.18 AELS ×σ s = × 400 = 252 MPa 141.42 AELU
This is a SLS calculation, considering this limitation: Calculating the moment resistance MRb for detecting the presence of compressed steel reinforcement:
x1 =
αe *σ c 15.82 * 15 *d = * 1.130 = 0.55m 15.82 * 15 + 252 αe *σ c + σ s
Fc =
1 1 * bw * x1 * σ c = * 0.65 * 0.55 *15 = 2.67 *106 N 2 2
zc = d −
x1 0.55 = 1.125 − = 0.94m 3 3
M rb = Fc * zc =
x 1 1 0.55 6 * bw * x1 * σ c * d − 1 = * 0.65 * 0.55 * 15 * 1.125 − = 2.52 * 10 Nm 2 3 2 3
According to the calculation above Mser,cq is grater then Mrb the compressed steel reinforcement is set. Theoretical section 1 (tensioned reinforcement only)
M 1 = M rb = 2.52 * 106 Nm
α1 =
x1 0.55 = = 0.488 d 1.125
zc = d − A1 =
x1 0.548 = 0.94m = 1.125 − 3 3
M1 2.52 = = 106.38cm² zc × σ s 0.94 × 252
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
M 2 = M ser , cq − M rb = (3.675 − 2.52) * 106 = 1.155 * 106 Nm Compressed reinforcement stresses:
σ sc = α e * σ c *
α1 *d − d ' α1 *d
σ sc = 15.82 *15 *
714
0.485 * 1.125 − 0.045 = 217.73MPa 0.485 * 1.125
ADVANCE DESIGN VALIDATION GUIDE
Compressed reinforcement area:
A' =
M2 1.155 = = 49.12cm² (d − d ' ) * σ sc (1.125 − 0.045) * 217.73
Complementary tensioned reinforcement area:
As = A' *
217.73 σ sc = 49.12 * = 42.44cm² 252 σs
Section area: Tensioned reinforcement: 106.38+42.44=148.82 cm2 Compressed reinforcement: 49.12 cm2 Reference solution for minimal reinforcement area The minimum percentage for a rectangular beam in pure bending is:
f ct , eff * bw * d 0.26 * = Max f yk 0.0013 * b * d w
As , min
According to: EC2 Part 1,1 EN 1992112002 Chapter 9.2.1.1(1); Note 2 Where:
f ct ,eff = f ctm = 2.56 MPa
from cracking conditions
Therefore:
As , min
2.56 * 0.65 * 1.125 = 9.73 * 10− 4 m² 0.26 * = max = 9.73cm² 500 0.0013 * 0.65 * 1.125 = 9.51 * 10− 4 m²
Finite elements modeling ■ ■ ■
Linear element: S beam, 15 nodes, 1 linear element.
715
ADVANCE DESIGN VALIDATION GUIDE
ULS and SLS load combinations(kNm) Simply supported beam subjected to bending ULS (reference value: 5255kNm)
SLS –Characteristic (reference value: 3675kNm)
SLS –Quasipermanent (reference value: 2303kNm)
Theoretical reinforcement area(cm2) (reference value: A’=148.82cm2 A=49.12cm2)
Minimum reinforcement area(cm2) (reference value: 9.73cm2)
716
ADVANCE DESIGN VALIDATION GUIDE
5.88.2.4 Reference results Result name
Result description
Reference value
My,ULS
My corresponding to the 101 combination (ULS) [kNm]
5255 kNm
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 [cm ]
148.82 cm2
Amin
Minimum reinforcement area [cm2]
9.73 cm2
5.88.3
2
Calculated results
Result name
Result description
Value
Error
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
0.0590 %
Amin
Amin
9.79662 cm²
0.1699 %
717
ADVANCE DESIGN VALIDATION GUIDE
5.89 EC2 / NF EN 199211/NA  France: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclined stressstrain diagram (evaluated by SOCOTEC France  ref. Test 4 I) Test ID: 4977 Test status: Passed
5.89.1
Description
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. The purpose of this test is to verify the software results for Pivot A efforts. For these tests, the constitutive law for reinforcement steel, on the inclined stressstrain 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
5.89.2
Background
This test was evaluated by the French control office SOCOTEC. This test performs the verification of the α 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:
xu ε cu 2 ε cu 2 .d =α = ⇒ x u = α .d = d ε ud + ε cu 2 ε ud + ε cu 2 The limit for α depends of the ductility class: ■
For a Class A steel: ε ud = 22.50 °
■
For a Class B steel: ε ud = 45 °
■
For a Class C steel: ε ud = 67.50 °
°°
°°
⇒ α u = 0.1346
⇒ α u = 0.072
°°
⇒ α u = 0.049
The purpose of this test is to verify the software results for Pivot A efforts. For these tests, it will be used the constitutive law for reinforcement steel, on the inclined stressstrain diagram.
S 500 A ⇒ σ su = 432,71 + 952,38.ε su ≤ 454 MPa
S 500 B ⇒ σ su = 432,71 + 727,27.ε su ≤ 466 MPa S 500C ⇒ σ su = 432,71 + 895,52.ε su ≤ 493MPa The Pivot efforts types are described below: ■ ■
718
Pivot A: Simple traction and simple bending or combined Pivot B: Simple or combined bending
ADVANCE DESIGN VALIDATION GUIDE
■
Pivot C: Combined bending with compression and simple compression
5.89.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992112002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■
Loadings from the structure: G = 25 kN/m+ dead load, Exploitation loadings (category A): Q = 50kN/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 ■ There will be considered a Class A reinforcement steel ■ The calculation will be made considering inclined stressstrain diagram The objective is to verify: ■ ■ ■
The stresses results The longitudinal reinforcement corresponding to Class A reinforcement steel ductility The minimum reinforcement percentage
719
ADVANCE DESIGN VALIDATION GUIDE
Simply supported beam
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■
Height: h = 0.90 m, Width: b = 0.50 m, Length: L = 5.80 m, 2 Section area: A = 0.45 m , Concrete cover: c = 4.00 cm Effective height: d = h  (0.6 * h + ebz)=0.806 m; d’ = ebz = 0.040 m
Materials properties Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■
Exposure class XC1 3 Concrete density: 25kN/m There will be considered a Class A reinforcement steel ductility The calculation will be made considering inclined stressstrain diagram Cracking calculation required
■
Concrete C25/30: fcd =
fck 25 = = 16,67MPa γc 1,5
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■
2/3 = 0.30 * 25 2 / 3 = 2.56MPa fctm = 0.30 * fck
According to: EC2 Part 1,1 EN 1992112002 Chapter 3.1.3(2); Table 3.1 ■
Steel S500 : f yd =
f yk γs
=
500 = 434,78MPa 1,15 According to: EC2 Part 1,1 EN 1992112002 Chapter 3.2
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.8) restrained in translation along Y and Z, and restrained rotation along X. Inner: None. ►
■
720
ADVANCE DESIGN VALIDATION GUIDE
Loading The beam is subjected to the following load combinations: 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
■
Load calculations:
M Ed =
123.94 * 5.80² = 521.16kNm 8
M Ecq =
86.25 * 5.80² = 362.68kNm 8
5.89.2.2 Reference results in calculating the concrete beam reduced moment limit For a S500B reinforcement steel, we have µ lu = 0.372 (since we consider no limit on the compression concrete to SLS). Reference reinforcement calculation: The calculation of the reinforcement is detailed below: ■ ■
Effective height: d=h(0.6*h+ebz)=0.806 m Calculation of reduced moment:
µ cu =
M Ed 0.52116 * 103 Nm = = 0,096 bw * d ² * f cd 0,50m * 0.806² m 2 * 16,67 MPa
■
The α value:
■
The condition: α cu < α u = 0.1346 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 ε ud = 22.50 ° , which gives the following available stress: °°
■
S500 A ⇒ σ su = 432,71 + 952,38 × 0.0225 = 454MPa
721
ADVANCE DESIGN VALIDATION GUIDE
The calculation of the concrete shortening:
ε cu = ε su *
0.127 α cu = 22.50 * = 3.27 ° °° 1 − α cu 1 − 0.127
It is therefore quite correct to consider a design stress of concrete equal to
f cd .
zc = d * (1 − 0,4 * α cu ) = 0.806 * (1 − 0.4 * 0.127) = 0.765m
Au =
M Ed 0.52116 = = 15.00cm² z c * f yd 0.765 * 454
Reference solution for minimal reinforcement area The minimum percentage for a rectangular beam in pure bending is:
f ct , eff * bw * d 0.26 * = Max f yk 0.0013 * b * d w
As , min
According to: EC2 Part 1,1 EN 1992112002 Chapter 9.2.1.1(1); Note 2 Where:
f ct ,eff = f ctm = 2.56 MPa
from cracking conditions
Therefore:
As , min
2.56 * 0.50 * 0.806 = 5.38 * 10− 4 m² 0.26 * = max = 5.38cm² 500 0.0013 * 0.50 * 0.806 = 5.24 * 10− 4 m²
Finite elements modeling ■ ■ ■
Linear element: S beam, 7 nodes, 1 linear element.
ULS and SLS load combinations (kNm) Simply supported beam subjected to bending
ULS (reference value: 521.16kNm)
722
ADVANCE DESIGN VALIDATION GUIDE
SLS (reference value: 362.68kNm)
Theoretical reinforcement area (cm2) For Class A reinforcement steel ductility (reference value: A=15.00cm2)
Minimum reinforcement area(cm2) (reference value: 5.38cm2)
5.89.2.3 Reference results Result name
Result description
Reference value
My,ULS
My corresponding to the 101 combination (ULS) [kNm]
521.16 kNm
My,SLS
My corresponding to the 102 combination (SLS) [kNm]
362.68 kNm
Az
Theoretical reinforcement area (Class A) [cm2]
15.00 cm2
Amin
Minimum reinforcement area [cm2]
5.38 cm2
5.89.3
Calculated results
Result name
Result description
Value
Error
My
My USL
521.125 kN*m
0.0000 %
My
My SLS
362.657 kN*m
0.0000 %
Az
Az Class A
14.9973 cm²
0.0000 %
Az
Az Class B
14.9973 cm²
0.7491 %
Amin
Amin
5.37514 cm²
0.0000 %
723
ADVANCE DESIGN VALIDATION GUIDE
5.90 EC2 / NF EN 199211  France: Reinforcement calculation on intermediate support on a continuous beam Test ID: 6277 Test status: Passed
5.90.1
Description
The purpose on this test is to calculate longitudinal and transversal reinforcement on intermediate support on a 2 spans beam.
5.90.2
Background
Verifies the reinforcement values on intermediate support for a continuous beam.
5.90.2.1 Model description ■ Analysis type: static linear (plane workspace); ■ Element type: linear. The following load cases and load combination are used: ■ ■ ■
Loadings from the structure: G = 20 kN/m + self weight Exploitation loadings (category A): Q1 = 30 kN on the first span and Q2 = 30 kN/m on the second span. The ultimate limit state (ULS) combination is: Cmax 1.35 x G + 1.5 x Q1 + 1.5 x Q2
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■
Height: h = 0.70 m, Width: b = 0.25 m, Length: L = 12.00 m, Concrete cover: cnom=3 cm
■
Reinforcement S500, Class: B, ss=500 MPa – Bilinear stressstrain diagram
■
Fck=25 MPa
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0 m) hinged connection, ► Support at end point (x=12 m) hinged connection, Inner: Hinged support at x = 6 m. ►
■
724
ADVANCE DESIGN VALIDATION GUIDE
5.90.2.2 Reference results in calculating the concrete beam Calculations of top longitudinal reinforcement area
Calculations of transversal reinforcement area
Finite elements modeling ■ ■ ■ ■
Linear element: S beam, 3 supports, 13 nodes, 1 linear element.
5.90.2.3 Reference results Result name
Result description
Reference value
Az
Top longitudinal reinforcement [cm2]
14.3 cm2
Atz
Transversal reinforcement [cm²]
11.8 cm²
5.90.3
Calculated results
Result name
Result description
Value
Error
Az
Top longitudinal reinforcement bars
14.4252 cm²
0.8755 %
Atz
Transversal reinforcement
11.8481 cm²
0.4076 %
725
ADVANCE DESIGN VALIDATION GUIDE
5.91 EC3 / CSN EN 199211  Czech Republic: Verification of stresses in steel and concrete on simply supported beam Test ID: 6309 Test status: Passed
5.91.1
Description
The objective of this test is to verify design conditions of stress limitation in reinforcement concrete beam.
5.91.2
Background
Verification of stresses in compressed concrete and steel reinforcement. Verifies values of stresses in compressed concrete for quasipermanent combination and steel reinforcement for characteristic combination on simple supported beam made of concrete C20/25 with defined reinforcement 5Φ16mm of steel B500B. The beam is in environment exposure class XC1. Bilinear stressstrain diagram for reinforcement.
5.91.2.1 Model description ■ Reference: Manually calculated example ■ Analysis type: static linear (plane problem) ■ Element type: linear. Loading and combinations: ■ ■ ■ ■ ■ ■ ■
726
Characteristic value of dead load (including self weight): g = 7.0 kN/m, Characteristic value of longterm live load: qk = 5.0 kN/m, Characteristic value of leading live load: qk1 = 14.5 kN/m, Characteristic value of accompanying live load: qk2 = 5.0 kN/m, Quasipermanent combination: ► qQP = 1.0*g + ψ2*qk = 7.0 + 0.8*5.0 = 11.0 kN/m Characteristic combination: ► qCQ = 1.0*g +qk1 + ψ0*qk2 = 7.0 + 14.5 + 0.7*7.0 = 26.4 kN/m Humidity RH = 50%; t0 = 28 days
ADVANCE DESIGN VALIDATION GUIDE
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■ ■
Height: h = 450 mm, Width: b = 250 mm, Length: L = 6000 mm, 3 2 Section area: AC = 112.5*10 mm , Considered reinforcement: 5*Φ16mm; As = 1006*106, Reinforcement center of gravity position: d´ = 38 mm, Effective height: d = 412 mm
Materials properties Concrete: C20/25 ■ ■ ■ ■
3 Weight: 2500 kg/m , fck = 20 MPa, fctm = 2,2 MPa, fcm = 28 MPa,
■
𝐸𝑐𝑚 = 30.0 𝐺𝑃𝑎
Reinforcement: B500B ■ ■
fyk = 500 MPa Es = 200 GPa
Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (Z=0), hinged connection, Support at end point (Z = 6.0) hinged connection restrained to rotate around Z Inner: none
► ►
■
Design loads In the calculation following design load are considered: 1 1 ∗ 𝑞 ∗ 𝐿2 = ∗ 11.0 ∗ 62 = 0.0495 𝑀𝑁𝑛 8 𝑄𝑃 8 1 1 = ∗ 𝑞𝐶𝑄 ∗ 𝐿2 = ∗ 26.4 ∗ 62 = 0.1188 𝑀𝑁𝑛 8 8
𝑀𝑄𝑃 = 𝑀𝐶𝑄
727
ADVANCE DESIGN VALIDATION GUIDE
5.91.2.2 Calculation of concrete creep coefficient 𝛽(𝑓𝑐𝑚 ) = 𝛽(𝑡0 ) =
ℎ0 =
16.8
�𝑓𝑐𝑚
=
16.8 √28
= 3.18 𝑀𝑃𝑎
According to: CSN EN199211 Chapter B.1(1), Formula (B.4)
1 1 = = 0.488 0.1 + 𝑡00.20 0.1 + 280.20
According to: CSN EN199211 Chapter B.1(1), Formula (B.5)
2 ∗ 𝐴𝑐 2 ∗ 0.45 ∗ 0.25 = = 0.1607 𝑚 = 160.7 𝑚𝑚 𝑢 2 ∗ (0.45 + 0.25)
According to: CSN EN199211 Chapter B.1(1), Formula (B.6)
𝜑𝑅𝐻 = 1 +
𝑓𝑐𝑚 = 28 𝑀𝑃𝑎 < 35.0 𝑀𝑃𝑎
1 − 𝑅𝐻/100 3
0.1 ∗ �ℎ0
=1+
1 − 0.5 3
0.1 ∗ √160.7
= 1.92
According to: CSN EN199211 Chapter B.1(1), Formula (B.3a)
𝜑(∞, 𝑡0 ) = 𝜑𝑅𝐻 ∗ 𝛽(𝑓𝑐𝑚 ) ∗ 𝛽(𝑡0 ) = 1.92 ∗ 3.18 ∗ 0.488 = 2.98
According to: CSN EN199211 Chapter B.1(1), Formula (B.2)
𝐸𝑐,𝑒𝑓𝑓 =
𝐸𝑐𝑚 30 = = 7.54 1 + 𝜑(∞, 𝑡0 ) 1 + 2.98
𝛼𝑒 =
According to: CSN EN199211 Chapter 7.4.3(3), Formula (7.20)
𝐸𝑠 200 = = 26.53 𝐸𝑐,𝑒𝑓𝑓 7.54
Neutral axis position 𝑥=
728
−𝛼𝑒 ∗ 𝐴𝑠 + �𝛼𝑒2 ∗ 𝐴2𝑠 + 2 ∗ 𝑏 ∗ 𝛼𝑒 ∗ 𝑑 ∗ 𝐴𝑠 𝑏 −26.53 ∗ 1005 ∗ 10−6 + �26.532 ∗ (1005 ∗ 10−6 )2 + 2 ∗ 0.25 ∗ 26.53 ∗ 0.412 ∗ 1005 ∗ 10−6 = = 0.2084 𝑚 0.25 1 1 𝐼𝑢 = ∗ 𝑏 ∗ 𝑥 3 + 𝐴𝑠 ∗ 𝛼𝑒 ∗ (𝑑 − 𝑥)2 = ∗ 0.25 ∗ 0.20843 + 1005 ∗ 10−6 ∗ 26.57 ∗ (0.412 − 0.2084)2 = 0.00186 𝑚4 3 3
ADVANCE DESIGN VALIDATION GUIDE
5.91.2.3 Verification of stresses Verification of stresses in compressed concrete For quasipermanent combination:
𝜎𝑐 =
𝜎𝑐 ≤ 0.45 ∗ 𝑓𝑐𝑘 = 0.45 ∗ 20 = 9.0 𝑀𝑝𝑎
𝑀𝑄𝑃 0.0495 ∗𝑥 = ∗ 0.2084 = 5.54 𝑀𝑃𝑎 < 9.0 𝑀𝑃𝑎 𝐼𝑢 0.00186
According to: CSN EN199211 Chapter 7.2(3)
Verification of stresses in reinforcement steel For characteristic combination:
𝜎𝑠 = 𝛼𝑒 ∗
𝜎𝑠 ≤ 0.8 ∗ 𝑓𝑦𝑘 = 0.8 ∗ 500 = 400 𝑀𝑃𝑎
𝑀𝐶𝑄 0.1188 ∗ (𝑑 − 𝑥) = 26.53 ∗ ∗ (0.412 − 0.2084) = 334.62 𝑀𝑃𝑎 𝐼𝑢 0.00186
According to: CSN EN199211 Chapter 7.2(5)
Finite elements modeling ■ ■ ■
Linear element: beam, 7 nodes, 1 linear element.
FEM Results Bending moment for Serviceability Limit State  Quasipermanent combination (reference value 49.5 kNm)
Bending moment for Serviceability Limit State  Characteristic combination (reference value 118.8 kNm)
Concrete design results Stress in compressed concrete for Quasipermanent combination (reference value 5.54 MPa)
729
ADVANCE DESIGN VALIDATION GUIDE
Stress in reinforcement steel for Characteristic combination (reference value 344.62 MPa)
5.91.2.4 Reference results Result name
Result description
Reference value
My
Bending moment for Quasipermanent combination [kNm]
49.5 kNm
My
Bending moment for Characteristic combination [kNm]
118.8 kNm
SCqp
Stress in compressed concrete for QP combination [MPa]
5.54 MPa
SScq
Stress in reinforcement for CQ combination [MPa]
344.62 MPa
5.91.3
730
Calculated results
Result name
Result description
Value
Error
My
Bending moment for quasipermanent combination
49.5 kN*m
0.0000 %
My
Bending moment for characteristic combination
118.8 kN*m
0.0000 %
Sc z QP
Stresses in concrete for quasipermanent combination
5.68905 MPa
2.6904 %
Ss z CQ
Stresses in steel for characteristic combination
337.858 MPa
1.9622 %
ADVANCE DESIGN VALIDATION GUIDE
5.92 EC2 / NF EN 199211  France: Tie reinforcement Test ID: 6260 Test status: Passed
5.92.1
Description
The purpose of this test is to find the minimal and longitudinal reinfocement values on a tie element.
5.92.2
Background
Reinforcement area on tie element Verifies the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction.
5.92.2.1 Model description ■ Reference: Pratique de l’Eurocode 2 – Jean Roux; ■ Analysis type: static linear (plane workspace); ■ Element type: linear. The following load cases and load combination are used: ■
Loadings from the structure: Fx,G = 100 kN The dead load is neglected
■ ■
Exploitation loadings (category A): Fx,Q = 40 kN The ultimate limit state (ULS) combination is: Ned = 1.35 x G + 1.5 x Q
Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■
Height: h = 0.20 m, Width: b = 0.20 m, Length: L = 2.00 m, 2 Section area: A = 0.04 m , Concrete cover: cnom=3 cm
■
Reinforcement S500, Class: A, ss=500 MPa – Inclinated stressstrain diagram
■ ■
Fck=25 MPa The load combinations will produce the following efforts:
731
ADVANCE DESIGN VALIDATION GUIDE
NEd=1.35*100+1.5*40=195 kN Nser=1*100+1*40=140 kN Boundary conditions The boundary conditions are described below: ■
Outer: Support at start point (x=0) fixed connection, Inner: None.
►
■
5.92.2.2 Reference results in calculating the concrete column Calculations according a inclined stressstrain diagram
S 500 − ClassA ⇒ σ s ,U = 454 MPa As ,U ≥
N Ed
σ s ,U
=
0.195 = 4.30 *10 −4 m² = 4.3cm² 454
Checking the condition of nonfragility:
As ≥ Ac *
f ctm f yk
f ctm = 0.30 * f ck2 / 3 = 0.30 * 252 / 3 = 2,56 MPa
Ac = 0.20 * 0.20 = 0.04m² As ≥ Ac *
f ctm 2.56 = 0.04 * = 2.05cm² 500 f yk
Finite elements modeling ■ ■ ■
732
Linear element: S beam, 4 nodes, 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
ULS load combinations Using the inclined stressstrain diagram, the reinforcement will result: (Ay=2.146 cm2=2*1.073 cm2)
Using the inclined stressstrain diagram, the reinforcement will result: (Az=2.146 cm2=2*1.073 cm2)
So total reinforcement area is 4.292 cm².
733
ADVANCE DESIGN VALIDATION GUIDE
5.92.2.3 Reference results Result name
Result description
Reference value
As
Longitudinal reinforcement obtained using the inclined stressstrain diagram [cm2]
4.30 cm2
5.92.3
734
Calculated results
Result name
Result description
Value
Error
Fx
Ned
195 kN
0.0000 %
Fx
Nser
140 kN
0.0000 %
Amin
As,min
2.05197 cm²
0.0961 %
Ay
Ayb
1.07346 cm²
0.1433 %
Az
Azb
1.07346 cm²
0.1433 %
ADVANCE DESIGN VALIDATION GUIDE
5.93 NTC 2008  Italy: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement Test ID: 6339 Test status: Passed
5.93.1
Description
Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram is generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max., will be calculated, along with the crosssectional area of the shear reinforcement, Asw, and the theoretical reinforcement. For the calculation, the reduced shear force values will be used
5.93.2
Background
5.93.2.1 Model description Below are described the beam cross section properties: ■ ■ ■ ■ ■ ■
Height: h = 0.70 m, Width: b = 0.35 m, Length: L = 5.75 m, 2 Section area: A = 0.24 m Concrete cover: ebz = 3.50 cm Effective height: d = h  (0.6 * h + ebz)=0.623 m; d’ = ehz = 0.035 m
■
Stirrup slope:
■
Strut slope:
Units Metric System Materials properties Rectangular solid concrete section C25/30 The design value is given by the formula 4.1.1 according NTC2008 Chapter 4, where the safety factor is 1.5, and the coefficient "alpha" represents the reduction factor for the long duration resistance; its value is 0.85.
735
ADVANCE DESIGN VALIDATION GUIDE
The steel reinforcement is B450C according NTC 2008 Chapter 11.3.2.1, with these properties:
fwyk the charateristic yelding strenght and γs = 1,15 according to formula 4.1.4 from NTC 2008 Chapter 4.1.2.1.1.2
Boundary conditions ■ ■
Hinged at the extremity (x = 0.00 m) Restrained in translation on y and z axes and restrained in rotation on x axis in the end (x = 5.75 m).
Loading The beam is stressed by the following loading condition: Dead load G1k is represented by a linear load of 40.00 kN/m The live load is considered from one linear load Q2k = 25 kN/m For ULS combination we have the following load combination according NTC 2008 Chapter 2.5.3: ► ►
Fd SLU = 1.3 x 40.00 kN/m + 1.5 x 25.00 kN/m = 89.5 kN/m Load calculation:
5.93.2.2 Reference results The maximum shear resistance value has to be bigger than the one which stresses the beam according the 4.1.17 equivalence from NTC 2008 Chapter 4.
The maximum shear resistance is given by the formula 4.1.19 from NTC 2008 Chapter 4.
736
ADVANCE DESIGN VALIDATION GUIDE
Where: “d” is the effective height of the section “bw” is the minimum width of the section is a coefficent defined as follow
“ f’cd ” rapresents the reduced design compressive strenght (f’cd = 0.5 x fcd) The maximum shear resistance will be VRd = 0.9 x 0.623 x 0.35 x 1 x (0.5 x 14.17) x (ctg90° + ctg45°) / (1 + ctg245°) = 695.19 kN VEd = 257.31 kN < VRd = 695.19 kN Calculation of the reduced shear force and transversal reinforcement When the shear force curve has no discontinuities, the EC2 allows to consider, as a reduction, the shear force to a horizontal axis
where z is considered 0.9 x d = 0.560.
In the case of a member subjected to a distributed load, the equation of the shear force is:
Therefore:
So the transversal reinforcement area will be (alpha = 90°):
Result name
Result description
Reference value
Fz
Shear force
257.31 kN
Atz
Transversal reinforcement area on z axes
9.40 cm2
5.93.3
Calculated results
Result name
Result description
Value
Error
Fz
Shear force
257.312 kN
0.0008 %
Atz
Transversal reinforcement area
9.38222 cm²
0.1891 %
737
ADVANCE DESIGN VALIDATION GUIDE
5.94 NTC 2008  Italy: Verifying a rectangular concrete beam subjected to a uniformly distributed load Test ID: 6311 Test status: Passed
5.94.1
Description
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage
5.94.2
Background
5.94.2.1 Model description Below are described the beam cross section properties: ■
Height: h = 0.90 m,
■
Width: b = 0.50 m,
■
Length: L = 5.80 m,
■
Section area: A = 0.45 m2
■
Concrete cover: c = 4.00 cm (for the concrete cover of the bottom / top external fiber measured along the local z axis, Advance Design is having the following notation: ebz = 4.00 cm; ehz = 4.00 cm Effective height: d = h  (0.06* h + ebz)=0.810 m; d’ = ehz= 0.040 m
■
Units Metric System Materials properties Concrete class C25/30 is used. Design compressive strength value is given by the formula 4.1.1 according NTC2008 Chapter 4 where the safety factor is 1.5, and the coefficient "alpha" represents the reduction factor for the long duration resistance, its value is 0.85.
738
ADVANCE DESIGN VALIDATION GUIDE
The steel reinforcement is B450C according NTC 2008 Chapter 11.3.2.1 with these properties:
fyk the characteristic yielding strength and γs = 1,15 according formula 4.1.4 NTC 2008 Chapter 4.1.2.1.1.2
Boundary conditions ■
Hinged at end (x = 0.00 m)
■
Restrained in translation on y and z axes and restrained in rotation on x axis (x = 5.80 m).
Loading The beam is stressed by the following loading condition: ■
3 Dead load G1k = 0.9 m x 0.5 m x 25 kN/m = 11.25 kN/m
■
Permanent load G2k = 25.00 kN/m
■
Live load Q3k = 13.20 kN/m
For ULS combination we have the following load combination according NTC 2008 Chapter 2.5.3:
Fd SLU = 1.3 x 11.25 kN/m + 1.5 x 25 kN/m + 1.5 x 50 kN/m = 127.12 kN/m For SLS charateristic combination we have the following load combination according NTC 2008 Chapter 2.5.3:
Fd SLS = 1 x 11.25 kN/m + 1 x 25 kN/m + 1 x 50 kN/m = 86.25 kN/m Load calculation:
739
ADVANCE DESIGN VALIDATION GUIDE
5.94.2.2 Reference results Calculation method used to obtain the reference solution We assumed the limit reduced moment (
) as 0.327 in the limit between the field 3 and field 4 with
= 0.625
Where: Effective height: d = h  (0.06* h + ebz) = 0.810 m
Mechanical reinforcement ratio
Reduced moment
Neutral axis position Calculation of the reduced moment MEd = 534 kNm
In our case we can find the
740
= 0.115 < 0.327, so we don't have to put compressed reinforcement, knowing the reduced moment parameter which defines the mechanical tensioned reinforcement ratio following the table:
ADVANCE DESIGN VALIDATION GUIDE
So the required reinforcement is given by the relationship
So we use 6 ø20 on the bottom of the beam > As = 18.85 cm2 We calculate the new mechanical reinforcement ratio with the real reinforcement
2 = 1885 mm x 391.30 / 500 x 810 x 14.17 = 0.128
The reinforcement ratio value resistance moment will be:
is 0.128
= 0.118 (found by interpolation on the table above) so the
= 0.118 x 0.50 x 0.8102 x 14.17 = 548.51 kNm > MEd = 534.53 kNm The minimum reinforcement area is defined by the formula (4.1.43) from NTC 2008:
Where:
Result name
Result description
Reference value
As
Longitudinal reinforcement area on the middle point of the beam Minimum reinforcement area
18.03 cm2
As,min
5.94.3
5.97 cm2
Calculated results
Result name
Result description
Value
Error
Az
Longitudinal reinforcement
18.0679 cm²
0.2102 %
Amin
Minimum reinforcement area
5.97238 cm²
0.0000 %
741
ADVANCE DESIGN VALIDATION GUIDE
5.95 EC2 / CSN EN 199211  Czech Republic: Stresses and cracks verification for simple supported beam Test ID: 6352 Test status: Passed
5.95.1
Description
Verifies the reinforced concrete rectangular crosssection to resist defined loads on simple bending. During this test are verified stresses in concrete section and steel reinforcement, the maximal spacing of cracks, difference between mean strain in concrete and steel and crack width.
5.95.2
Background
Verifies stresses and crack width on simple supported beam made of concrete C20/25, considering real longitudinal reinforcement 5 Φ 16 mm. Profile´s length is 6m, cross section dimensions are 0.25 x 0.45 m. Self weight is not taken into account.
5.95.2.1 Model description ■ ■ ■ ■
Reference: manually calculated reference example Analysis type: static linear (plane problem) Element type: linear Load cases: Dead load: g = 7.0 kN/m Live load: q = 12.0 kN/m (Category E: Storages; ψ0 = 1.0, ψ2 = 0.8)
Units Metric System Geometry Below are described beam´s cross section characteristics: ■ ■ ■ ■ ■ ■ ■
742
Height: h = 0.45 m Width: b = 0.25 m Section area: Ac = 0.1125 m2 Length: l = 6.0 m 6 2 Reinforcement 5 Φ 16 mm, As = 1005*10 m Concrete cover: c = 0.03 m Effective height: d = 0.038 m
ADVANCE DESIGN VALIDATION GUIDE
Materials properties Concrete C20/25 and reinforcement steel B500A are used. Following characteristics are used in relation to these material(s): ■ ■ ■ ■ ■ ■ ■ ■ ■
Characteristic compressive cylinder strength of concrete at 28 days: fck = 20 MPa Mean axial tensile strength: fctm = 2.2 MPa Mean secant modulus of elasticity of concrete: Ecm = 30000 MPa Characteristic yield strength of reinforcement: fyk = 500 MPa Young´s modulus of reinforcement: Es = 200000 MPa Exposure class: XD1 Stressstrain law for reinforcement: Bilinear stressstrain diagram Concrete age: t0 = 28 days Relative humidity: RH = 50 % Boundary conditions Boundary conditions are described below: ■ Outer: Hinge at start point (X = 0); restrained translation along X, Y and Z axis Hinge at end point (X = 6.0 m); restrained in translation along Y and Z, in rotation around X axis ■ Inner: None. ► ►
Loading The beam is subjected to the following load combinations and actions: ■
The ultimate limit state (ULS) combination is: CSTR = 1.35 * g + 1.5 * q = 1.35 * 7.0 kN/m + 1.5 * 12.0 kN/m = 27.45 kN/m MSTR =
1 1 C * l 2 = 27.45 * 6 2 = 123.53 kNm 8 STR 8
■ The serviceability limit state (SLS) quasipermanent combination is: CQP = 1 * g + ψ2 * q = 1 * 7.0 kN/m + 0.8 * 12.0kN/m = 10.0 kN/m MQP =
1 1 C * l 2 = 10.0 * 6 2 = 74.70 kNm 8 QP 8
■ The serviceability limit state (SLS) characteristic combination is: CCQ = 1 * g + ψ0 * q = 1 * 7.0 kN/m + 1 * 12.0kN/m = 19.0 kN/m MCQ =
1 1 C CQ * l 2 = 19.0 * 6 2 = 85.50 kNm 8 8
5.95.2.2 Concrete final value of creep coefficient β( fcm ) =
16.8 fcm
=
16.8 28
= 3.17 MPa According to: CSN EN199211 Chapter B.1(1), Formula (B.4)
fcm = fck + 8 = 20 + 8 = 28 MPa
β( t 0 ) =
1
0.1 + t 00.20
=
1
0.1 + 28 0.20
= 0.488 According to: CSN EN199211 Chapter B.1(1), Formula (B.5)
Fcm = 28 MPa < 35.0 MPa
φRH
RH 50 1100 100 = 1+ = 1+ = 1.92 0 .1 * 3 h 0 0.1 * 3 160.71 1
According to: CSN EN199211 Chapter B.1(1), Formula (B.3a)
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ADVANCE DESIGN VALIDATION GUIDE
h0 =
2 * A c 2 * 0.25 * 0.45 = 160.71 mm = 2(0.25 + 0.45 ) u
According to: CSN EN199211 Chapter B.1(1), Formula (B.6)
φ(∞ , t 0 ) = φRH * β( fcm ) * β( t 0 ) = 1.92 * 3.17 * 0.488 = 2.97 According to: CSN EN199211 Chapter B.1(1), Formula (B.2)
αe =
Es 200000 = = 26.47 E cm 30000 1+ φ(∞ , t 0 ) 1+ 2.97
5.95.2.3 Geometric characteristics of noncracked crosssection Ideal crosssection calculation Ideal crosssection area:
A i = A c + α e * A s = 0.1125 + 26.47 * 1005 * 10 6 = 0.1391 m 2 Ideal crosssection centre of gravity distance from bottom surface:
a gi =
A c * a c + α e * A s * d 0.1125 * 0.45 * 0.225 + 26.47 * 1005 * 10 6 * 0.412 = = 0.2310 m Ai 0.1391
Ideal crosssection moment of inertia to its center of gravity: Ii = Ic + A c (a gi  a c ) 2 + α e * A s (d  a gi ) 2 Ii =
1 0.25 * 0.45 3 + 0.1125 (0.231  0.225) 2 + 26.47 * 1005 * 10  6 (0.412  0.231) 2 = 0.0067 m 4 12
Bending moment for cracks verification
Mcr,lt = fctm *
Ii 0.0067 = 2 .2 * = 67.5 kNm < MQP = 74.7 kNm => cracks are expected h  a gi 0.45  0.231
5.95.2.4 Geometric characteristics cracked crosssection Distance of neutral axis from compressed surface
x=
αe 2 * b * As * d 26.47 2 * 0.25 * 1005 * 10 6 * 0.412 A s (1+ 1+ )= 1005 * 10  6 [1+ 1+ ] = 0.208 m b αe * A s 0.25 26.47 * (1005 * 10  6 ) 2
Moment of inertia of cracked cross/section Iu =
1 1 b * x 3 + α e * A s (d  x)2 = 0.25 * 0.208 3 + 26.47 * 1005 * 10  6 (0.412  0.208) 2 = 0.001857 m 4 3 3
5.95.2.5 Reference results in calculating stresses in concrete and reinforcement Verification of stress in compressed concrete for quasipermanent combination
σc =
MQP 0.0704 = = 8.37 MPa < 0.45 * fck = 0.45 * 20 = 9.0 MPa Iu 0.001857
According to: CSN EN199211 Chapter 7.2(3) Verification of stress in reinforcement steel for characteristic combination
σ s = αe
MCQ 0.0855 (0.412  0.208 ) = 248.62 MPa < 0.8 * f yk = 0.8 * 500 = 400.0 MPa (d  x ) = 26.47 0.001857 Iu According to: CSN EN199211 Chapter 7.2(5)
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ADVANCE DESIGN VALIDATION GUIDE
5.95.2.6 Reference results in calculating crack width Calculation of maximal spacing of cracks Longitudinal bottom reinforcement is 5 Φ 16 mm, As = 1005*106 m2.
A c,eff
2.5 * (h  d) 2.5 * (0.45  0.412) = 0.095 m (h  x) (0.45  0.412) = b * min = 0.25 * min = 0.087 m = 0.25 * 0.087 = 0.020167 m 2 3 3 0.5 * h 0.5 * 0.45 = 0.225 m
According to: CSN EN199211 Chapter 7.3.2(3) ρp,eff =
6
As 1005 * 10 = = 0.0498 A c,eff 0.020167
According to: CSN EN199211 Chapter 7.3.4(2), Formula (7.10) Concrete cover c = 0.036 m (considering stirrups Φ 6 mm). k1 = 0.8; k2 = 0.5; k3 = 3.4; k4 = 0.425; kT = 0.4
s r,max = k 3 * c +
k 4 * k1 * k 2 * Φ 0.425 * 0.8 * 0.5 * 0.016 = 3.4 * 0.036 + = 0.1770 m ρp,eff 0.0498 According to: CSN EN199211 Chapter 7.3.4(3), Formula (7.11)
Calculation of average strain Reinforcement stress for quasipermanent combination:
σ s = αe
αe =
MQP 0.0747 (0.412  0.208 ) = 217.22 MPa (d  x ) = 26.47 0.001857 Iu
Es 200 = = 6.67 E cm 30 σs  kt
ε sm  ε cm =
fct,eff ρp,eff
(1+ α e * ρp,eff ) Es
=
217.22  0.4
2.2 (1+ 6.67 * 0.0498) 0.0498 = 9.68 * 10 4 200 * 10 3
According to: CSN EN199211 Chapter 7.3.4(2), Formula (7.9) Calculation of crack width w k = s r,max (ε sm  ε cm ) = 0.1770 * 9.68 * 10 4 = 0.1716 mm According to: CSN EN199211 Chapter 7.3.4(1), Formula (7.8) According to CSN EN199211 Chapter 7.3.1(5), Table 7.1N, for exposure class XD1 the value of crack width must be smaller than 0.3 mm.
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ADVANCE DESIGN VALIDATION GUIDE
Finite elements modeling ■ ■ ■
Linear element: Beam, 7 nodes, 1 linear element.
FEM calculation – bending moment results ULS combination (reference value: MSTR = 123.53 kNm)
SLS quasipermanent combination (reference value: MQP = 74.70 kNm)
SLS characteristic combination (reference value: MCQ = 85.50 kNm)
Compressing stresses in concrete and reinforcement Compressing stress in concrete for quasipermanent combination (reference value: σc =8.37 MPa)
Compressing stress in reinforcement for characteristic combination (reference value: σscq =248.62 MPa)
Compressing stress in reinforcement for quasipermanent combination (reference value: σsqp =217.22 MPa)
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ADVANCE DESIGN VALIDATION GUIDE
Crack calculation results Maximal spacing of cracks (reference value: sr,max = 0.177 m)
4 Average strain in steel and concrete (reference value: εsm – εcm = 9.68 * 10 )
Crack width (reference value: wk = 0.1716 mm)
5.95.2.7 Reference results Result name
Result description
Reference value
MySTR
Bending moment from ULS load combination [kNm]
123.53 kNm
MyQP
Bending moment from quasipermanent combination [kNm]
74.70 kNm
MyCQ
Bending moment from characteristic combination [kNm]
85.50 kNm
σc
Stress in concrete for quasipermanent combination [MPa]
8.37 MPa
σscq
Stress in reinforcement for quasiperm. combination [MPa]
248.62 MPa
σsqp
Stress in reinforcement for characteristic combination [MPa]
217.22 MPa
sr,max
Maximal spacing of cracks [m]
0.177 m
εsm – εcm
Average strain in steel and concrete []
9.68 * 104
wk
Crack width [mm]
0.1716 mm
5.95.3
Calculated results
Result name
Result description
Value
Error
Sc z QP
Concrete stress QP
8.58529 MPa
2.5722 %
Ss z CQ
Steel stress CQ
250.516 MPa
0.7626 %
Sr,max z
Max crack
177.306 mm
0.1729 %
Îµsm  Îµcm z
Stresses difference
0.000982881 adim
1.5373 %
wk z
Crack width
0.174271 mm
1.5565 %
My
My STR
123.525 kN*m
0.0000 %
My
My QP
74.7 kN*m
0.0000 %
My
My CQ
85.5 kN*m
0.0000 %
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6 General applications
ADVANCE DESIGN VALIDATION GUIDE
6.1
Verifying element creation using commas for coordinates (TTAD #11141) Test ID: 4554 Test status: Passed
6.1.1 Description Verifies element creation using commas for coordinates in the command line.
6.2
Modifying the "Design experts" properties for concrete linear elements (TTAD #12498) Test ID: 4542 Test status: Passed
6.2.1 Description Defines the "Design experts" properties for a concrete (EC2) linear element in analysis model and verifies properties from descriptive model.
6.3
Verifying the precision of linear and planar concrete covers (TTAD #12525) Test ID: 4547 Test status: Passed
6.3.1 Description Verifies the linear and planar concrete covers precision.
6.4
Defining the reinforced concrete design assumptions (TTAD #12354) Test ID: 4528 Test status: Passed
6.4.1 Description Verifies the definition of the reinforced concrete design assumptions.
6.5
Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238) Test ID: 4480 Test status: Passed
6.5.1 Description Performs the finite elements calculation on a model with 2 joined vertical elements with the clipping option enabled.
6.6
Creating system trees using the copy/paste commands (DEV2012 #1.5) Test ID: 4164 Test status: Passed
6.6.1 Description Creates system trees using the copy/paste commands in the Pilot. The source system consists of several subsystems. The target system is in the source system.
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ADVANCE DESIGN VALIDATION GUIDE
6.7
Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724) Test ID: 4361 Test status: Passed
6.7.1 Description Generates liquid pressure on a concrete structure with horizontal and vertical surfaces. Generates the loadings description report.
6.8
Changing the default material (TTAD #11870) Test ID: 4436 Test status: Passed
6.8.1 Description Selects a different default material for concrete elements.
6.9
Creating and updating model views and postprocessing views (TTAD #11552) Test ID: 3820 Test status: Passed
6.9.1 Description Creates and updates the model view and the postprocessing views for a simple steel frame.
6.10 Importing a cross section from the Advance Steel profiles library (TTAD #11487) Test ID: 3719 Test status: Passed
6.10.1
Description
Imports the Canam Z 203x10.0 section from the Advance Steel Profiles library in the Advance Design list of available cross sections.
6.11 Verifying the synthetic table by type of connection (TTAD #11422) Test ID: 3646 Test status: Passed
6.11.1
Description
Performs the finite elements calculation and the steel calculation on a structure with four types of connections. Generates the "Synthetic table by type of connection" report. The structure consists of linear steel elements (S275) with CE505, IPE450, IPE140 and IPE500 cross section. The model connections: columns base plates, beam  column fixed connections, beam  beam fixed connections and gusset plate. Live loads, snow loads and wind loads are applied.
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ADVANCE DESIGN VALIDATION GUIDE
6.12 Verifying mesh, CAD and climatic forces  LPM meeting Test ID: 4079 Test status: Passed
6.12.1
Description
Generates the "Description of climatic loads" report for a model consisting of a concrete structure with dead loads, wind loads, snow loads and seism loads. Performs the model verification, meshing and finite elements calculation. Generates the "Displacements of linear elements by element" report.
6.13 Verifying the appearance of the local x orientation legend (TTAD #11737) Test ID: 4109 Test status: Passed
6.13.1
Description
Verifies the color legend display. On a model with a planar element, the color legend is enabled; it displays the elements by the local x orientation color.
6.14 Creating system trees using the copy/paste commands (DEV2012 #1.5) Test ID: 4162 Test status: Passed
6.14.1
Description
Creates system trees using the copy/paste commands in the Pilot. The source system consists of several subsystems. The target system is on the same level as the source system.
6.15 Verifying the objects rename function (TTAD #12162) Test ID: 4229 Test status: Passed
6.15.1
Description
Verifies the renaming function from the properties list of a linear element. The name contains the "_" character.
6.16 Launching the verification of a model containing steel connections (TTAD #12100) Test ID: 4096 Test status: Passed
6.16.1
Description
Launches the verification function for a model containing a beamcolumn steel connection.
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ADVANCE DESIGN VALIDATION GUIDE
6.17 Creating a new Advance Design file using the "New" command from the "Standard" toolbar (TTAD #12102) Test ID: 4095 Test status: Passed
6.17.1
Description
Creates a new .fto file using the "New" command from the "Standard" toolbar. Creates a rigid punctual support and tests the CAD coordinates.
6.18 Verifying geometry properties of elements with compound cross sections (TTAD #11601) Test ID: 3546 Test status: Passed
6.18.1
Description
Verifies the geometry properties of a steel structure with elements which have compound cross sections (CS4 IPE330 UPN240). Generates the geometrical data report.
6.19 Verifying material properties for C25/30 (TTAD #11617) Test ID: 3571 Test status: Passed
6.19.1
Description
Verifies the material properties on a model with a concrete (C25/30) bar. Generates the material description report.
6.20 Verifying the overlapping when a planar element is built in a hole (TTAD #13772) Test ID: 6214 Test status: Passed
6.20.1
Description
Verifies that when a planar element is built in a hole of another planar element by checking the mesh.
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7 Import / Export
ADVANCE DESIGN VALIDATION GUIDE
7.1
Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952) Test ID: 3628 Test status: Passed
7.1.1 Description Exports a linear element to GTC.
7.2
Exporting an Advance Design model to DO4 format (DEV2012 #1.10) Test ID: 4101 Test status: Passed
7.2.1 Description Launches the "Export > Text file" command and saves the current project as a .do4 archive file. The model contains all types of structural elements, loads and geometric objects.
7.3
Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172) Test ID: 4297 Test status: Passed
7.3.1 Description Imports a GTC file from SuperSTRESS. The file contains steel linear elements with haunch sections.
7.4
Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) Test ID: 4195 Test status: Passed
7.4.1 Description Exports the analysis model to ADA (through GTC) with:  Export results: disabled  Export meshed model: enabled
7.5
Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) Test ID: 4193 Test status: Passed
7.5.1 Description Exports the analysis model to ADA (through GTC) with:  Export results: enabled  Export meshed model: disabled
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ADVANCE DESIGN VALIDATION GUIDE
7.6
Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197) Test ID: 4388 Test status: Passed
7.6.1 Description Imports a GTC file from SuperSTRESS. The file contains elements with circular hollow section. Verifies if the cross sections are imported from the attached "UK Steel Sections" database.
7.7
System stability when importing AE files with invalid geometry (TTAD #12232) Test ID: 4479 Test status: Passed
7.7.1 Description Imports a complex model containing elements with invalid geometry.
7.8
Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197) Test ID: 4389 Test status: Passed
7.8.1 Description Imports a GTC file from SuperSTRESS when the "UK Steel Sections" database is not attached. The file contains elements with circular hollow section. Verifies the cross sections definition.
7.9
Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9) Test ID: 4445 Test status: Passed
7.9.1 Description Verifies the GTC files exchange (import/export) between Advance Design and SuperSTRESS.
7.10 Exporting linear elements to IFC format (TTAD #10561) Test ID: 4530 Test status: Passed
7.10.1
Description
Exports to IFC format a model containing linear elements having sections of type "I symmetric" and "I asymmetric".
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ADVANCE DESIGN VALIDATION GUIDE
7.11 Verifying the load case properties from models imported as GTC files (TTAD #12306) Test ID: 4515 Test status: Passed
7.11.1
Description
Performs the finite elements calculation on a model with dead load cases and exports the model to GTC. Imports the GTC file to verify the load case properties.
7.12 Importing IFC files containing continuous foundations (TTAD #12410) Test ID: 4531 Test status: Passed
7.12.1
Description
Imports an IFC file containing a continuous foundation (linear support) and verifies the element display.
7.13 Verifying the releases option of the planar elements edges after the model was exported and imported via GTC format (TTAD #12137) Test ID: 4506 Test status: Passed
7.13.1
Description
Exports to GTC a model with planar elements on which the edges releases were defined. Imports the GTC file to verify the planar elements releases option.
7.14 Importing GTC files containing "PH.RDC" system (TTAD #12055) Test ID: 4548 Test status: Passed
7.14.1
Description
Imports a GTC file exported from Advance Design. The file contains the automatically created system "PH.RDC".
7.15 Exporting a meshed model to GTC (TTAD #12550) Test ID: 4552 Test status: Passed
7.15.1
Description
Exports a meshed model to GTC. The meshed planar element from the model contains a triangular mesh.
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8 Joint design
ADVANCE DESIGN VALIDATION GUIDE
8.1
Deleting a welded tube connection  1 gusset bar (TTAD #12630) Test ID: 4561 Test status: Passed
8.1.1 Description Deletes a welded tube connection  1 gusset bar after the joint was exported to ADSC.
8.2
Creating connections groups (TTAD #11797) Test ID: 4250 Test status: Passed
8.2.1 Description Verifies the connections groups function.
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