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Transcript of Ad validation guide 2013 en parte1
Advance Design
Validation Guide
INTRODUCTION Before being officially released, each version of GRAITEC software, including Advance Design, undergoes a series of validation tests. This validation is performed in parallel and in addition to manual testing and beta testing, in order to obtain the "operational version" status. This document contains a description of the automatic tests, highlighting the theoretical background and the results we have obtained using the current software release.
Usually, a test is made of a reference (independent from the specific software version tested), a transformation (a calculation or a data processing scenario), a result (given by the specific software version tested) and a difference usually measured in percentage as a drift from a set of reference values. Depending on the cases, the used reference is either a theoretical calculation done manually, a sample taken from the technical literature, or the result of a previous version considered as good by experience.
Starting with version 2012, Graitec Advance has made significant steps forward in term of quality management by extending the scope and automating the testing process. While in previous versions, the tests were always about the calculation results which were compared to a reference set, starting with version 2012, tests have been extended to user interface behavior, import/export procedures, etc. The next major improvement is the capacity to pass the tests automatically. These current tests have obviously been passed on the “operational version”, but they are actually passed on a daily basis during the development process, which helps improve the daily quality by solving potential issues, immediately after they have been introduced in the code.
In the field of structural analysis and design, software users must keep in mind that the results highly depend on the modeling (especially when dealing with finite elements) and on the settings of the numerous assumptions and options available in the software. A software package cannot replace engineers experience and analysis. Despite all our efforts in term of quality management, we cannot guaranty the correct behavior and the validity of the results issued by Advance Design in any situation. With this validation guide, we are providing a set of concrete test cases showing the behavior of Advance Design in various areas and various conditions. The tests cover a wide field of expertise: modeling, climatic load generation according to Eurocode 1, combinations management, meshing, finite element calculation, reinforced concrete design according to Eurocode 2, steel member design according to Eurocode 3, steel connection design according to Eurocode 3, timber member design according to Eurocode 5, seismic analysis according to Eurocode 8, report generation, import / export procedures and user interface behavior.
We hope that this guide will highly contribute to the knowledge and the confidence you are placing in Advance Design.
Manuel LIEDOT
Chief Product Officer
ADVANCE DESIGN VALIDATION GUIDE
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CONTENTS
1 FINITE ELEMENTS ANALYSIS ..........................................................................................................21 1.1 System of two bars with three hinges (01-0002SSLLB_FEM)......................................................................................22 1.2 Slender beam with variable section (fixed-free) (01-0004SDLLB_FEM) ......................................................................25 1.3 Thin lozenge-shaped plate fixed on one side (alpha = 15 °) (01-0008SDLSB_FEM)...................................................28 1.4 Thin circular ring fixed in two points (01-0006SDLLB_FEM) ........................................................................................31 1.5 Thin lozenge-shaped plate fixed on one side (alpha = 45 °) (01-0010SDLSB_FEM)...................................................35 1.6 Vibration mode of a thin piping elbow in plane (case 2) (01-0012SDLLB_FEM)..........................................................38 1.7 Thin circular ring hanged on an elastic element (01-0014SDLLB_FEM)......................................................................41 1.8 Tied (sub-tensioned) beam (01-0005SSLLB_FEM) .....................................................................................................45 1.9 Circular plate under uniform load (01-0003SSLSB_FEM)............................................................................................50 1.10 Double fixed beam (01-0016SDLLB_FEM) ..................................................................................................................53 1.11 Short beam on simple supports (eccentric) (01-0018SDLLB_FEM).............................................................................57 1.12 Thin lozenge-shaped plate fixed on one side (alpha = 30 °) (01-0009SDLSB_FEM)...................................................61 1.13 Vibration mode of a thin piping elbow in plane (case 3) (01-0013SDLLB_FEM)..........................................................64 1.14 Short beam on simple supports (on the neutral axis) (01-0017SDLLB_FEM)..............................................................67 1.15 Thin lozenge-shaped plate fixed on one side (alpha = 0 °) (01-0007SDLSB_FEM).....................................................71 1.16 Vibration mode of a thin piping elbow in plane (case 1) (01-0011SDLLB_FEM)..........................................................74 1.17 Double fixed beam with a spring at mid span (01-0015SSLLB_FEM)..........................................................................77 1.18 Rectangular thin plate simply supported on its perimeter (01-0020SDLSB_FEM) .......................................................80 1.19 Slender beam on two fixed supports (01-0024SSLLB_FEM) .......................................................................................84 1.20 Annular thin plate fixed on a hub (repetitive circular structure) (01-0022SDLSB_FEM) ...............................................89 1.21 Bimetallic: Fixed beams connected to a stiff element (01-0026SSLLB_FEM)..............................................................91 1.22 Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM) ......................................................................................94 1.23 Slender beam on three supports (01-0025SSLLB_FEM) .............................................................................................96 1.24 Fixed thin arc in out of plane bending (01-0028SSLLB_FEM) ...................................................................................100 1.25 Portal frame with lateral connections (01-0030SSLLB_FEM) ....................................................................................103 1.26 Double hinged thin arc in planar bending (01-0029SSLLB_FEM)..............................................................................106 1.27 Thin square plate fixed on one side (01-0019SDLSB_FEM)......................................................................................109 1.28 Bending effects of a symmetrical portal frame (01-0023SDLLB_FEM) ......................................................................113 1.29 Fixed thin arc in planar bending (01-0027SSLLB_FEM) ............................................................................................116 1.30 Beam on elastic soil, hinged ends (01-0034SSLLB_FEM).........................................................................................119 1.31 Square plate under planar stresses (01-0039SSLSB_FEM) ......................................................................................123 1.32 Beam on elastic soil, free ends (01-0032SSLLB_FEM) .............................................................................................126 1.33 EDF Pylon (01-0033SFLLA_FEM) .............................................................................................................................129 1.34 Thin cylinder under a uniform radial pressure (01-0038SSLSB_FEM).......................................................................133
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1.35 Caisson beam in torsion (01-0037SSLSB_FEM)....................................................................................................... 135 1.36 Beam on two supports considering the shear force (01-0041SSLLB_FEM) .............................................................. 138 1.37 Thin cylinder under a hydrostatic pressure (01-0043SSLSB_FEM)........................................................................... 141 1.38 Thin cylinder under a uniform axial load (01-0042SSLSB_FEM)............................................................................... 144 1.39 Truss with hinged bars under a punctual load (01-0031SSLLB_FEM) ...................................................................... 147 1.40 Simply supported square plate (01-0036SSLSB_FEM) ............................................................................................. 150 1.41 Stiffen membrane (01-0040SSLSB_FEM) ................................................................................................................. 153 1.42 Torus with uniform internal pressure (01-0045SSLSB_FEM) .................................................................................... 156 1.43 Spherical dome under a uniform external pressure (01-0050SSLSB_FEM).............................................................. 159 1.44 Pinch cylindrical shell (01-0048SSLSB_FEM) ........................................................................................................... 162 1.45 Simply supported rectangular plate under a uniform load (01-0052SSLSB_FEM) .................................................... 164 1.46 Spherical shell under internal pressure (01-0046SSLSB_FEM) ................................................................................ 166 1.47 Simply supported square plate under a uniform load (01-0051SSLSB_FEM) ........................................................... 169 1.48 Simply supported rectangular plate loaded with punctual force and moments (01-0054SSLSB_FEM)..................... 171 1.49 Triangulated system with hinged bars (01-0056SSLLB_FEM) .................................................................................. 173 1.50 Shear plate perpendicular to the medium surface (01-0055SSLSB_FEM) ................................................................ 176 1.51 Thin cylinder under its self weight (01-0044SSLSB_MEF) ........................................................................................ 178 1.52 Spherical shell with holes (01-0049SSLSB_FEM) ..................................................................................................... 180 1.53 Simply supported rectangular plate under a uniform load (01-0053SSLSB_FEM) .................................................... 183 1.54 A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0058SSLSB_FEM)................ 185 1.55 A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0060SSLSB_FEM)...................... 187 1.56 A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (01-0064SSLSB_FEM).......................... 189 1.57 A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (01-0062SSLSB_FEM).......................... 191 1.58 A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0059SSLSB_FEM)...................... 194 1.59 A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (01-0063SSLSB_FEM).................... 196 1.60 A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (01-0066SSLSB_FEM)............................ 199 1.61 Vibration mode of a thin piping elbow in space (case 2) (01-0068SDLLB_FEM)....................................................... 201 1.62 Vibration mode of a thin piping elbow in space (case 1) (01-0067SDLLB_FEM)....................................................... 204 1.63 A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0057SSLSB_FEM)...................... 207 1.64 A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0061SSLSB_FEM)........................ 209 1.65 A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (01-0065SSLSB_FEM).......................... 211 1.66 Reactions on supports and bending moments on a 2D portal frame (Rafters) (01-0077SSLPB_FEM) ..................... 213 1.67 Slender beam of variable rectangular section (fixed-fixed) (01-0086SDLLB_FEM)................................................... 215 1.68 Short beam on two hinged supports (01-0084SSLLB_FEM) ..................................................................................... 218 1.69 Double fixed beam in Eulerian buckling with a thermal load (01-0091HFLLB_FEM)................................................. 220 1.70 Reactions on supports and bending moments on a 2D portal frame (Columns) (01-0078SSLPB_FEM) .................. 222 1.71 Plane portal frame with hinged supports (01-0089SSLLB_FEM)............................................................................... 224 1.72 A 3D bar structure with elastic support (01-0094SSLLB_FEM) ................................................................................. 226
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1.73 Fixed/free slender beam with eccentric mass or inertia (01-0096SDLLB_FEM) ........................................................233 1.74 Fixed/free slender beam with centered mass (01-0095SDLLB_FEM)........................................................................237 1.75 Vibration mode of a thin piping elbow in space (case 3) (01-0069SDLLB_FEM) .......................................................242 1.76 Slender beam of variable rectangular section with fixed-free ends (ß=5) (01-0085SDLLB_FEM) .............................245 1.77 Cantilever beam in Eulerian buckling with thermal load (01-0092HFLLB_FEM) ........................................................250 1.78 Simple supported beam in free vibration (01-0098SDLLB_FEM)...............................................................................252 1.79 Membrane with hot point (01-0099HSLSB_FEM) ......................................................................................................255 1.80 Beam on 3 supports with T/C (k = -10000 N/m) (01-0102SSNLB_FEM) ...................................................................258 1.81 Beam on 3 supports with T/C (k = 0) (01-0100SSNLB_FEM) ....................................................................................261 1.82 Non linear system of truss beams (01-0104SSNLB_FEM) ........................................................................................264 1.83 Linear system of truss beams (01-0103SSLLB_FEM) ...............................................................................................267 1.84 Linear element in combined bending/tension - without compressed reinforcements - Partially tensioned section
(02-0158SSLLB_B91) ................................................................................................................................................270 1.85 Design of a Steel Structure according to CM66 (03-0206SSLLG_CM66) ..................................................................275 1.86 Linear element in simple bending - without compressed reinforcement (02-0162SSLLB_B91) .................................284 1.87 Double cross with hinged ends (01-0097SDLLB_FEM) .............................................................................................288 1.88 Beam on 3 supports with T/C (k -> infinite) (01-0101SSNLB_FEM)...........................................................................291 1.89 Study of a mast subjected to an earthquake (02-0112SMLLB_P92)..........................................................................294 1.90 Design of a concrete floor with an opening (03-0208SSLLG_BAEL91) .....................................................................300 1.91 Design of a 2D portal frame (03-0207SSLLG_CM66) ................................................................................................308 1.92 Cantilever rectangular plate (01-0001SSLSB_FEM)..................................................................................................315 1.93 Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontal force (TTAD #13175) ...318 1.94 Verifying torsors on a single story coupled walls subjected to horizontal forces ........................................................318 1.95 Verifying diagrams for Mf Torsors on divided walls (TTAD #11557)...........................................................................318 1.96 Verifying the level mass center (TTAD #11573, TTAD #12315).................................................................................318 1.97 Generating results for Torsors NZ/Group (TTAD #11633) .........................................................................................318 1.98 Verifying Sxx results on beams (TTAD #11599).........................................................................................................319 1.99 Verifying forces results on concrete linear elements (TTAD #11647).........................................................................319 1.100 Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD #11854) ........................319 1.101 Verifying stresses in beam with "extend into wall" property (TTAD #11680) ..............................................................319 1.102 Verifying constraints for triangular mesh on planar elements (TTAD #11447) ...........................................................319 1.103 Verifying the displacement results on linear elements for vertical seism (TTAD #11756) ..........................................320 1.104 Verifying forces for triangular meshing on planar element (TTAD #11723)................................................................320 1.105 Generating planar efforts before and after selecting a saved view (TTAD #11849) ...................................................320 1.106 Verifying tension/compression supports on nonlinear analysis (TTAD #11518).........................................................320 1.107 Verifying tension/compression supports on nonlinear analysis (TTAD #11518).........................................................321 1.108 Verifying the display of the forces results on planar supports (TTAD #11728)...........................................................321 1.109 Verifying results on punctual supports (TTAD #11489) ..............................................................................................321 1.110 Generating a report with torsors per level (TTAD #11421).........................................................................................321
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1.111 Verifying nonlinear analysis results for frames with semi-rigid joints and rigid joints (TTAD #11495) ........................ 322 1.112 Verifying forces on a linear elastic support which is defined in a user workplane (TTAD #11929) ............................ 322 1.113 Verifying the internal forces results for a simple supported steel beam ..................................................................... 322 1.114 Verifying the main axes results on a planar element (TTAD #11725) ........................................................................ 322
2 CAD, RENDERING AND VISUALIZATION ...................................................................................... 323 2.1 Verifying annotation on selection (TTAD #12700)...................................................................................................... 324 2.2 Verifying rotation for steel beam with joint (TTAD #12592)........................................................................................ 324 2.3 Verifying hide/show elements command (TTAD #11753) .......................................................................................... 324 2.4 System stability during section cut results verification (TTAD #11752)...................................................................... 324 2.5 Verifying the grid text position (TTAD #11704) .......................................................................................................... 324 2.6 Verifying the grid text position (TTAD #11657) .......................................................................................................... 324 2.7 Generating combinations (TTAD #11721) ................................................................................................................. 325 2.8 Verifying the coordinates system symbol (TTAD #11611) ......................................................................................... 325 2.9 Verifying descriptive actors after creating analysis (TTAD #11589)........................................................................... 325 2.10 Creating a circle (TTAD #11525) ............................................................................................................................... 325 2.11 Creating a camera (TTAD #11526)............................................................................................................................ 325 2.12 Verifying the local axes of a section cut (TTAD #11681) ........................................................................................... 326 2.13 Verifying the snap points behavior during modeling (TTAD #11458) ......................................................................... 326 2.14 Verifying the representation of elements with HEA cross section (TTAD #11328)..................................................... 326 2.15 Verifying the descriptive model display after post processing results in analysis mode (TTAD #11475) ................... 326 2.16 Verifying holes in horizontal planar elements after changing the level height (TTAD #11490) .................................. 326 2.17 Verifying the display of elements with compound cross sections (TTAD #11486) ..................................................... 327 2.18 Modeling using the tracking snap mode (TTAD #10979) ........................................................................................... 327 2.19 Turning on/off the "ghost" rendering mode (TTAD #11999) ....................................................................................... 327 2.20 Moving a linear element along with the support (TTAD #12110) ............................................................................... 327 2.21 Verifying the "ghost" display after changing the display colors (TTAD #12064)......................................................... 327 2.22 Verifying the "ghost display on selection" function for saved views (TTAD #12054).................................................. 327 2.23 Verifying the steel connections modeling (TTAD #11698) ......................................................................................... 328 2.24 Verifying the fixed load scale function (TTAD #12183). ............................................................................................. 328 2.25 Verifying the dividing of planar elements which contain openings (TTAD #12229).................................................... 328 2.26 Verifying the program behavior when trying to create lintel (TTAD #12062).............................................................. 328 2.27 Verifying the program behavior when launching the analysis on a model with overlapped loads (TTAD #11837) .... 328 2.28 Verifying the display of punctual loads after changing the load case number (TTAD #11958) .................................. 329 2.29 Verifying the display of a beam with haunches (TTAD #12299)................................................................................ 329 2.30 Creating base plate connections for non-vertical columns (TTAD #12170) ............................................................... 329 2.31 Verifying drawing of joints in y-z plan (TTAD #12453) ............................................................................................... 329
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3 CLIMATIC GENERATOR ..................................................................................................................331 3.1 EC1: generating snow loads on a 3 slopes 3D portal frame with parapets (NF EN 1991-1-3/NA) (TTAD #11111) ...332 3.2 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.3 - Wind - Example C) ....332 3.3 EC1: generating wind loads on a 3D portal frame with one slope roof (NF EN 1991-1-4/NA)
(VT : 3.2 - Wind - Example B) ....................................................................................................................................332 3.4 EC1: generating wind loads on a triangular based lattice structure with compound profiles and automatic
calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12276)............................................................................................332 3.5 EC1: generating snow loads on a 2 slopes 3D portal frame (NF EN 1991-1-3/NA) (VT : 3.4 - Snow - Example A)...333 3.6 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.1 - Wind - Example A).....333 3.7 EC1: wind loads on a triangular based lattice structure with compound profiles and user defined "n"
(NF EN 1991-1-4/NA) (TTAD #12276) .......................................................................................................................333 3.8 EC1: Verifying the geometry of wind loads on an irregular shed. (TTAD #12233) .....................................................333 3.9 EC1: Verifying the wind loads generated on a building with protruding roof (TTAD #12071, #12278) .......................333 3.10 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11531) ..............................................................334 3.11 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11569) ....334 3.12 Generating the description of climatic loads report according to EC1 Romanian standards (TTAD #11688).............334 3.13 EC1: generating snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height
(TTAD #11943)...........................................................................................................................................................334 3.14 EC1: generating wind loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11687) .....335 3.15 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11699) .............................................................335 3.16 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11570) ....335 3.17 NV2009: generating wind loads and snow loads on a simple structure with planar support (TTAD #11380).............335 3.18 EC1: verifying the snow loads generated on a monopitch frame (TTAD #11302) ......................................................336 3.19 EC1: generating wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls (TTAD #11937) ...............336 3.20 EC1: generating snow loads on two side by side roofs with different heights, according to German standards
(DIN EN 1991-1-3/NA) (DEV2012 #3.13)...................................................................................................................336 3.21 EC1: generating wind loads on a 55m high structure according to German standards (DIN EN 1991-1-4/NA)
(DEV2012 #3.12)........................................................................................................................................................336 3.22 EC1: generating wind loads on double slope 3D portal frame according to Czech standards (CSN EN 1991-1-4)
(DEV2012 #3.18)........................................................................................................................................................337 3.23 EC1: generating snow loads on duopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA)
(DEV2012 #3.13)........................................................................................................................................................337 3.24 EC1: generating snow loads on two close roofs with different heights according to Czech standards
(CSN EN 1991-1-3) (DEV2012 #3.18) .......................................................................................................................337 3.25 EC1: generating snow loads on monopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA)
(DEV2012 #3.13)........................................................................................................................................................337 3.26 EC1: snow on a 3D portal frame with horizontal roof and parapet with height reduction (TTAD #11191) ..................338 3.27 EC1: generating snow loads on a 2 slopes 3D portal frame with gutter (TTAD #11113)............................................338 3.28 EC1: generating snow loads on a 3D portal frame with horizontal roof and gutter (TTAD #11113) ...........................338 3.29 EC1: generating snow loads on a 3D portal frame with a roof which has a small span (< 5m) and a parapet
(TTAD #11735)...........................................................................................................................................................338 3.30 EC1: generating wind loads on double slope 3D portal frame with a fully opened face (DEV2012 #1.6)...................339 3.31 EC1: generating wind loads on an isolated roof with two slopes (TTAD #11695) ......................................................339
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3.32 EC1: generating wind loads on duopitch multispan roofs with pitch < 5 degrees (TTAD #11852) ............................. 339 3.33 EC1: wind load generation on a simple 3D structure with horizontal roof .................................................................. 339 3.34 EC1 NF: generating wind loads on a 3D portal frame with 2 slopes roof (TTAD #11932) ......................................... 340 3.35 EC1: wind load generation on simple 3D portal frame with 4 slopes roof (TTAD #11604)......................................... 340 3.36 EC1: wind load generation on a signboard ................................................................................................................ 340 3.37 EC1: wind load generation on a simple 3D portal frame with 2 slopes roof (TTAD #11602)...................................... 340 3.38 EC1: wind load generation on a building with multispan roofs ................................................................................... 341 3.39 EC1: wind load generation on a high building with horizontal roof ............................................................................. 341 3.40 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ......................................................... 341 3.41 EC1: Generating snow loads on a single slope with lateral parapets (TTAD #12606) ............................................... 341 3.42 EC1: generating wind loads on a square based lattice structure with compound profiles and automatic
calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12744) ........................................................................................... 342 3.43 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ......................................................... 342 3.44 EC1: Generating snow loads on two side by side buildings with gutters (TTAD #12806) .......................................... 342 3.45 EC1: Generating wind loads on a square based structure according to UK standards (BS EN 1991-1-4:2005)
(TTAD #12608) .......................................................................................................................................................... 342 3.46 EC1: Generating snow loads on a 4 slopes with gutters building. (TTAD #12719) .................................................... 342 3.47 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12808)....................................................... 343 3.48 EC1: Generating snow loads on a 4 slopes with gutters building (TTAD #12716) ..................................................... 343 3.49 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12835)....................................................... 343 3.50 EC1: Generating snow loads on a 2 slope building with gutters and parapets. (TTAD #12878)................................ 343 3.51 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12841)....................................................... 343 3.52 NV2009: Verifying wind and snow reports for a protruding roof (TTAD #11318) ....................................................... 344 3.53 NV2009: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604) ................................ 344
4 COMBINATIONS............................................................................................................................... 345 4.1 Verifying combinations for CZ localization (TTAD #12542)........................................................................................ 346 4.2 Generating combinations (TTAD #11673) ................................................................................................................. 346 4.3 Defining concomitance rules for two case families (TTAD #11355) ........................................................................... 346 4.4 Generating load combinations with unfavorable and favorable/unfavorable predominant action (TTAD #11357) ..... 346 4.5 Generating combinations for NEWEC8.cbn (TTAD #11431) ..................................................................................... 346 4.6 Generating load combinations after changing the load case number (TTAD #11359)............................................... 347 4.7 Generating the concomitance matrix after adding a new dead load case (TTAD #11361) ........................................ 347 4.8 Generating a set of combinations with seismic group of loads (TTAD #11889) ......................................................... 347 4.9 Generating the concomitance matrix after switching back the effect for live load (TTAD #11806) ............................ 347 4.10 Performing the combinations concomitance standard test no. 5 (DEV2012 #1.7) ..................................................... 348 4.11 Performing the combinations concomitance standard test no.9 (DEV2012 #1.7) ...................................................... 348 4.12 Performing the combinations concomitance standard test no.4 (DEV2012 #1.7) ..................................................... 349 4.13 Performing the combinations concomitance standard test no.6 (DEV2012 #1.7) ...................................................... 349 4.14 Performing the combinations concomitance standard test no.8 (DEV2012 #1.7) ...................................................... 350
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4.15 Performing the combinations concomitance standard test no.10 (DEV2012 #1.7) ....................................................350 4.16 Performing the combinations concomitance standard test no.7 (DEV2012 #1.7) ......................................................351 4.17 Generating a set of combinations with Q group of loads (TTAD #11960) ..................................................................351 4.18 Performing the combinations concomitance standard test no.1 (DEV2010#1.7) .......................................................351 4.19 Generating a set of combinations with different Q "Base" types (TTAD #11806) .......................................................352 4.20 Performing the combinations concomitance standard test no.2 (DEV2012 #1.7) ......................................................352 4.21 Performing the combinations concomitance standard test no.3 (DEV2012 #1.7) ......................................................353
5 CONCRETE DESIGN ........................................................................................................................355 5.1 EC2: Verifying the minimum reinforcement area for a simply supported beam..........................................................356 5.2 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load.....................................................356 5.3 EC2: Verifying the longitudinal reinforcement area for a beam subjected to point loads............................................356 5.4 Modifying the "Design experts" properties for concrete linear elements (TTAD #12498) ...........................................356 5.5 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - horizontal level behavior law.....356 5.6 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - bilinear stress-strain diagram....357 5.7 Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular cross section........................357 5.8 Verifying the reinforced concrete results on a structure with 375 load cases combinations (TTAD #11683) .............357 5.9 Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678) ..........................................357 5.10 Verifying the reinforced concrete results on a fixed beam (TTAD #11836) ................................................................358 5.11 Verifying the longitudinal reinforcement for a fixed linear element (TTAD #11700)....................................................358 5.12 Verifying the longitudinal reinforcement for linear elements (TTAD #11636) .............................................................358 5.13 EC2 : calculation of a square column in traction (TTAD #11892)...............................................................................358 5.14 Verifying Aty and Atz for a fixed concrete beam (TTAD #11812) ...............................................................................359 5.15 Verifying concrete results for planar elements (TTAD #11583) ..................................................................................359 5.16 Verifying concrete results for linear elements (TTAD #11556) ...................................................................................359 5.17 Verifying the reinforcement of concrete columns (TTAD #11635) ..............................................................................359 5.18 Verifying the minimum transverse reinforcement area results for an articulated beam (TTAD #11342).....................360 5.19 Verifying the minimum transverse reinforcement area results for articulated beams (TTAD #11342)........................360 5.20 EC2: column design with “Nominal Stiffness method” square section (TTAD #11625) ..............................................360 5.21 EC2: Verifying the transverse reinforcement area for a beam subjected to linear loads ............................................360 5.22 EC2 Test 1: Verifying a rectangular cross section beam made from concrete C25/30 to resist simple bending
- Bilinear stress-strain diagram...................................................................................................................................361 5.23 EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinear stress-strain diagram ......362 5.24 EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement
- Inclined stress-strain diagram ..................................................................................................................................367 5.25 EC2 Test 10: Verifying a T concrete section, without compressed reinforcement - Inclined stress-strain diagram....377 5.26 EC2 Test 12: Verifying a rectangular concrete beam subjected to uniformly distributed load, without
compressed reinforcement - Bilinear stress-strain diagram (Class XD3) ...................................................................383 5.27 EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .....389 5.28 EC2 Test 16: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .....395
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5.29 EC2 Test 11: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ....................................................................................... 401
5.30 EC2 Test 19: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ...................................................... 401
5.31 EC2 Test 20: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ...................................................... 402
5.32 EC2 Test 14: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram (Class XD1) ........................................................................................ 409
5.33 EC2 Test 18: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ....................................................................................... 415
5.34 EC2 Test 24: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 420
5.35 EC2 Test 25: Verifying the shear resistance for a rectangular concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 420
5.36 EC2 Test 23: Verifying the shear resistance for a rectangular concrete - Bilinear stress-strain diagram (Class XC1)................................................................................................................................................................ 420
5.37 EC2 Test 13: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ....................................................................................... 420
5.38 EC2 Test 17: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Inclined stress-strain diagram (Class XD1)....................................................................................... 421
5.39 EC2 Test28: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class X0)................................................................................................................................ 427
5.40 EC2 Test32: Verifying a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (Class XC1) .................................................................. 431
5.41 EC2 Test29: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Inclined stress-strain diagram (Class XC1) ............................................................................................................................. 441
5.42 EC2 Test33: Verifying a square concrete column subjected to compression by nominal rigidity method- Bilinear stress-strain diagram (Class XC1) ............................................................................................................................. 445
5.43 EC2 Test 27: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 452
5.44 EC2 Test31: Verifying a square concrete column subjected to compression and rotation moment to the top - Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 457
5.45 EC2 Test35: Verifying a rectangular concrete column subjected to compression to top Based on nominal rigidity method - Bilinear stress-strain diagram (Class XC1) ........................................................ 471
5.46 EC2 Test36: Verifying a rectangular concrete column using the method based on nominal curvature Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 481
5.47 EC2 Test 37: Verifying a square concrete column using the simplified method – Professional rules Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 482
5.48 EC2 Test 26: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 485
5.49 EC2 Test30: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 489
5.50 EC2 Test34: Verifying a rectangular concrete column subjected to compression on the top – Method based on nominal curvature - Bilinear stress-strain diagram (Class XC1)................................................................................. 489
5.51 EC2 Test 38: Verifying a rectangular concrete column using the simplified method – Professional rules Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 498
5.52 EC2 Test 40: Verifying a square concrete column subjected to a small compression force and significant rotation moment to the top - Bilinear stress-strain diagram (Class XC1) ................................................................................ 501
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5.53 EC2 Test 41: Verifying a square concrete column subjected to a significant compression force and small rotation moment to the top - Bilinear stress-strain diagram (Class XC1).................................................................................508
5.54 EC2 Test 44: Verifying a rectangular concrete beam subjected to eccentric loading - Bilinear stress-strain diagram (Class X0) ..................................................................................................................................................................520
5.55 EC2 Test 45: Verifying a rectangular concrete beam supporting a balcony - Bilinear stress-strain diagram (Class XC1) ................................................................................................................................................................521
5.56 EC2 Test 47: Verifying a rectangular concrete beam subjected to a tension distributed load - Bilinear stress-strain diagram (Class XD2) ..................................................................................................................................................528
5.57 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load Inclined stress strain behavior law..............................................................................................................................528
5.58 EC2 Test 3: Verifying a rectangular concrete beam subjected to uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram.............................................................................................................529
5.59 EC2 Test 2: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram............................................................................................................538
5.60 EC2 Test 6: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .......539 5.61 EC2 Test 7: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .......543 5.62 EC2 Test 8: Verifying a rectangular concrete beam without compressed reinforcement – Inclined stress-strain
diagram ......................................................................................................................................................................547 5.63 EC2 Test 4: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclined stress-strain diagram.....554 5.64 EC2 Test 39: Verifying a circular concrete column using the simplified method – Professional rules
Bilinear stress-strain diagram (Class XC1).................................................................................................................555 5.65 EC2 Test 43: Verifying a square concrete column subjected to a small rotation moment and significant compression
force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1)...................................559 5.66 EC2 Test 46 I: Verifying a square concrete beam subjected to a normal force of traction - Inclined stress-strain
diagram (Class X0).....................................................................................................................................................568 5.67 EC2 Test 42: Verifying a square concrete column subjected to a significant rotation moment and small compression
force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1)...................................571 5.68 EC2 Test 46 II: Verifying a square concrete beam subjected to a normal force of traction - Bilinear stress-strain
diagram (Class X0).....................................................................................................................................................580
6 GENERAL APPLICATION ................................................................................................................583 6.1 Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238)................................................584 6.2 Verifying the precision of linear and planar concrete covers (TTAD #12525).............................................................584 6.3 Defining the reinforced concrete design assumptions (TTAD #12354) ......................................................................584 6.4 Verifying the synthetic table by type of connection (TTAD #11422) ...........................................................................584 6.5 Importing a cross section from the Advance Steel profiles library (TTAD #11487) ....................................................584 6.6 Creating and updating model views and post-processing views (TTAD #11552).......................................................585 6.7 Creating system trees using the copy/paste commands (DEV2012 #1.5)..................................................................585 6.8 Creating system trees using the copy/paste commands (DEV2012 #1.5)..................................................................585 6.9 Creating a new Advance Design file using the "New" command from the "Standard" toolbar (TTAD #12102) ..........585 6.10 Verifying mesh, CAD and climatic forces - LPM meeting ...........................................................................................585 6.11 Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724) ........................................................586 6.12 Changing the default material (TTAD #11870)...........................................................................................................586 6.13 Verifying the objects rename function (TTAD #12162)...............................................................................................586
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6.14 Launching the verification of a model containing steel connections (TTAD #12100) ................................................. 586 6.15 Verifying the appearance of the local x orientation legend (TTAD #11737) ............................................................... 586 6.16 Verifying geometry properties of elements with compound cross sections (TTAD #11601) ...................................... 587 6.17 Verifying material properties for C25/30 (TTAD #11617) ........................................................................................... 587 6.18 Verifying element creation using commas for coordinates (TTAD #11141) ............................................................... 587
7 IMPORT / EXPORT ........................................................................................................................... 589 7.1 Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952) ....................................................... 590 7.2 Exporting an Advance Design model to DO4 format (DEV2012 #1.10) ..................................................................... 590 7.3 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ....................................................................... 590 7.4 Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172)................................. 590 7.5 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ....................................................................... 590 7.6 Exporting linear elements to IFC format (TTAD #10561) ........................................................................................... 591 7.7 Importing IFC files containing continuous foundations (TTAD #12410) ..................................................................... 591 7.8 Importing GTC files containing "PH.RDC" system (TTAD #12055)............................................................................ 591 7.9 Exporting a meshed model to GTC (TTAD #12550) .................................................................................................. 591 7.10 Verifying the load case properties from models imported as GTC files (TTAD #12306) ............................................ 591 7.11 Verifying the releases option of the planar elements edges after the model was exported and imported via GTC
format (TTAD #12137) ............................................................................................................................................... 592 7.12 System stability when importing AE files with invalid geometry (TTAD #12232)........................................................ 592 7.13 Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9)........................... 592 7.14 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197)........... 592 7.15 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197)........... 592
8 CONNECTION DESIGN .................................................................................................................... 593 8.1 Deleting a welded tube connection - 1 gusset bar (TTAD #12630)............................................................................ 594 8.2 Creating connections groups (TTAD #11797)............................................................................................................ 594
9 MESH................................................................................................................................................. 595 9.1 Verifying the mesh for a model with generalized buckling (TTAD #11519)................................................................ 596 9.2 Verifying mesh points (TTAD #11748) ....................................................................................................................... 596 9.3 Creating triangular mesh for planar elements (TTAD #11727) .................................................................................. 596
10 REPORTS GENERATOR.................................................................................................................. 597 10.1 Verifying the modal analysis report (TTAD #12718) .................................................................................................. 598 10.2 Verifying the shape sheet strings display (TTAD #12622) ......................................................................................... 598 10.3 Verifying the shape sheet for a steel beam (TTAD #12455) ...................................................................................... 598 10.4 Verifying the global envelope of linear elements stresses (on the start point of super element) (TTAD #12230) ...... 598 10.5 Verifying the Max row on the user table report (TTAD #12512) ................................................................................. 598 10.6 Verifying the steel shape sheet display (TTAD #12657) ............................................................................................ 599 10.7 Verifying the EC2 calculation assumptions report (TTAD #11838) ............................................................................ 599
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10.8 Verifying the shape sheet report (TTAD #12353) .......................................................................................................599 10.9 Verifying the model geometry report (TTAD #12201).................................................................................................599 10.10 Verifying the global envelope of linear elements forces result (on end points and middle of super element)
(TTAD #12230)...........................................................................................................................................................599 10.11 Verifying the global envelope of linear elements forces result (on start and end of super element) (TTAD #12230) .600 10.12 Verifying the global envelope of linear elements stresses (on the end point of super element)
(TTAD #12230, TTAD #12261) ..................................................................................................................................600 10.13 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element) (TTAD #12230) .....600 10.14 Verifying the global envelope of linear elements displacements (on all quarters of super element) (TTAD #12230) .600 10.15 Verifying the global envelope of linear elements forces result (on all quarters of super element) (TTAD #12230) ....601 10.16 Verifying the global envelope of linear elements displacements (on the start point of super element)
(TTAD #12230)...........................................................................................................................................................601 10.17 Verifying the global envelope of linear elements displacements (on the end point of super element) (TTAD #12230,
TTAD #12261)............................................................................................................................................................601 10.18 Verifying the global envelope of linear elements forces result (on the end point of super element)
(TTAD #12230, #12261).............................................................................................................................................601 10.19 Verifying the global envelope of linear elements displacements (on start and end of super element)
(TTAD #12230)...........................................................................................................................................................602 10.20 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element) (TTAD #12230).........602 10.21 Verifying the global envelope of linear elements forces result (on the start point of super element)
(TTAD #12230)...........................................................................................................................................................602 10.22 Verifying the global envelope of linear elements displacements (on end points and middle of super element)
(TTAD #12230)...........................................................................................................................................................602 10.23 Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD #12230)...............603 10.24 Verifying the global envelope of linear elements stresses (on start and end of super element) (TTAD #12230) .......603 10.25 Verifying the global envelope of linear elements stresses (on end points and middle of super element)
(TTAD #12230)...........................................................................................................................................................603 10.26 Verifying the Min/Max values from the user reports (TTAD# 12231)..........................................................................603 10.27 Verifying the global envelope of linear elements stresses (on all quarters of super element) (TTAD #12230)...........604 10.28 Creating the rules table (TTAD #11802).....................................................................................................................604 10.29 Creating the steel materials description report (TTAD #11954) .................................................................................604 10.30 System stability when the column releases interfere with support restraints (TTAD #10557) ....................................604 10.31 Generating the critical magnification factors report (TTAD #11379)...........................................................................605 10.32 Modal analysis: eigen modes results for a structure with one level ............................................................................605 10.33 Generating a report with modal analysis results (TTAD #10849) ...............................................................................605
11 SEISMIC ANALYSIS .........................................................................................................................607 11.1 Verifying the spectrum results for EC8 seism (TTAD #11478) ...................................................................................608 11.2 Verifying the spectrum results for EC8 seism (TTAD #12472) ...................................................................................608 11.3 Verifying the combinations description report (TTAD #11632) ...................................................................................608 11.4 EC8 : Verifying the displacements results of a linear element according to Czech seismic standards (CSN EN 1998-1)
(DEV2012 #3.18)........................................................................................................................................................608 11.5 Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517) .............................................608
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11.6 EC8 French Annex: verifying torsors on walls ........................................................................................................... 609 11.7 EC8 French Annex: verifying torsors on grouped walls from a multi-storey concrete structure ................................. 609 11.8 Verifying the damping correction influence over the efforts in supports (TTAD #13011). .......................................... 609 11.9 EC8 French Annex: verifying seismic results when a design spectrum is used (TTAD #13778) ............................... 609 11.10 EC8: verifying the sum of actions on supports and nodes restraints (TTAD #12706) ................................................ 609 11.11 Seismic norm PS92: verifying efforts and torsors on planar elements (TTAD #12974) ............................................. 609 11.12 EC8 Fr Annex: Generating forces results per modes on linear and planar elements (TTAD #13797) ....................... 610
12 STEEL DESIGN................................................................................................................................. 611 12.1 EC3 Test 2: Class section classification and share verification of an IPE300 beam subjected to linear
uniform loading .......................................................................................................................................................... 612 12.2 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 column................ 619 12.3 EC3 Test 3: Class section classification and share and bending moment verification of an IPE300 column............. 626 12.4 EC3 Test 5: Class section classification and combined axial force with bending moment verification of
an IPE300 column...................................................................................................................................................... 633 12.5 EC3 test 4: Class section classification and bending moment verification of an IPE300 column............................... 639 12.6 EC3 Test 1: Class section classification and compression verification of an IPE300 column .................................... 645 12.7 Generating the shape sheet by system (TTAD #11471) ............................................................................................ 651 12.8 Verifying the calculation results for steel cables (TTAD #11623) ............................................................................... 651 12.9 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545) .......................................................... 651 12.10 Verifying the shape sheet results for a column (TTAD #11550)................................................................................. 651 12.11 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) ................................................ 651 12.12 Verifying the cross section optimization according to EC3 (TTAD #11516) ............................................................... 652 12.13 Verifying shape sheet on S275 beam (TTAD #11731)............................................................................................... 652 12.14 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) ........................................ 652 12.15 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533) ............................................ 652 12.16 Changing the steel design template for a linear element (TTAD #12491).................................................................. 652 12.17 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation
(TTAD #12389) .......................................................................................................................................................... 653 12.18 EC3: Verifying the buckling length results (TTAD #11550) ........................................................................................ 653 12.19 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975)..... 653 12.20 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method ........................................... 653 12.21 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method .............................................. 653 12.22 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method................................................. 654 12.23 Verifying the steel shape optimization when using sections from Advance Steel Profiles database
(TTAD #11873) .......................................................................................................................................................... 654 12.24 Verifying the buckling coefficient Xy on a class 2 section .......................................................................................... 654
13 TIMBER DESIGN .............................................................................................................................. 655 13.1 Modifying the "Design experts" properties for timber linear elements (TTAD #12259) .............................................. 656 13.2 Verifying the timber elements shape sheet (TTAD #12337) ...................................................................................... 656
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13.3 Verifying the units display in the timber shape sheet (TTAD #12445) ........................................................................656 13.4 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending ......................................................656 13.5 EC5: Verifying a C24 timber beam subjected to shear force......................................................................................657 13.6 EC5: Verifying a timber column subjected to compression forces..............................................................................661 13.7 EC5: Verifying a timber beam subjected to combined bending and axial tension ......................................................665 13.8 EC5: Verifying a timber column subjected to tensile forces........................................................................................671 13.9 EC5: Shear verification for a simply supported timber beam......................................................................................674 13.10 EC5: Verifying a timber purlin subjected to oblique bending ......................................................................................675 13.11 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression ...679 13.12 EC5: Verifying a timber beam subjected to simple bending .......................................................................................680 13.13 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression ...................................................684 13.14 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes ..............................................689
14 XML TEMPLATE FILES ....................................................................................................................693 14.1 Loading a template with the properties of a planar element (DEV2012 #1.4) ............................................................694 14.2 Loading a template with the properties of a linear element (DEV2012 #1.4)..............................................................694 14.3 Saving the properties of a planar element as a template (DEV2012 #1.4).................................................................694 14.4 Saving the properties of a linear element as a template (DEV2012 #1.4) ..................................................................694
1 Finite Elements Analysis
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1.1 System of two bars with three hinges (01-0002SSLLB_FEM)
Test ID: 2434
Test status: Passed
1.1.1 Description On a system of two bars (AC and BC) with three hinges, a punctual load in applied in point C. The vertical displacement in point C and the tensile stress on the bars are verified.
1.1.2 Background
1.1.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 09/89; ■ Analysis type: linear static; ■ Element type: linear.
System of two bars with three hinges Scale =1/33 0002SSLLB_FEM
4.500 m
30° 30°
4.500 m
AA BB
CC
FF
X
Y
Z X
Y
Z
Units
I. S.
Geometry
■ Bars angle relative to horizontal: θ = 30°, ■ Bars length: l = 4.5 m, ■ Bar section: A = 3 x 10-4 m2.
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa.
Boundary conditions
■ Outer: Hinged in A and B, ■ Inner: Hinge on C
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Loading
■ External: Punctual load in C: F = -21 x 103 N. ■ Internal: None.
1.1.2.2 Displacement of the model in C
Reference solution
uc = -3 x 10-3 m
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 21 nodes, ■ 20 linear elements.
Displacement shape
System of two bars with three hinges Scale =1/33 Displacement in C 0002SSLLB_FEM
1.1.2.3 Bars stresses
Reference solutions
σAC bar = 70 MPa
σBC bar = 70 MPa
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 21 nodes, ■ 20 linear elements.
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1.1.2.4 Shape of the stress diagram
System of two bars with three hinges Scale =1/34 Bars stresses 0002SSLLB_FEM
1.1.2.5 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
1.1.3 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%
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1.2 Slender beam with variable section (fixed-free) (01-0004SDLLB_FEM)
Test ID: 2436
Test status: Passed
1.2.1 Description Verifies the first eigen mode frequencies for a slender beam with variable section, subjected to its own weight.
1.2.2 Background
1.2.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 09/89; ■ Analysis type: modal analysis; ■ Element type: linear.
Slender beam with variable section (fixed-free) Scale =1/4 01-0004SDLLB_FEM
Units
I. S.
Geometry
■ Beam length: l = 1 m, ■ Initial section (in A):
► Height: h1 = 0.04 m, ► Width: b1 = 0.04 m, ► Section: A1 = 1.6 x 10-3 m2, ► Flexure moment of inertia relative to z-axis: Iz1 = 2.1333 x 10-7 m4,
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■ Final section (in B): ► Height: h2 = 0.01 m, ► Width: b2 = 0.01 m, ► Section: A2 = 10-4 m2, ► Flexure moment of inertia relative to z-axis: Iz2 = 8.3333 x 10-10 m4.
Materials properties
■ Longitudinal elastic modulus: E = 2 x 1011 Pa, ■ Density: 7800 kg/m3.
Boundary conditions
■ Outer: Fixed in A, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.2.2.2 Eigen mode frequencies
Reference solutions
Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories):
∂2
∂x2 (EIz ∂2v∂x2 ) = -ρA
∂2v∂x2 where Iz and A vary with the abscissa.
The result is: fi = 12π λi
h2l2
E12ρ
λ1 λ2 λ3 λ4 λ5 23.289 73.9 165.23 299.7 478.1
Finite elements modeling
■ Linear element: variable beam, imposed mesh, ■ 31 nodes, ■ 30 linear elements.
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Eigen mode shapes
1.2.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 54.18 CM2 Eigen mode Eigen mode 2 frequency [Hz] 171.94 CM2 Eigen mode Eigen mode 3 frequency [Hz] 384.4 CM2 Eigen mode Eigen mode 4 frequency [Hz] 697.24 CM2 Eigen mode Eigen mode 5 frequency [Hz] 1112.28
1.2.3 Calculated results
Result name
Result description Value Error
Eigen mode 1 frequency [Hz] 54.01 Hz -0.31% Eigen mode 2 frequency [Hz] 170.58 Hz -0.79% Eigen mode 3 frequency [Hz] 378.87 Hz -1.44% Eigen mode 4 frequency [Hz] 681.31 Hz -2.28% Eigen mode 5 frequency [Hz] 1075.7 Hz -3.29%
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1.3 Thin lozenge-shaped plate fixed on one side (alpha = 15 °) (01-0008SDLSB_FEM)
Test ID: 2440
Test status: Passed
1.3.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.
1.3.2 Background
1.3.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; ■ Analysis type: modal analysis; ■ Element type: planar.
Thin lozenge-shaped plate fixed on one side Scale =1/10 01-0008SDLSB_FEM
Units
I. S.
Geometry
■ Thickness: t = 0.01 m, ■ Side: a = 1 m, ■ α = 15° ■ Points coordinates:
► A ( 0 ; 0 ; 0 ) ► B ( a ; 0 ; 0 ) ► C ( 0.259a ; 0.966a ; 0 ) ► D ( 1.259a ; 0.966a ; 0 )
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: AB side fixed, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.3.2.2 Eigen modes frequencies function by α angle
Reference solution
M. V. Barton formula for a lozenge of side "a" leads to the frequencies:
fj = ⋅⋅π 2a21
λi2
)1(12Et
2
2
ν−ρ where i = 1,2, or λi
2 = g(α).
α = 15° λ1
2 3.601 λ2
2 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
1.3.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 8.999 CM2 Eigen mode Eigen mode 2 frequency [Hz] 22.1714
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1.3.3 Calculated results
Result name
Result description Value Error
Eigen mode 1 frequency [Hz] 8.95 Hz -0.54% Eigen mode 2 frequency [Hz] 21.69 Hz -2.17%
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1.4 Thin circular ring fixed in two points (01-0006SDLLB_FEM)
Test ID: 2438
Test status: Passed
1.4.1 Description Verifies the first eigen modes frequencies for a thin circular ring fixed in two points, subjected to its own weight only.
1.4.2 Background
1.4.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 12/89; ■ Analysis type: modal analysis, plane problem; ■ Element type: linear.
Thin circular ring fixed in two points Scale =1/2 01-0006SDLLB_FEM
Units
I. S.
Geometry
■ Average radius of curvature: OA = OB = R = 0.1 m, ■ Angular spacing between points A and B: 120° ; ■ Rectangular straight section:
► Thickness: h = 0.005 m, ► Width: b = 0.010 m, ► Section: A = 5 x 10-5 m2, ► Flexure moment of inertia relative to the vertical axis: I = 1.042 x 10-10 m4,
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■ Point coordinates: ► O (0 ;0),
► A (-0.05 3 ; -0.05),
► B (0.05 3 ; -0.05).
Materials properties
■ Longitudinal elastic modulus: E = 7.2 x 1010 Pa ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 2700 kg/m3.
Boundary conditions
■ Outer: Fixed at A and B, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.4.2.2 Eigen mode frequencies
Reference solutions
The deformation of the fixed ring is calculated from the deformations of the free-free thin ring
■ Symmetrical mode: ► u’i = i cos(iθ) ► v’i = sin (iθ)
► θ’i = 1-i2R sin (iθ)
■ Antisymmetrical mode: ► u’i = i sin(iθ) ► v’i = -cos (iθ)
► θ’i = 1-i2R cos (iθ)
From Green’s method results:
fj = π21
λj ⋅2R
h 12E
ρ with a support angle of 120°.
i 1 2 3 4 Symmetrical mode 4.8497 14.7614 23.6157
Antisymmetrical mode 1.9832 9.3204 11.8490 21.5545
Finite elements modeling
■ Linear element: beam, without meshing, ■ 32 nodes, ■ 32 linear elements.
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Eigen mode shapes
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1.4.2.3 Theoretic results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency - 1 antisymmetric 1 [Hz] 235.3 CM2 Eigen mode Eigen mode 2 frequency - 2 symmetric 1 [Hz] 575.3 CM2 Eigen mode Eigen mode 3 frequency - 3 antisymmetric 2 [Hz] 1105.7 CM2 Eigen mode Eigen mode 4 frequency - 4 antisymmetric 3 [Hz] 1405.6 CM2 Eigen mode Eigen mode 5 frequency - 5 symmetric 2 [Hz] 1751.1 CM2 Eigen mode Eigen mode 6 frequency - 6 antisymmetric 4 [Hz] 2557 CM2 Eigen mode Eigen mode 7 frequency - 7 symmetric 3 [Hz] 2801.5
1.4.3 Calculated results
Result name Result description 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%
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1.5 Thin lozenge-shaped plate fixed on one side (alpha = 45 °) (01-0010SDLSB_FEM)
Test ID: 2442
Test status: Passed
1.5.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.
1.5.2 Background
1.5.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; ■ Analysis type: modal analysis; ■ Element type: planar.
Thin lozenge-shaped plate fixed on one side Scale =1/10 01-0010SDLSB_FEM
Units
I. S.
Geometry
■ Thickness: t = 0.01 m, ■ Side: a = 1 m, ■ α = 45°
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■ Points coordinates: ► A ( 0 ; 0 ; 0 ) ► B ( a ; 0 ; 0 )
► C ( 22
a ; 22
a ; 0 )
► D (2
22 +a ;
22
a ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: AB side fixed, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.5.2.2 Eigen mode frequencies relative to the α angle
Reference solution
M. V. Barton formula for a lozenge of side "a" leads to the frequencies:
fj = ⋅⋅π 2a21
λi2
)1(12Et
2
2
ν−ρ where i = 1,2, or λi
2 = g(α).
α = 45° λ1
2 4.4502 λ2
2 10.56 M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 961 nodes, ■ 900 surface quadrangles.
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Eigen mode shapes
1.5.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 11.1212 CM2 Eigen mode Eigen mode 2 frequency [Hz] 26.3897
1.5.3 Calculated results
Result name Result description Value Error Eigen mode 1 frequency [Hz] 11.28 Hz 1.43% Eigen mode 2 frequency [Hz] 28.08 Hz 6.41%
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1.6 Vibration mode of a thin piping elbow in plane (case 2) (01-0012SDLLB_FEM)
Test ID: 2444
Test status: Passed
1.6.1 Description Verifies the vibration modes of a thin piping elbow (1 m radius) extended by two straight elements of length L, subjected to its self weight only.
1.6.2 Background
1.6.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; ■ Analysis type: modal analysis (plane problem); ■ Element type: linear.
Vibration mode of a thin piping elbow Scale = 1/11 Case 2 01-0012SDLLB_FEM
Units
I. S.
Geometry
■ Average radius of curvature: OA = R = 1 m, ■ L = 0.6 m, ■ Straight circular hollow section: ■ Outer diameter de = 0.020 m, ■ Inner diameter di = 0.016 m, ■ Section: A = 1.131 x 10-4 m2, ■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4, ■ Polar inertia: Ip = 9.274 x 10-9 m4.
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■ Points coordinates (in m): ► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 ) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: ► Fixed at points C and D ► At A: translation restraint along y and z, ► At B: translation restraint along x and z,
■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.6.2.2 Eigen mode frequencies
Reference solution
The Rayleigh method applied to a thin curved beam is used to determine parameters such as:
■ in plane bending:
fj = 2
2i
R2 ⋅π
λ
AEIz
ρ where i = 1,2,
Finite elements modeling
■ Linear element: beam, ■ 23 nodes, ■ 22 linear elements.
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Eigen mode shapes
1.6.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode frequency in plane 1 [Hz] 94 CM2 Eigen mode Eigen mode frequency in plane 2 [Hz] 180
1.6.3 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 100%
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1.7 Thin circular ring hanged on an elastic element (01-0014SDLLB_FEM)
Test ID: 2446
Test status: Passed
1.7.1 Description Verifies the first eigen modes frequencies of a circular ring hanged on an elastic element, subjected to its self weight only.
1.7.2 Background
1.7.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 13/89; ■ Analysis type: modal analysis, plane problem; ■ Element type: linear.
Thin circular ring hang from an elastic element Scale = 1/1 01-0014SDLLB_FEM
Units
I. S.
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Geometry
■ Average radius of curvature: OB = R = 0.1 m, ■ Length of elastic element: AB = 0.0275 m ; ■ Straight rectangular section:
► Ring Thickness: h = 0.005 m,
Width: b = 0.010 m,
Section: A = 5 x 10-5 m2,
Flexure moment of relative to the vertical axis: I = 1.042 x 10-10 m4,
► Elastic element Thickness: h = 0.003 m,
Width: b = 0.010 m,
Section: A = 3 x 10-5 m2,
Flexure moment of inertia relative to the vertical axis: I = 2.25 x 10-11 m4,
■ Points coordinates: ► O ( 0 ; 0 ), ► A ( 0 ; -0.0725 ), ► B ( 0 ; -0.1 ).
Materials properties
■ Longitudinal elastic modulus: E = 7.2 x 1010 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 2700 kg/m3.
Boundary conditions
■ Outer: Fixed in A, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.7.2.2 Eigen mode frequencies
Reference solutions
The reference solution was established from experimental results of a mass manufactured aluminum ring.
Finite elements modeling
■ Linear element: beam, ■ 43 nodes, ■ 43 linear elements.
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Eigen mode shapes
1.7.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 Asymmetrical frequency [Hz] 28.80 CM2 Eigen mode Eigen mode 2 Symmetrical frequency [Hz] 189.30 CM2 Eigen mode Eigen mode 3 Asymmetrical frequency [Hz] 268.80 CM2 Eigen mode Eigen mode 4 Asymmetrical frequency [Hz] 641.00 CM2 Eigen mode Eigen mode 5 Symmetrical frequency [Hz] 682.00 CM2 Eigen mode Eigen mode 6 Asymmetrical frequency [Hz] 1063.00
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1.7.3 Calculated results
Result name
Result description Value Error
Eigen mode 1 Asymmetrical frequency [Hz] 28.81 Hz 0.03% Eigen mode 2 Symmetrical frequency [Hz] 189.69 Hz 0.21% Eigen mode 3 Asymmetrical frequency [Hz] 269.38 Hz 0.22% Eigen mode 4 Asymmetrical frequency [Hz] 642.15 Hz 0.18% Eigen mode 5 Symmetrical frequency [Hz] 683.9 Hz 0.28% Eigen mode 6 Asymmetrical frequency [Hz] 1065.73 Hz 0.26%
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1.8 Tied (sub-tensioned) beam (01-0005SSLLB_FEM)
Test ID: 2437
Test status: Passed
1.8.1 Description Verifies the tension force on a beam reinforced by a system of hinged bars, subjected to a uniform linear load.
1.8.2 Background
1.8.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 13/89; ■ Analysis type: static, thermoelastic (plane problem); ■ Element type: linear.
Tied (sub-tensioned) beam Scale =1/37 01-0005SSLLB_FEM
Units
I. S.
Geometry
■ Length: ► AD = FB = a = 2 m, ► DF = CE = b = 4 m, ► CD = EF = c = 0.6 m, ► AC = EB = d = 2.088 m, ► Total length: L = 8 m,
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■ AD, DF, FB Beams: ► Section: A = 0.01516 m2, ► Shear area: Ar = A / 2.5, ► Inertia moment: I = 2.174 x 10-4 m4,
■ CE Bar: ► Section: A1 = 4.5 x 10-3 m2,
■ AC, EB bar: ► Section: A2 = 4.5 x 10-3 m2,
■ CD, EF bars: ► Section: A3 = 3.48 x 10-3 m2.
Materials properties
■ Isotropic linear elastic material, ■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Shearing module: G = 0.4x E.
Boundary conditions
■ Outer: Hinged in A, support connection in B (blocked vertical translation), ■ Inner: Hinged at bar ends: AC, CD, EF, EB.
Loading
■ External: Uniform linear load p = -50000 N/ml, ■ Internal: Shortening of the CE tie of δ = 6.52 x 10-3 m (dilatation coefficient: αCE = 1 x 10-5 /°C and temperature
variation ΔT = -163°C).
1.8.2.2 Compression force in CE bar
Reference solution
The solution is established by considering the deformation effects due to the shear force and normal force:
μ = 1 - 43 x
aL
k = AAr
= 2.5
t = IA
γ = (L/c)2 x (1+ (A/A1) x (b/L) + 2 x (A/A2) x (d/a)2 x (d/L) + 2 x (A/A3) (c/a)2 x (c/L)
τ = k x [(2Et2) / (GaL)]
ρ = μ + γ + τ
μ0 = 1 – (a/L)2 x (2 – a/L)
τ0 = 6k x (E/G) x (t/L)2 x (1 + b/L)
ρ0 = μ0 + τ0
NCE = - (1/12) x (pL2/c) x (ρ0 /ρ) + (EI/(Lc2)) x (δ/ρ) = 584584 N
Finite elements modeling
Linear element: without meshing,
■ AD, DF, FB: S beam (considering the shear force deformations), ■ AC, CD, EF, EB: bar, ■ CE: beam, ■ 6 nodes.
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Force diagrams
Tied (sub-tensioned) beam Scale =1/31 Compression force in CE bar
1.8.2.3 Bending moment at point H
Reference solution
MH = - (1/8) x pL2 x [1- (2/3) x (ρ0/ρ)] – (EI/(Lc)) x (δ/p) = 49249.5 N
Finite elements modeling
Linear element: without meshing,
■ AD, DF, FB: S beam (considering the shear force deformations), ■ AC, CD, EF, EB: bar, ■ CE: beam, ■ 6 nodes.
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Shape of the bending moment diagram
Tied (sub-tensioned) beam Scale =1/31 Mz bending moment
1.8.2.4 Vertical displacement at point D
Reference solution
The reference displacement vD provided by AFNOR is determined by averaging the results of several software with implemented finite elements method.
vD = -0.5428 x 10-3 m
Finite elements modeling
■ Linear element: without meshing, ► AD, DF, FB: S beam (considering the shear force deformations), ► AC, CD, EF, EB: bar, ► CE: beam,
■ 6 nodes.
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Deformed shape
Tied (sub-tensioned) beam Scale =1/31 Deformed
1.8.2.5 Theoretical results
Solver Result name Result description Reference value CM2 FX Tension force on CE bar [N] 584584
1.8.3 Calculated results
Result name Result description Value Error Fx Tension force on CE bar [N] 584580 N 0.00%
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1.9 Circular plate under uniform load (01-0003SSLSB_FEM)
Test ID: 2435
Test status: Passed
1.9.1 Description On a circular plate of 5 mm thickness and 2 m diameter, an uniform load, perpendicular on the plan of the plate, is applied. The vertical displacement on the plate center is verified.
1.9.2 Background
1.9.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 03/89; ■ Analysis type: linear static; ■ Element type: planar.
Circular plate under uniform load Scale =1/10 01-0003SSLSB_FEM
Units
I. S.
Geometry
■ Circular plate radius: r = 1m, ■ Circular plate thickness: h = 0.005 m.
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: Plate fixed on the side (in all points of its perimeter), For the modeling, we consider only a quarter of the plate and we impose symmetry conditions on some nodes (see the following model; yz plane symmetry condition):translation restrained nodes along x and rotation restrained nodes along y and z: translation restrained nodes along x and rotation restrained nodes along y and z:
■ Inner: None.
Loading
■ External: Uniform loads perpendicular on the plate: pZ = -1000 Pa, ■ Internal: None.
1.9.2.2 Vertical displacement of the model at the center of the plate
Reference solution
Circular plates form:
u = pr4
64D = -1000 x 14
64 x 2404 = - 6.50 x 10-3 m
with the plate radius coefficient: D = Eh3
12(1-ν2) = 2.1 x 1011 x 0.0053
12(1-0.32)
D = 2404
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 70 nodes, ■ 58 planar elements.
Circular plate under uniform load Scale =1.5 Meshing 01-0003SSLSB_FEM
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Deformed shape
Circular plate under uniform load Scale =1.5 Deformed 01-0003SSLSB_FEM
1.9.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement on the plate center [mm] -6.50
1.9.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement on the plate center [mm] -6.47032 mm 0.46%
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1.10 Double fixed beam (01-0016SDLLB_FEM)
Test ID: 2448
Test status: Passed
1.10.1 Description Verifies the eigen modes frequencies and the vertical displacement on the middle of a beam consisting of eight elements of length "l", having identical characteristics. A punctual load of -50000 N is applied.
1.10.2 Background
1.10.2.1 Model description ■ Reference: internal GRAITEC test (beams theory); ■ Analysis type: static linear, modal analysis; ■ Element type: linear.
Units
I. S.
Geometry
■ Length: l = 16 m, ■ Axial section: S=0.06 m2 ■ Inertia I = 0.0001 m4
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 N/m2, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7850 kg/m3
Boundary conditions
■ Outer: Fixed at both ends x = 0 and x = 8 m, ■ Inner: None.
Loading
■ External: Punctual load P = -50000 N at x = 4m, ■ Internal: None.
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1.10.2.2 Displacement of the model in the linear elastic range
Reference solution
The reference vertical displacement v5, is calculated at the middle of the beam at x = 2 m.
m 05079.00001.0111.2192
1650000192
33
5 =××
×==
EEIPlv
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 9 nodes, ■ 8 elements.
Deformed shape
Double fixed beam Deformed
1.10.2.3 Eigen mode frequencies of the model in the linear elastic range
Reference solution
Knowing that the first four eigen mode frequencies of a double fixed beam are given by the following formula:
SIE
Lf nn .
...2 2
2
ρπχ
= where for the first 4 eigen modes frequencies
⎪⎪
⎩
⎪⎪
⎨
⎧
→=
→=
→=
→=
Hz 26.228=f 8.199
Hz 15.871=f 9.120
Hz 8.095=f 67.61
Hz 2.937=f 37.22
424
323
222
121
χ
χ
χ
χ
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 9 nodes, ■ 8 elements.
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Modal deformations
Double fixed beam Mode 1
Double fixed beam Mode 2
Double fixed beam Mode 3
Double fixed beam Mode 4
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1.10.2.4 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
1.10.3 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%
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1.11 Short beam on simple supports (eccentric) (01-0018SDLLB_FEM)
Test ID: 2450
Test status: Passed
1.11.1 Description Verifies the first eigen mode frequencies of a short beam on simple supports (the supports are eccentric relative to the neutral axis).
1.11.2 Background
1.11.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; ■ Analysis type: modal analysis, (plane problem); ■ Element type: linear.
Short beam on simple supports (eccentric) Scale = 1/5 01-0018SDLLB_FEM
Units
I. S.
Geometry
■ Height: h = 0.2m, ■ Length: l = 1 m, ■ Width: b = 0.1 m, ■ Section: A = 2 x 10-2 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 6.667 x 10-5 m4.
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: ► Hinged at A (null horizontal and vertical displacements), ► Simple support at B.
■ Inner: None.
Loading
■ External: None. ■ Internal: None.
1.11.2.2 Eigen modes frequencies
Reference solution
The problem has no analytical solution, the solution is determined by averaging several software: Timoshenko model with shear force deformation effects and rotation inertia. The bending modes and the traction-compression are coupled.
Finite elements modeling
■ Linear element: S beam, imposed mesh, ■ 10 nodes, ■ 9 linear elements.
Eigen modes shape
Short beam on simple supports (eccentric) Mode 1
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Short beam on simple supports (eccentric) Mode 2
Short beam on simple supports (eccentric) Mode 3
Short beam on simple supports (eccentric) Mode 4
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Short beam on simple supports (eccentric) Mode 5
1.11.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 392.8 CM2 Eigen mode Eigen mode 2 frequency [Hz] 902.2 CM2 Eigen mode Eigen mode 3 frequency [Hz] 1591.9 CM2 Eigen mode Eigen mode 4 frequency [Hz] 2629.2 CM2 Eigen mode Eigen mode 5 frequency [Hz] 3126.2
1.11.3 Calculated results
Result name
Result description Value Error
Eigen mode 1 frequency [Hz] 393.7 Hz 0.23% Eigen mode 2 frequency [Hz] 945.35 Hz 4.78% Eigen mode 3 frequency [Hz] 1595.94 Hz 0.25% Eigen mode 4 frequency [Hz] 2526.22 Hz -3.92% Eigen mode 5 frequency [Hz] 3118.91 Hz -0.23%
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1.12 Thin lozenge-shaped plate fixed on one side (alpha = 30 °) (01-0009SDLSB_FEM)
Test ID: 2441
Test status: Passed
1.12.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.
1.12.2 Background
1.12.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; ■ Analysis type: modal analysis; ■ Element type: planar.
Thin lozenge-shaped plate fixed on one side Scale =1/10 01-0009SDLSB_FEM
Units
I. S.
Geometry
■ Thickness: t = 0.01 m, ■ Side: a = 1 m, ■ α = 30° ■ Points coordinates:
► A ( 0 ; 0 ; 0 ) ► B ( a ; 0 ; 0 )
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► C ( 0.5a ; 3 2 a ; 0 )
► D ( 1.5a ; 3 2 a ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: AB side fixed, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.12.2.2 Eigen mode frequencies relative to the α angle
Reference solution
M. V. Barton formula for a lozenge of side "a" leads to the frequencies:
fj = ⋅⋅π 2a21
λi2
)1(12Et
2
2
ν−ρ where i = 1,2, or λi
2 = g(α).
α = 30° λ1
2 3.961 λ2
2 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
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1.12.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 9.8987 CM2 Eigen mode Eigen mode 2 frequency [Hz] 25.4651
1.12.3 Calculated results
Result name
Result description Value Error
Eigen mode 1 frequency [Hz] 9.82 Hz -0.80% Eigen mode 2 frequency [Hz] 23.44 Hz -7.95%
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1.13 Vibration mode of a thin piping elbow in plane (case 3) (01-0013SDLLB_FEM)
Test ID: 2445
Test status: Passed
1.13.1 Description Verifies the vibration modes of a thin piping elbow (1 m radius) extended by two straight elements of length L, subjected to its self weight only.
1.13.2 Background
1.13.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; ■ Analysis type: modal analysis (plane problem); ■ Element type: linear.
Vibration mode of a thin piping elbow Scale = 1/12 Case 3 01-0013SDLLB_FEM
Units
I. S.
Geometry
■ Average radius of curvature: OA = R = 1 m, ■ Straight circular hollow section: ■ Outer diameter: de = 0.020 m, ■ Inner diameter: di = 0.016 m, ■ Section: A = 1.131 x 10-4 m2, ■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4, ■ Polar inertia: Ip = 9.274 x 10-9 m4.
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■ Points coordinates (in m): ► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 ) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: ► Fixed at points C and Ds, ► At A: translation restraint along y and z, ► At B: translation restraint along x and z,
■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.13.2.2 Eigen mode frequencies
Reference solution
The Rayleigh method applied to a thin curved beam is used to determine parameters such as:
■ in plane bending:
fj = 2
2i
R2 ⋅π
λ
AEIz
ρ where i = 1,2,
Finite elements modeling
■ Linear element: beam, ■ 41 nodes, ■ 40 linear elements.
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Eigen mode shapes
1.13.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode frequency in plane 1 [Hz] 25.300 CM2 Eigen mode Eigen mode frequency in plane 2 [Hz] 27.000
1.13.3 Calculated results
Result name
Result description Value Error
Eigen mode frequency in plane 1 [Hz] 24.96 Hz -1.34% Eigen mode frequency in plane 2 [Hz] 26.71 Hz 100%
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1.14 Short beam on simple supports (on the neutral axis) (01-0017SDLLB_FEM)
Test ID: 2449
Test status: Passed
1.14.1 Description Verifies the first eigen mode frequencies of a short beam on simple supports (the supports are located on the neutral axis), subjected to its own weight only.
1.14.2 Background ■ Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; ■ Analysis type: modal analysis (plane problem); ■ Element type: linear.
Short beam on simple supports on the neutral axis Scale = 1/6 01-0017SDLLB_FEM
Units
I. S.
Geometry
■ Height: h = 0.2 m, ■ Length: l = 1 m, ■ Width: b = 0.1 m, ■ Section: A = 2 x 10-2 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 6.667 x 10-5 m4.
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: ► Hinged at A (null horizontal and vertical displacements), ► Simple support in B.
■ Inner: None.
Loading
■ External: None. ■ Internal: None.
1.14.2.1 Eigen modes frequencies
Reference solution
The bending beams equation gives, when superimposing, the effects of simple bending, shear force deformations and rotation inertia, Timoshenko formula.
The reference eigen modes frequencies are determined by a numerical simulation of this equation, independent of any software.
The eigen frequencies in tension-compression are given by:
fi = ⋅π
λl2
i ρE
where λi = 2
)1i2( −
Finite elements modeling
■ Linear element: S beam, imposed mesh, ■ 10 nodes, ■ 9 linear elements.
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Eigen mode shapes
1.14.2.2 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 431.555 CM2 Eigen mode Eigen mode 2 frequency [Hz] 1265.924 CM2 Eigen mode Eigen mode 3 frequency [Hz] 1498.295 CM2 Eigen mode Eigen mode 4 frequency [Hz] 2870.661 CM2 Eigen mode Eigen mode 5 frequency [Hz] 3797.773 CM2 Eigen mode Eigen mode 6 frequency [Hz] 4377.837
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1.14.3 Calculated results
Result name
Result description Value Error
Eigen mode 1 frequency [Hz] 437.12 Hz 1.29% Eigen mode 2 frequency [Hz] 1264.32 Hz -0.13% Eigen mode 3 frequency [Hz] 1537.16 Hz 2.59% Eigen mode 4 frequency [Hz] 2911.46 Hz 1.42% Eigen mode 5 frequency [Hz] 3754.54 Hz -1.14% Eigen mode 6 frequency [Hz] 4281.23 Hz -2.21%
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1.15 Thin lozenge-shaped plate fixed on one side (alpha = 0 °) (01-0007SDLSB_FEM)
Test ID: 2439
Test status: Passed
1.15.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.
1.15.2 Background
1.15.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; ■ Analysis type: modal analysis; ■ Element type: planar.
Thin lozenge-shaped plate fixed on one side Scale =1/10 01-0007SDLSB_FEM
Units
I. S.
Geometry
■ Thickness: t = 0.01 m, ■ Side: a = 1 m, ■ α = 0° ■ Points coordinates:
► A ( 0 ; 0 ; 0 ) ► B ( a ; 0 ; 0 ) ► C ( 0 ; a ; 0 ) ► D ( a ; a ; 0 )
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: AB side fixed, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.15.2.2 Eigen mode frequencies relative to the α angle
Reference solution
M. V. Barton formula for a side "a" lozenge, leads to the frequencies:
fj = ⋅⋅π 2a21
λi2
)1(12Et
2
2
ν−ρ where i = 1,2, and λi
2 = g(α).
α = 0 λ1
2 3.492 λ2
2 8.525 M.V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 61 nodes, ■ 900 surface quadrangles.
Eigen mode shapes
1.15.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 8.7266 CM2 Eigen mode Eigen mode 2 frequency [Hz] 21.3042
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1.15.3 Calculated results
Result name Result description Value Error Eigen mode 1 frequency [Hz] 8.67 Hz -0.65% Eigen mode 2 frequency [Hz] 21.21 Hz -0.44%
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1.16 Vibration mode of a thin piping elbow in plane (case 1) (01-0011SDLLB_FEM)
Test ID: 2443
Test status: Passed
1.16.1 Description Verifies the vibration modes of a thin piping elbow (1 m radius) with fixed ends and subjected to its self weight only.
1.16.2 Background
1.16.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; ■ Analysis type: modal analysis (plane problem); ■ Element type: linear.
Vibration mode of a thin piping elbow in plane Scale = 1/7 Case 1 01-0011SDLLB_FEM
Units
I. S.
Geometry
■ Average radius of curvature: OA = R = 1 m, ■ Straight circular hollow section: ■ Outer diameter: de = 0.020 m, ■ Inner diameter: di = 0.016 m, ■ Section: A = 1.131 x 10-4 m2, ■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4,
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■ Polar inertia: Ip = 9.274 x 10-9 m4. ■ Points coordinates (in m):
► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: Fixed at points A and B , ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.16.2.2 Eigen mode frequencies
Reference solution
The Rayleigh method applied to a thin curved beam is used to determine parameters such as:
■ in plane bending:
fj = 2
2i
R2 ⋅π
λ
AEIz
ρ where i = 1,2,
Finite elements modeling
■ Linear element: beam, ■ 11 nodes, ■ 10 linear elements.
Eigen mode shapes
1.16.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode frequency in plane 1 [Hz] 119 CM2 Eigen mode Eigen mode frequency in plane 2 [Hz] 227
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1.16.3 Calculated results
Result name
Result description Value Error
Eigen mode frequency in plane 1 [Hz] 120.09 Hz 0.92% Eigen mode frequency in plane 2 [Hz] 227.1 Hz 100%
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1.17 Double fixed beam with a spring at mid span (01-0015SSLLB_FEM)
Test ID: 2447
Test status: Passed
1.17.1 Description Verifies the vertical displacement on the middle of a beam consisting of four elements of length "l", having identical characteristics. A punctual load of -10000 N is applied.
1.17.2 Background
1.17.2.1 Model description ■ Reference: internal GRAITEC test; ■ Analysis type: linear static; ■ Element type: linear.
Units
I. S.
Geometry
■ l = 1 m ■ S = 0.01 m2 ■ I = 0.0001 m4
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: ■ Fixed at ends x = 0 and x = 4 m, ■ Elastic support with k = EI/l rigidity ■ Inner: None.
Loading
■ External: Punctual load P = -10000 N at x = 2m, ■ Internal: None.
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1.17.2.2 Displacement of the model in the linear elastic range
Reference solution
The reference vertical displacement v3, is calculated at the middle of the beam at x = 2 m.
Rigidity matrix of a plane beam:
Given the symmetry / X and load of the structure, it is unnecessary to consider the degrees of freedom associated with normal work (u2, u3, u4).
The same symmetry allows the deduction of:
■ v2 = v4 ■ β2 = -β4 ■ β3 = 0
( )( )( )( )( )( )654321
000
00
4626
612612
268026
612024612
268026
6120124612
268026
612024612
2646
612612
5
5
1
1
5
5
4
4
3
3
2
2
1
1
22
22
22
22
22
22
22
22
22
22
⎪⎪⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
−=
⎪⎪⎪⎪⎪⎪⎪
⎭
⎪⎪⎪⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−−
−
−−−
−
−⎟⎠
⎞⎜⎝
⎛ +−−
−
−−−
−
−
MR
P
MR
v
v
v
v
v
EI
β
β
β
β
β
llll
llll
lllll
lllll
lllll
llllll
lllll
lllll
llll
llll
33
333
333
333
33
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The elementary rigidity matrix of the spring in its local axis system, [ ])()(
1111
6
35 U
UEIk ⎥⎦
⎤⎢⎣
⎡−
−=
l, must be expressed in
the global axis system by means of the rotation matrix (90° rotation):
[ ]
( )( )( )( )( )( )6
6
6
3
3
3
5
000000010010000000000000010010000000
β
β
vu
vu
EIK
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
=l
→ 344332 43 0826 vvllll
−=⇒=++ βββ
→ 344332332 024612 vvvv =⇒=+−−
lllβ
→ y)unnecessar(usually 026826244423222 vvvv =⇒=+−++ βββ
lllll
(3) →
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 6 nodes, ■ 4 linear elements + 1 spring,
Deformed shape
Double fixed beam with a spring at mid span Deformed
Note: the displacement is expressed here in μm
1.17.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement on the middle of the beam [mm] -0.11905
1.17.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement on the middle of the beam [mm] -0.119048 mm 0.00%
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1.18 Rectangular thin plate simply supported on its perimeter (01-0020SDLSB_FEM)
Test ID: 2452
Test status: Passed
1.18.1 Description Verifies the first eigen mode frequencies of a thin rectangular plate simply supported on its perimeter.
1.18.2 Background
1.18.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLS 03/89; ■ Analysis type: modal analysis; ■ Element type: planar.
Rectangular thin plate simply supported on its perimeter Scale = 1/8 01-0020SDLSB_FEM
Units
I. S.
Geometry
■ Length: a = 1.5 m, ■ Width: b = 1 m, ■ Thickness: t = 0.01 m, ■ Points coordinates in m:
► A (0 ;0 ;0) ► B (0 ;1.5 ;0) ► C (1 ;1.5 ;0) ► D (1 ;0 ;0)
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: ► Simple support on all sides, ► For the modeling: hinged at A, B and D.
■ Inner: None.
Loading
■ External: None. ■ Internal: None.
1.18.2.2 Eigen modes frequencies
Reference solution
M. V. Barton formula for a rectangular plate with supports on all four sides, leads to:
fij = 2π
[ ( ai
)2 + ( bj
)2] )1(12
Et2
2
ν−ρ
where:
i = number of half-length of wave along y ( dimension a)
j = number of half-length of wave along x ( dimension b)
Finite elements modeling
■ Planar element: shell, ■ 496 nodes, ■ 450 planar elements.
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Eigen mode shapes
1.18.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 1; j = 1. [Hz] 35.63 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 2; j = 1. [Hz] 68.51 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 1; j = 2. [Hz] 109.62 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 3; j = 1. [Hz] 123.32 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 2; j = 2. [Hz] 142.51 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 3; j = 2. [Hz] 197.32
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1.18.3 Calculated results
Result name 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%
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1.19 Slender beam on two fixed supports (01-0024SSLLB_FEM)
Test ID: 2456
Test status: Passed
1.19.1 Description A straight slender beam with fixed ends is loaded with a uniform load, several punctual loads and a torque. The shear force, bending moment, vertical displacement and horizontal reaction are verified.
1.19.2 Background
1.19.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 01/89; ■ Analysis type: linear static; ■ Element type: linear.
Slender beam on two fixed supports Scale = 1/4 01-0024SSLLB_FEM
Units
I. S.
Geometry
■ Length: L = 1 m, ■ Beam inertia: I = 1.7 x 10-8 m4.
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa.
Boundary conditions
■ Outer: Fixed at A and B, ■ Inner: None.
Loading
■ External: ► Uniformly distributed load from A to B: py = p = -24000 N/m, ► Punctual load at D: Fx = F1 = 30000 N, ► Torque at D: Cz = C = -3000 Nm, ► Punctual load at E: Fx = F2 = 10000 N, ► Punctual load at E: Fy = F = -20000 N.
■ Internal: None.
1.19.2.2 Shear force at G
Reference solution
Analytical solution:
■ Shear force at G: VG
VG = 0.216F – 1.26 LC
Finite elements modeling
■ Linear element: beam, ■ 5 nodes, ■ 4 linear elements.
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Results shape
Slender beam on two fixed supports Scale = 1/5 Shear force
1.19.2.3 Bending moment in G
Reference solution
Analytical solution:
■ Bending moment at G: MG
MG = pL2
24 - 0.045LF – 0.3C
Finite elements modeling
■ Linear element: beam, ■ 5 nodes, ■ 4 linear elements.
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Results shape
Slender beam on two fixed supports Scale = 1/5 Bending moment
1.19.2.4 Vertical displacement at G
Reference solution
Analytical solution:
■ Vertical displacement at G: vG
vG = pl4
384EI + 0.003375FL3
EI + 0.015CL2
EI
Finite elements modeling
■ Linear element: beam, ■ 5 nodes, ■ 4 linear elements.
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Results shape
Slender beam on two fixed supports Scale = 1/4 Deformed
1.19.2.5 Horizontal reaction at A
Reference solution
Analytical solution:
■ Horizontal reaction at A: HA HA = -0.7F1 –0.3F2
Finite elements modeling
■ Linear element: beam, ■ 5 nodes, ■ 4 linear elements.
1.19.2.6 Theoretical results
Solver Result name Result description Reference value CM2 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
1.19.3 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%
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1.20 Annular thin plate fixed on a hub (repetitive circular structure) (01-0022SDLSB_FEM)
Test ID: 2454
Test status: Passed
1.20.1 Description Verifies the eigen mode frequencies of a thin annular plate fixed on a hub.
1.20.2 Background
1.20.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLS 04/89; ■ Analysis type: modal analysis; ■ Element type: planar element.
Annular thin plate fixed on a hub (repetitive circular structure) Scale = 1/3 01-0022SDLSB_FEM
Units
I. S.
Geometry
■ Inner radius: Ri = 0.1 m, ■ Outer radius: Re = 0.2 m, ■ Thickness: t = 0.001 m.
Material properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
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Boundary conditions
■ Outer: Fixed on a hub at any point r = Ri. ■ Inner: None.
Loading
■ External: None. ■ Internal: None.
1.20.2.2 Eigen modes frequencies
Reference solution
The solution of determining the frequency based on Bessel functions leads to the following formula:
fij = 1
2πRe2 λij
2 Et2
12ρ(1-ν2)
where:
i = the number of nodal diameters
j = the number of nodal circles
and λij2 such as:
j \ i 0 1 2 3 0 13.0 13.3 14.7 18.5 1 85.1 86.7 91.7 100
Finite elements modeling
■ Planar element: plate, ■ 360 nodes, ■ 288 planar elements.
1.20.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 0; j = 0. [Hz] 79.26 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 0; j = 1. [Hz] 518.85 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 1; j = 0. [Hz] 81.09 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 1; j = 1. [Hz] 528.61 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 2; j = 0. [Hz] 89.63 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 2; j = 1. [Hz] 559.09 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 3; j = 0. [Hz] 112.79 CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 3; j = 1. [Hz] 609.70
1.20.3 Calculated results
Result name Result description 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%
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1.21 Bimetallic: Fixed beams connected to a stiff element (01-0026SSLLB_FEM)
Test ID: 2458
Test status: Passed
1.21.1 Description Two beams fixed at one end and rigidly connected to an undeformable beam is loaded with a punctual load. The deflection, vertical reaction and bending moment are verified in several points.
1.21.2 Background
1.21.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 05/89; ■ Analysis type: linear static; ■ Element type: linear.
Fixed beams connected to a stiff element Scale = 1/10 01-0026SSLLB_FEM
Units
I. S.
Geometry
■ Lengths: ► L = 2 m, ► l = 0.2 m,
■ Beams inertia moment: I = (4/3) x 10-8 m4, ■ The beam sections are squared, of side: 2 x 10-2 m.
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Materials properties
■ Longitudinal elastic modulus: E = 2 x 1011 Pa.
Boundary conditions
■ Outer: Fixed in A and C, ■ Inner: The tangents to the deflection of beams AB and CD at B and D remain horizontal; practically, we
restraint translations along x and z at nodes B and D.
Loading
■ External: In D: punctual load F = Fy = -1000N. ■ Internal: None.
1.21.2.2 Deflection at B and D
Reference solution
The theory of slender beams bending (Euler-Bernouilli formula) leads to a deflection at B and D:
The resolution of the hyperstatic system of the slender beam leads to:
vB = vD = FL3
24EI
Finite elements modeling
■ Linear element: beam, ■ 4 nodes, ■ 3 linear elements.
Results shape
Fixed beams connected to a stiff element Scale = 1/10 Deformed
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1.21.2.3 Vertical reaction at A and C
Reference solution
Analytical solution.
Finite elements modeling
■ Linear element: beam, ■ 4 nodes, ■ 3 linear elements.
1.21.2.4 Bending moment at A and C
Reference solution
Analytical solution.
Finite elements modeling
■ Linear element: beam, ■ 4 nodes, ■ 3 linear elements
1.21.2.5 Theoretical results
Solver Result name Result description Reference value CM2 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
1.21.3 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%
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1.22 Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM)
Test ID: 2453
Test status: Passed
1.22.1 Description Verifies the critical load result on node 5 of a cantilever beam in Eulerian buckling. A punctual load of -100000 is applied.
1.22.2 Background
1.22.2.1 Model description ■ Reference: internal GRAITEC test (Euler theory); ■ Analysis type: Eulerian buckling; ■ Element type: linear.
Units
I. S.
Geometry
■ L = 10 m ■ S=0.01 m2 ■ 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.
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1.22.2.2 Critical load on node 5
Reference solution
The reference critical load established by Euler is:
98696.010000098696N 98696
L4EIP 2
2
critique ==λ⇒=π
=
Finite elements modeling
■ Planar element: beam, imposed mesh, ■ 5 nodes, ■ 4 elements.
Deformed shape
1.22.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Fx Critical load on node 5. [N] -98696
1.22.3 Calculated results
Result name
Result description Value Error
Fx Critical load on node 5 (mode 1) [N] -100000 N -1.32%
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1.23 Slender beam on three supports (01-0025SSLLB_FEM)
Test ID: 2457
Test status: Passed
1.23.1 Description A straight slender beam on three supports is loaded with two punctual loads. The bending moment, vertical displacement and reaction on the center are verified.
1.23.2 Background
1.23.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 03/89; ■ Analysis type: static (plane problem); ■ Element type: linear.
Slender beam on three supports Scale = 1/49 01-0025SSLLB_FEM
Units
I. S.
Geometry
■ Length: L = 3 m, ■ Beam inertia: I = 6.3 x 10-4 m4.
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Materials properties
Longitudinal elastic modulus: E = 2.1 x 1011 Pa.
Boundary conditions
■ Outer: ► Hinged at A, ► Elastic support at B (Ky = 2.1 x 106 N/m), ► Simple support at C.
■ Inner: None.
Loading
■ External: 2 punctual loads F = Fy = -42000N. ■ Internal: None.
1.23.2.2 Bending moment at B
Reference solution
The resolution of the hyperstatic system of the slender beam leads to:
k = Ky3LEI6
■ Bending moment at B: MB
MB = ± 2L
)k8(F)k26(
++−
Finite elements modeling
■ Linear element: beam, ■ 5 nodes, ■ 4 linear elements.
Results shape
Slender beam on three supports Scale = 1/49 Bending moment
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1.23.2.3 Reaction in B
Reference solution
■ Compression force in the spring: VB
VB = -11F8 + k
Finite elements modeling
■ Linear element: beam, ■ 5 nodes, ■ 4 linear elements.
1.23.2.4 Vertical displacement at B
Reference solution
■ Deflection at the spring location: vB
vB = 11F
Ky(8 + k)
Finite elements modeling
■ Linear element: beam, ■ 5 nodes, ■ 4 linear elements.
Results shape
Slender beam on three supports Deformed
1.23.2.5 Theoretical results
Solver Result name Result description Reference value CM2 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
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1.23.3 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%
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1.24 Fixed thin arc in out of plane bending (01-0028SSLLB_FEM)
Test ID: 2460
Test status: Passed
1.24.1 Description An arc of a circle fixed at one end is loaded with a punctual force at its free end, perpendicular to the plane. The out of plane displacement, torsion moment and bending moment are verified.
1.24.2 Background
1.24.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 07/89; ■ Analysis type: static linear; ■ Element type: linear.
Fixed thin arc in out of plane bending Scale = 1/6 01-0028SSLLB_FEM
Units
I. S.
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Geometry
■ Medium radius: R = 1 m , ■ Circular hollow section:
► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.
Materials properties
■ Longitudinal elastic modulus: E = 2 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: Fixed at A. ■ Inner: None.
Loading
■ External: Punctual force in B perpendicular on the plane: Fz = F = 100 N. ■ Internal: None.
1.24.2.2 Displacements at B
Reference solution
Displacement out of plane at point B:
uB = FR3
EIx [ π4 +
EIx KT
(3π4 - 2)]
where KT is the torsional rigidity for a circular section (torsion constant is 2Ix).
KT = 2GIx = EIx
1 + ν ⇒ uB = FR3
EIx [ π4 + (1 + ν) (
3π4 - 2)]
Finite elements modeling
■ Linear element: beam, ■ 46 nodes, ■ 45 linear elements.
1.24.2.3 Moments at θ = 15°
Reference solution
■ Torsion moment: Mx’ = Mt = FR(1 - sinθ) ■ Bending moment: Mz’ = Mf = -FRcosθ
Finite elements modeling
■ Linear element: beam, ■ 46 nodes, ■ 45 linear elements.
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1.24.2.4 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
1.24.3 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%
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1.25 Portal frame with lateral connections (01-0030SSLLB_FEM)
Test ID: 2462
Test status: Passed
1.25.1 Description Verifies the rotation about z-axis and the bending moment on a portal frame with lateral connections.
1.25.2 Background
1.25.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 10/89; ■ Analysis type: static linear; ■ Element type: linear.
Portal frame with lateral connections Scale = 1/21 01-0030SSLLB_FEM
Units
I. S.
Geometry
Beam Length Moment of inertia AB lAB = 4 m IAB =
643 x 10-8 m4
AC lAC = 1 m IAC = 112 x 10-8 m4
AD lAD = 1 m IAD = 112 x 10-8 m4
AE lAE = 2 m IAE = 43 x 10-8 m4
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■ G is in the middle of DA. ■ The beams have square sections:
► AAB = 16 x 10-4 m ► AAD = 1 x 10-4 m ► AAC = 1 x 10-4 m ► AAE = 4 x 10-4 m
Materials properties
Longitudinal elastic modulus: E = 2 x 1011 Pa,
Boundary conditions
■ Outer: ► Fixed at B, D and E, ► Hinge at C,
■ Inner: None.
Loading
■ External: ► Punctual force at G: Fy = F = - 105 N, ► Distributed load on beam AD: p = - 103 N/m.
■ Internal: None.
1.25.2.2 Displacements at A
Reference solution
Rotation at A about z-axis:
We say: kAn = EIAnlAn
where n = B, C, D or E
K = kAB + kAD + kAE + 34 kAC
rAn = kAnK
C1 = FlAD
8 - plAB
2
12
θ = C14K
Finite elements modeling
■ Linear element: beam, ■ 6 nodes, ■ 5 linear elements.
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Displacements shape
Portal frame with lateral connections Deformed
1.25.2.3 Moments in A
Reference solution
■ MAB = plAB
2
12 + rAB x C1
■ MAD = - FlAD
8 + rAD x C1
■ MAE = rAE x C1
■ MAC = rAC x C1
Finite elements modeling
■ Linear element: beam, ■ 6 nodes, ■ 5 linear elements
1.25.2.4 Theoretical results
Solver Result name Result description Reference value CM2 RY Rotation θ about z-axis 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
1.25.3 Calculated results
Result name Result description Value Error RY Rotation Theta about z-axis in point A [rad] -0.227401 Rad -0.12% My Bending moment in point A (Moment AB) [Nm] 11021 N*m -0.02% My Bending moment in point A (Moment AC) [Nm] 113.704 N*m 0.13% My Bending moment in point A (Moment AD) [Nm] 12347.5 N*m -0.01% My Bending moment in point A (Moment AE) [Nm] 1212.77 N*m 0.12%
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1.26 Double hinged thin arc in planar bending (01-0029SSLLB_FEM)
Test ID: 2461
Test status: Passed
1.26.1 Description Verifies the rotation about Z-axis, the vertical displacement and the horizontal displacement on several points of a double hinged thin arc in planar bending.
1.26.2 Background
1.26.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 08/89; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Double hinged thin arc in planar bending Scale = 1/8 01-0029SSLLB_FEM
Units
I. S.
Geometry
■ Medium radius: R = 1 m , ■ Circular hollow section:
► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.
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Materials properties
■ Longitudinal elastic modulus: E = 2 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: ► Hinge at A, ► At B: allowed rotation along z, vertical displacement restrained along y.
■ Inner: None.
Loading
■ External: Punctual load at C: Fy = F = - 100 N. ■ Internal: None.
1.26.2.2 Displacements at A, B and C
Reference solution
■ Rotation about z-axis
θA = - θB = ( π2 - 1)
FR22EI
■ Displacement;
Vertical at C: vC = π8
FREA + (
3π4 - 2)
FR3
2EI
Horizontal at B: uB = FR
2EA - FR3
2EI
Finite elements modeling
■ Linear element: beam, ■ 37 nodes, ■ 36 linear elements.
Displacements shape
Fixed thin arc in planar bending Scale = 1/11 Deformed
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1.26.2.3 Theoretical results
Solver Result name Result description Reference value CM2 RY Rotation about Z-axis in point A [rad] 0.030774 CM2 RY Rotation about Z-axis in point B [rad] -0.030774 CM2 DZ Vertical displacement in point C [cm] -1.9206 CM2 DX Horizontal displacement in point B [cm] 5.3912
1.26.3 Calculated results
Result name
Result description Value Error
RY Rotation about Z-axis in point A [rad] 0.0307785 Rad 0.01% RY Rotation about Z-axis in point B [rad] -0.0307785 Rad -0.01% DZ Vertical displacement in point C [cm] -1.92019 cm 0.02% DX Horizontal displacement in point B [cm] 5.386 cm -0.10%
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1.27 Thin square plate fixed on one side (01-0019SDLSB_FEM)
Test ID: 2451
Test status: Passed
1.27.1 Description Verifies the first eigen modes frequencies of a thin square plate fixed on one side.
1.27.2 Background
1.27.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLS 01/89; ■ Analysis type: modal analysis; ■ Element type: planar.
Thin square plate fixed on one side Scale = 1/6 01-0019SDLSB_FEM
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: t = 1 m, ■ Points coordinates in m:
► A (0 ;0 ;0) ► B (1 ;0 ;0) ► C (1 ;1 ;0) ► D (0 ;1 ;0)
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: Edge AD fixed. ■ Inner: None.
Loading
■ External: None. ■ Internal: None.
1.27.2.2 Eigen modes frequencies
Reference solution
M. V. Barton formula for a square plate with side "a", leads to:
fj = 2a2
1⋅π
λi2
)1(12Et
2
2
ν−ρ where i = 1,2, . . .
i 1 2 3 4 5 6 λi 3.492 8.525 21.43 27.33 31.11 54.44
Finite elements modeling
■ Planar element: shell, ■ 959 nodes, ■ 900 planar elements.
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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
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1.27.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 frequency [Hz] 8.7266 CM2 Eigen mode Eigen mode 2 frequency [Hz] 21.3042 CM2 Eigen mode Eigen mode 3 frequency [Hz] 53.5542 CM2 Eigen mode Eigen mode 4 frequency [Hz] 68.2984 CM2 Eigen mode Eigen mode 5 frequency [Hz] 77.7448 CM2 Eigen mode Eigen mode 6 frequency [Hz] 136.0471
1.27.3 Calculated results
Result name
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%
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1.28 Bending effects of a symmetrical portal frame (01-0023SDLLB_FEM)
Test ID: 2455
Test status: Passed
1.28.1 Description Verifies the first eigen mode frequencies of a symmetrical portal frame with fixed supports.
1.28.2 Background
1.28.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; ■ Analysis type: modal analysis; ■ Element type: linear.
Bending effects of a symmetrical portal frame Scale = 1/5 01-0023SDLLB_FEM
Units
I. S.
Geometry
■ Straight rectangular sections for beams and columns: ■ Thickness: h = 0.0048 m, ■ Width: b = 0.029 m, ■ Section: A = 1.392 x 10-4 m2, ■ Flexure moment of inertia relative to z-axis: Iz = 2.673 x 10-10 m4,
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■ Points coordinates in m: A B C D E F x -0.30 0.30 -0.30 0.30 -0.30 0.30 y 0 0 0.36 0.36 0.81 0.81
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: Fixed at A and B, ■ Inner: None.
Loading
■ External: None. ■ Internal: None.
1.28.2.2 Eigen modes frequencies
Reference solution
Dynamic radius method (slender beams theory).
Finite elements modeling
■ Linear element: beam, ■ 60 nodes, ■ 60 linear elements.
Deformed shape
Bending effects of a symmetrical portal frame Scale = 1/7 Mode 13
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1.28.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode 1 antisymmetric frequency [Hz] 8.8 CM2 Eigen mode Eigen mode 2 antisymmetric frequency [Hz] 29.4 CM2 Eigen mode Eigen mode 3 symmetric frequency [Hz] 43.8 CM2 Eigen mode Eigen mode 4 symmetric frequency [Hz] 56.3 CM2 Eigen mode Eigen mode 5 antisymmetric frequency [Hz] 96.2 CM2 Eigen mode Eigen mode 6 symmetric frequency [Hz] 102.6 CM2 Eigen mode Eigen mode 7 antisymmetric frequency [Hz] 147.1 CM2 Eigen mode Eigen mode 8 symmetric frequency [Hz] 174.8 CM2 Eigen mode Eigen mode 9 antisymmetric frequency [Hz] 178.8 CM2 Eigen mode Eigen mode 10 antisymmetric frequency [Hz] 206 CM2 Eigen mode Eigen mode 11 symmetric frequency [Hz] 266.4 CM2 Eigen mode Eigen mode 12 antisymmetric frequency [Hz] 320 CM2 Eigen mode Eigen mode 13 symmetric frequency [Hz] 335
1.28.3 Calculated results
Result name
Result description 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%
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1.29 Fixed thin arc in planar bending (01-0027SSLLB_FEM)
Test ID: 2459
Test status: Passed
1.29.1 Description Arc of a circle fixed at one end, subjected to two punctual loads and a torque at its free end. The horizontal displacement, vertical displacement and rotation about Z-axis are verified.
1.29.2 Background
1.29.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 06/89; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Fixed thin arc in planar bending Scale = 1/24 01-0027SSLLB_FEM
Units
I. S.
Geometry
■ Medium radius: R = 3 m , ■ Circular hollow section:
► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.
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Materials properties
Longitudinal elastic modulus: E = 2 x 1011 Pa.
Boundary conditions
■ Outer: Fixed in A. ■ Inner: None.
Loading
■ External: At B:
► punctual load F1 = Fx = 10 N, ► punctual load F2 = Fy = 5 N, ► bending moment about Oz, Mz = 8 Nm.
■ Internal: None.
1.29.2.2 Displacements at B
Reference solution
At point B:
■ displacement parallel to Ox: u = R2
4EI [F1πR + 2F2R + 4Mz]
■ displacement parallel to Oy: v = R2
4EI [2F1πR + (3π - 8)F2R + 2(π - 2)Mz]
■ rotation around Oz: θ = R
4EI [4F1R + 2(π - 2)F2R + 2πMz]
Finite elements modeling
■ Linear element: beam, ■ 31 nodes, ■ 30 linear elements.
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Results shape
Fixed thin arc in planar bending Scale = 1/19 Deformed
1.29.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DX Horizontal displacement in point B [m] 0.3791 CM2 DZ Vertical displacement in point B [m] 0.2417 CM2 RY Rotation about Z-axis in point B [rad] -0.1654
1.29.3 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 Z-axis in point B [rad] -0.165362 Rad 0.02%
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1.30 Beam on elastic soil, hinged ends (01-0034SSLLB_FEM)
Test ID: 2466
Test status: Passed
1.30.1 Description A beam under a punctual load, a distributed load and two torques lays on a soil of constant linear stiffness. The rotation around z-axis, the vertical reaction, the vertical displacement and the bending moment are verified in several points.
1.30.2 Background
1.30.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 16/89; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Beam on elastic soil, hinged ends Scale = 1/27 01-0034SSLLB_FEM
Units
I. S.
Geometry
■ L = (π 10 )/2, ■ I = 10-4 m4.
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Materials properties
Longitudinal elastic modulus: E = 2.1 x 1011 Pa.
Boundary conditions
■ Outer: ► Free A and B ends, ► Soil with a constant linear stiffness ky = K = 840000 N/m2.
■ Inner: None.
Loading
■ External: ► Punctual force at D: Fy = F = - 10000 N, ► Uniformly distributed force from A to B: fy = p = - 5000 N/m, ► Torque at A: Cz = -C = -15000 Nm, ► Torque at B: Cz = C = 15000 Nm.
■ Internal: None.
1.30.2.2 Displacement and support reaction at A
Reference solution
β = 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 1
2βλ
■ Rotation about z-axis:
θA = p(sh(2ϕ) – sin(2ϕ)) + 2βFsh(ϕ)sin(ϕ) - 2β2C(sh(2ϕ) + sin(2ϕ)) x 1
(K/β)λ
Finite elements modeling
■ Linear element: beam, ■ 50 nodes, ■ 49 linear elements.
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Deformed shape
Beam on elastic soil, hinged ends Scale = 1/20 Deformed
1.30.2.3 Displacement and bending moment at D
Reference solution
■ Vertical displacement:
vD = 2p(λ - 2ch(ϕ)cos(ϕ)) + βF(sh(2ϕ) – sin(2ϕ)) - 8β2Csh(ϕ)sin(ϕ) x 1
2Kλ
■ Bending moment:
MD = 4psh(ϕ)sin(ϕ) + βF(sh(2ϕ) + sin(2ϕ)) - 8β2Cch(ϕ)cos(ϕ) x 1
4β2λ
Finite elements modeling
■ Linear element: beam, ■ 50 nodes, ■ 49 linear elements.
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Bending moment diagram
Beam on elastic soil, hinged ends Scale = 1/20 Bending moment
1.30.2.4 Theoretical results
Solver Result name Result description Reference value CM2 RY Rotation around z-axis in point A [rad] 0.003045 CM2 Fz Vertical reaction in point A [N] -11674 CM2 Dz Vertical displacement in point D [cm] -0.423326 CM2 My Bending moment in point D [Nm] -33840
1.30.3 Calculated results
Result name
Result description Value Error
RY Rotation around z-axis in point A [rad] 0.00304333 Rad -0.05% Fz Vertical reaction in point A [N] -11709 N -0.30% Dz Vertical displacement in point D [cm] -0.423297 cm 0.01% My Bending moment in point D [Nm] -33835.9 N*m 0.01%
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1.31 Square plate under planar stresses (01-0039SSLSB_FEM)
Test ID: 2470
Test status: Passed
1.31.1 Description Verifies the vertical displacement and the stresses on a square plate of 2 x 2 m, fixed on 3 sides with a uniform surface load on its surface.
1.31.2 Background
1.31.2.1 Model description ■ Reference: Internal GRAITEC test; ■ Analysis type: static linear; ■ Element type: planar (membrane).
Square plate under planar stresses Scale = 1/19 Modeling
[ ]1;1, −∈ηξ
Units
I. S.
Geometry
■ Thickness: e = 1 m, ■ 4 square elements of side h = 1 m.
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: Fixed on 3 sides, ■ Inner: None.
Loading
■ External: Uniform load p = -1. 108 N/ml on the upper surface, ■ Internal: None.
1.31.2.2 Displacement of the model in the linear elastic range
Reference solution
The reference displacements are calculated on nodes 7 and 9.
v9 = -6ph(3 + ν)(1 - ν2)
E(8(3 - ν)2 - (3 + ν)2) = -0.1809 x 10-3 m,
v7 = 4(3 - ν)3 + ν v9 = -0.592 x 10-3 m,
For element 1.4:
(For the stresses calculated above, the abscissa point (x = 0; y = 0) corresponds to node 8.)
σyy = E
1 - ν2 (v9 - v7)
2h (1 + ξ) for
σxx = νσyy for
σxy = E
1 + ν (v9 + v7) + η(v9 - v7)
4h (1 + ξ) for
Finite elements modeling
■ Planar element: membrane, imposed mesh, ■ 9 nodes, ■ 4 surface quadrangles.
ξ = -1 ; σxx = 0 ξ = 0 ; σxx = -14.23 MPa ξ = 1 ; σxx = -28.46 MPa
η = -1 ; ξ = 0 ; σxy = -47.82 MPa η = 0 ; ξ = 0 ; σxy = -31.21 MPa η= 1 ; ξ = 0 ; σxy = -14.61 MPa
ξ = -1 ; σyy = 0 ξ = 0 ; σyy = -47.44 MPa ξ = 1 ; σyy = -94.88 MPa
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Deformed shape
1.31.2.3 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
1.31.3 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] 7.45058e-015 MPa 0.00% sxx_mid Sigma xx stresses on Element 1.4 in x = 0.5 m [MPa] -14.2315 MPa -0.01% sxx_mid Sigma xx stresses on Element 1.4 in x = 1 m [MPa] -28.463 MPa -0.01% syy_mid Sigma yy stresses on Element 1.4 in x = 0 m [MPa] 1.49012e-014 MPa 0.00% syy_mid Sigma yy stresses on Element 1.4 in x = 0.5 m [MPa] -47.4383 MPa 0.00% syy_mid Sigma yy stresses on Element 1.4 in x = 1 m [MPa] -94.8767 MPa 0.00% sxy_mid Sigma xy stresses on Element 1.4 in y = 0 m [MPa] -14.611 MPa 0.33% sxy_mid Sigma xy stresses on Element 1.4 in y = 1 m [MPa] -47.8178 MPa 0.00%
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1.32 Beam on elastic soil, free ends (01-0032SSLLB_FEM)
Test ID: 2464
Test status: Passed
1.32.1 Description A beam under 3 punctual loads lays on a soil of constant linear stiffness. The bending moment, vertical displacement and rotation about z-axis on several points of the beam are verified.
1.32.2 Background
1.32.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 15/89; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Beam on elastic soil, free ends Scale = 1/21 01-0032SSLLB_FEM
Units
I. S.
Geometry
■ L = (π 10 )/2, ■ I = 10-4 m4.
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Materials properties
Longitudinal elastic modulus: E = 2.1 x 1011 Pa.
Boundary conditions
■ Outer: ► Free A and B extremities, ► Constant linear stiffness of soil ky = K = 840000 N/m2.
■ Inner: None.
Loading
■ External: Punctual load at A, C and B: Fy = F = - 10000 N. ■ Internal: None.
1.32.2.2 Bending moment and displacement at C
Reference solution
β = 4
K/(4EI)
ϕ = βL/2
λ = sh (2ϕ) + sin (2ϕ)
■ Bending moment: MC = (F/(4β))(ch(2ϕ) - cos (2ϕ) – 8sh(ϕ)sin(ϕ))/λ
■ Vertical displacement: vC = - (Fβ/(2K))( ch(2ϕ) + cos (2ϕ) + 8ch(ϕ)cos(ϕ) + 2)/λ
Finite elements modeling
■ Linear element: beam, ■ 72 nodes, ■ 71 linear elements.
Bending moment diagram
Beam on elastic soil, free ends Scale = 1/20 Bending moment
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1.32.2.3 Displacements at A
Reference solution
■ Vertical displacement: vA = (2Fβ/K)( ch(ϕ)cos(ϕ) + ch(2ϕ) + cos(2ϕ))/λ
■ Rotation about z-axis θA = (-2Fβ2/K)( sh(ϕ)cos(ϕ) - sin(ϕ)ch(ϕ) + sh(2ϕ) - sin(2ϕ))/λ
Finite elements modeling
■ Linear element: beam, ■ 72 nodes, ■ 71 linear elements
1.32.2.4 Theoretical results
Solver Result name Result description Reference value CM2 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 z-axis in point A [rad] -0.000706
1.32.3 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 z-axis in point A [rad] -0.000707427 Rad -0.20%
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1.33 EDF Pylon (01-0033SFLLA_FEM)
Test ID: 2465
Test status: Passed
1.33.1 Description Verifies the displacement at the top of an EDF Pylon and the dominating buckling results. Three punctual loads corresponding to wind loads are applied on the main arms, on the upper arm and on the lower horizontal frames of the pylon.
1.33.2 Background
1.33.2.1 Model description ■ Reference: Internal GRAITEC test; ■ Analysis type: static linear, Eulerian buckling; ■ Element type: linear
Units
I. S.
Geometry
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: ► Hinged support, ► For the modeling, a fixed restraint and 4 beams were added at the pylon supports level.
■ Inner: None.
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Loading
■ External: Punctual loads corresponding to a wind load.
► FX = 165550 N, FY = - 1240 N, FZ = - 58720 N on the main arms, ► FX = 50250 N, FY = - 1080 N, FZ = - 12780 N on the upper arm, ► FX = 11760 N, FY = 0 N, FZ = 0 N on the lower horizontal frames
■ Internal: None.
1.33.2.2 Displacement of the model in the linear elastic range
Reference solution
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Software ANSYS 5.3 NE/NASTRAN 7.0 Max deflection (m) 0.714 0.714 λ dominating mode 2.77 2.77
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 402 nodes, ■ 1034 elements.
Deformed shape
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Buckling modal deformation (dominating mode)
1.33.2.3 Theoretical results
Solver Result name Result description Reference value CM2 D Displacement at the top of the pylon [m] 0.714 CM2 Dominating buckling - critical λ, mode 4 [Hz] 2.77
1.33.3 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%
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1.34 Thin cylinder under a uniform radial pressure (01-0038SSLSB_FEM)
Test ID: 2469
Test status: Passed
1.34.1 Description Verifies the stress, the radial deformation and the longitudinal deformation of a cylinder loaded with a uniform internal pressure.
1.34.2 Background
1.34.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 06/89; ■ Analysis type: static elastic; ■ Element type: planar.
Thin cylinder under a uniform radial pressure Scale = 1/18 01-0038SSLSB_FEM
Units
I. S.
Geometry
■ Length: L = 4 m, ■ Radius: R = 1 m, ■ Thickness: h = 0.02 m.
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: ► Free conditions ► For the modeling, only ¼ of the cylinder is considered and the symmetry conditions are applied. On the
other side, we restrained the displacements at a few nodes in order to make the model stable. ■ Inner: None.
Loading
■ External: Uniform internal pressure: p = 10000 Pa, ■ Internal: None.
1.34.2.2 Stresses in all points
Reference solution
Stresses in the planar elements coordinate system (x axis is parallel with the length of the cylinder):
■ σxx = 0
■ σyy = pRh
Finite elements modeling
■ Planar element: shell, ■ 209 nodes, ■ 180 planar elements.
1.34.2.3 Cylinder deformation in all points ■ Radial deformation:
δR = pR2Eh
■ Longitudinal deformation:
δL = -pRνL
Eh
1.34.2.4 Theoretical results
Solver Result name Result description Reference value CM2 syy_mid σyy stress in all points [Pa] 500000.000000 CM2 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
1.34.3 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%
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1.35 Caisson beam in torsion (01-0037SSLSB_FEM)
Test ID: 2468
Test status: Passed
1.35.1 Description A torsion moment is applied on the free end of a caisson beam fixed on one end. For both ends, the displacement, the rotation about Z-axis and the stress are verified.
1.35.2 Background
1.35.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 05/89; ■ Analysis type: static linear; ■ Element type: planar.
Caisson beam in torsion Scale = 1/4 01-0037SSLSB_FEM
Units
I. S.
Geometry
■ Length; L = 1m, ■ Square section of side: b = 0.1 m, ■ Thickness = 0.005 m.
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: Beam fixed at end x = 0; ■ Inner: None.
Loading
■ External: Torsion moment M = 10N.m applied to the free end (for modeling, 4 forces of 50 N). ■ Internal: None.
1.35.2.2 Displacement and stress at two points
Reference solution
The reference solution is determined by averaging the results of several calculation software with implemented finite elements method.
Points coordinates:
■ A (0,0.05,0.5) ■ B (-0.05,0,0.8) Note: point O is the origin of the coordinate system (x,y,z).
Finite elements modeling
■ Planar element: shell, ■ 90 nodes, ■ 88 planar elements.
Deformed shape
Caisson beam in torsion Scale = 1/4 Deformed
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1.35.2.3 Theoretical results
Solver Result name Result description Reference value CM2 D Displacement in point A [m] -0.617 x 10-6 CM2 D Displacement in point B [m] -0.987 x 10-6 CM2 RY Rotation about Z-axis in point A [rad] 0.123 x 10-4 CM2 RY Rotation about Z-axis in point B [rad] 0.197 x 10-4 CM2 sxy_mid σxy stress in point A [MPa] -0.11 CM2 sxy_mid σxy stress in point B [MPa] -0.11
1.35.3 Calculated results
Result name
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 Z-axis in point A [rad] -1.23211e-005 Rad -0.17% RY Rotation about Z-axis in point B [rad] -1.97172e-005 Rad -0.09% sxy_mid Sigma xy stress in point A [MPa] -0.100037 MPa -0.04% sxy_mid Sigma xy stress in point B [MPa] -0.100212 MPa -0.21%
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1.36 Beam on two supports considering the shear force (01-0041SSLLB_FEM)
Test ID: 2472
Test status: Passed
1.36.1 Description Verifies the vertical displacement on a 300 cm long beam, consisting of an I shaped profile of a total height of 20.04 cm, a 0.96 cm thick web and 20.04 cm wide / 1.46 cm thick flanges.
1.36.2 Background
1.36.2.1 Model description ■ Reference: Internal GRAITEC test; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Units
I. S.
Geometry
l = 300 cm h = 20.04 cm b= 20.04 cm tw = 1.46 cm tf = 0.96 cm Sx= 74.95 cm2 Iz = 5462 cm4 Sy = 16.43 cm2
Materials properties
■ Longitudinal elastic modulus: E = 2285938 daN/cm2, ■ Transverse elastic modulus G = 879207 daN/cm2 ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: ► Simple support on node 11, ► For the modeling, put an hinge at node 1 (instead of a simple support).
■ Inner: None.
Loading
■ External: Vertical punctual load P = -20246 daN at node 6, ■ Internal: None.
1.36.2.2 Vertical displacement of the model in the linear elastic range
Reference solution
The reference displacement is calculated in the middle of the beam, at node 6.
( )cm 017.1105.0912.0
43.163.012
22859384
300202465462228593848
30020246448
33
6 −=−−=
+
−+
−=+=
xx
xxx
xGSPl
EIPlv
shear
y
flexion
z
876876
Finite elements modeling
■ Planar element: S beam, imposed mesh, ■ 11 nodes, ■ 10 linear elements.
Deformed shape
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1.36.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement at node 6 [cm] -1.017
1.36.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement at node 6 [cm] -1.01722 cm -0.02%
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1.37 Thin cylinder under a hydrostatic pressure (01-0043SSLSB_FEM)
Test ID: 2474
Test status: Passed
1.37.1 Description Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder under a hydrostatic pressure.
1.37.2 Background
1.37.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 08/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
Thin cylinder under a hydrostatic pressure Scale = 1/25 01-0043SSLSB_FEM
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.
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Boundary conditions
■ Outer: For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on the nodes that are parallel with the cylinder’s axis.
■ Inner: None.
Loading
■ External: Radial internal pressure varies linearly with the "p" height, p = p0 zL ,
■ Internal: None.
1.37.2.2 Stresses
Reference solution
x axis of the local coordinate system of planar elements is parallel to the cylinders axis.
σxx = 0
σyy = p0RzLh
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 209 nodes, ■ 180 surface quadrangles.
1.37.2.3 Cylinder deformation
Reference solution
■ δL longitudinal deformation of the cylinder:
δL = -p0Rνz2
2ELh
■ δL radial deformation of the cylinder:
δR = p0R2zELh
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 209 nodes, ■ 180 surface quadrangles.
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Deformation shape
Thin cylinder under a hydrostatic pressure Deformed
1.37.2.4 Theoretical results
Solver Result name Result description Reference value CM2 syy_mid 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
1.37.3 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%
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1.38 Thin cylinder under a uniform axial load (01-0042SSLSB_FEM)
Test ID: 2473
Test status: Passed
1.38.1 Description Verifies the stress, the longitudinal deformation and the radial deformation of a cylinder under a uniform axial load.
1.38.2 Background
1.38.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 07/89; ■ Analysis type: static elastic; ■ Element type: planar.
Thin cylinder under a uniform axial load Scale = 1/19 01-0042SSLSB_FEM
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.
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Boundary conditions
■ Outer: ► Null axial displacement at the left end: vz = 0, ► For the modeling, only a ¼ of the cylinder is considered.
■ Inner: None.
Loading
■ External: Uniform axial load q = 10000 N/m ■ Inner: None.
1.38.2.2 Stress in all points
Reference solution
x axis of the local coordinate system of planar elements is parallel to the cylinders axis.
σxx = qh
σyy = 0
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 697 nodes, ■ 640 surface quadrangles.
1.38.2.3 Cylinder deformation at the free end
Reference solution
■ δL longitudinal deformation of the cylinder:
δL = qLEh
■ δR radial deformation of the cylinder:
δR = -qνREh
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 697 nodes, ■ 640 surface quadrangles.
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Deformation shape
Thin cylinder under a uniform axial load Scale = 1/22 Deformation shape
1.38.2.4 Theoretical results
Solver Result name Result description Reference value CM2 sxx_mid 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] 9.52 x 10-6 CM2 Dz δR radial deformation at the free end [m] -7.14 x 10-7
1.38.3 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] 1.05305e-009 Pa 0.00% DY Delta L longitudinal deformation at the free end [mm] -0.00952381 mm -0.04% Dz Delta R radial deformation at the free end [mm] 0.000710887 mm -0.44%
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1.39 Truss with hinged bars under a punctual load (01-0031SSLLB_FEM)
Test ID: 2463
Test status: Passed
1.39.1 Description Verifies the horizontal and the vertical displacement in several points of a truss with hinged bars, subjected to a punctual load.
1.39.2 Background ■ Reference: Structure Calculation Software Validation Guide, test SSLL 11/89; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
1.39.2.1 Model description
Truss with hinged bars under a punctual load Scale = 1/10 01-0031SSLLB_FEM
Units
I. S.
Geometry
Elements Length (m) Area (m2) AC 0.5 2 2 x 10-4 CB 0.5 2 2 x 10-4 CD 2.5 1 x 10-4 BD 2 1 x 10-4
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Materials properties
Longitudinal elastic modulus: E = 1.962 x 1011 Pa.
Boundary conditions
■ Outer: Hinge at A and B, ■ Inner: None.
Loading
■ External: Punctual force at D: Fy = F = - 9.81 x 103 N. ■ Internal: None.
1.39.2.2 Displacements at C and D
Reference solution
Displacement method.
Finite elements modeling
■ Linear element: beam, ■ 4 nodes, ■ 4 linear elements.
Displacements shape
Truss with hinged bars under a punctual load Scale = 1/9 Deformed
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1.39.2.3 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
1.39.3 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%
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1.40 Simply supported square plate (01-0036SSLSB_FEM)
Test ID: 2467
Test status: Passed
1.40.1 Description Verifies the vertical displacement in the center of a simply supported square plate.
1.40.2 Background
1.40.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 02/89; ■ Analysis type: static linear; ■ Element type: planar.
Simply supported square plate Scale = 1/9 01-0036SSLSB_FEM
Units
I. S.
Geometry
■ Side = 1 m, ■ Thickness h = 0.01m.
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7950 kg/m3.
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Boundary conditions
■ Outer: ► Simple support on the plate perimeter, ► For the modeling, we add a fixed support at B.
■ Inner: None.
Loading
■ External: Self weight (gravity = 9.81 m/s2). ■ Internal: None.
1.40.2.2 Vertical displacement at O
Reference solution
According to Love- Kirchhoff hypothesis, the displacement w at a point (x,y):
w(x,y) = Σ wmnsinmπxsinnπy
where wmn = 192ρg(1 - ν2)
mn(m2 + n2)π6Eh2
Finite elements modeling
■ Planar element: shell, ■ 441 nodes, ■ 400 planar elements.
Deformed shape
Simply supported square plate Scale = 1/6 Deformed
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1.40.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point O [μm] -0.158
1.40.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement in point O [µm] -0.164901 µm -4.37%
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1.41 Stiffen membrane (01-0040SSLSB_FEM)
Test ID: 2471
Test status: Passed
1.41.1 Description Verifies the horizontal displacement and the stress on a plate (8 x 12 cm) fixed in the middle on 3 supports with a punctual load at its free node.
1.41.2 Background
1.41.2.1 Model description ■ Reference: Klaus-Jürgen Bathe - Finite Element Procedures in Engineering Analysis, Example 5.13; ■ Analysis type: static linear; ■ Element type: planar (membrane).
[ ]1;1, −∈ηξ
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.
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Loading
■ External: Uniform load Fx = F = 6000 N at A, ■ Internal: None.
1.41.2.2 Results of the model in the linear elastic range
Reference solution
Point B is the origin of the coordinate system used for the results positions.
( ) ( )
( )
( )⎪⎩
⎪⎨
⎧
−=−==−=−==
=−=+
+−=
⎪⎩
⎪⎨
⎧
==−====
===
⎪⎩
⎪⎨
⎧
==−====
==−
−=
=+
=+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
+−
== −
MPa 96.17N/cm 1796 ;1MPa 98.8N/cm 898 ;0
0 ;1for 1
81
MPa 55.11N/cm 1155 ;1MPa 77.5N/cm 577 ;0
0 ;1for
MPa 49.38N/cm 3849 ;1MPa 24.19N/cm 1924 ;0
0 ;1for 1
21
3410.97510.367410.2
6000
211
12
3
2xy1
2xy1
xy1
1
2yy1
2yy1
yy1
11
2xx1
2xx1
xx1
21
466
222
σξσξ
σξξ
νσ
σηση
σηνσσ
σηση
σηη
νσ
νν
buE
auE
cm
aES
baeabE
FKFu
Axy
xxyy
Axx
A
Finite elements modeling
■ Planar element: membrane, imposed mesh, ■ 6 nodes, ■ 2 quadrangle planar elements and 1 bar.
Deformed shape
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1.41.2.3 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
1.41.3 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] 3.63798e-015 MPa 0.00% syy_mid Sigma yy stress Element 1 in y = 0 cm [MPa] 11.5467 MPa -0.03% syy_mid Sigma yy stress Element 1 in y = 6 cm [MPa] -9.09495e-016 MPa 0.00% sxy_mid Sigma xy stress Element 1 in x = 0 cm [MPa] -2.96059e-015 MPa 0.00% sxy_mid Sigma xy stress Element 1 in x = 4 cm [MPa] -8.98076 MPa -0.01% sxy_mid Sigma xy stress Element 1 in x = 8 cm [MPa] -17.9615 MPa -0.01%
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1.42 Torus with uniform internal pressure (01-0045SSLSB_FEM)
Test ID: 2476
Test status: Passed
1.42.1 Description Verifies the stress and the radial deformation of a torus with uniform internal pressure.
1.42.2 Background
1.42.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 10/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
Torus with uniform internal pressure
01-0045SSLSB_FEM
Units
I. S.
Geometry
■ Thickness: h = 0.02 m, ■ Transverse section radius: b = 1 m, ■ Average radius of curvature: a = 2 m.
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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.
1.42.2.2 Stresses
Reference solution
(See stresses description on the first scheme of the overview)
If a – b ≤ r ≤ a + b
σ11 = pb2h
r + ar
σ22 = pb2h
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 361 nodes, ■ 324 surface quadrangles.
1.42.2.3 Cylinder deformation
Reference solution
■ δR radial deformation of the torus:
δR = pb
2Eh (r - ν(r + a))
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 361 nodes, ■ 324 surface quadrangles.
1.42.2.4 Theoretical results
Solver Result name Result description Reference value CM2 syy_mid σ11 stresses for r = a - b [Pa] 7.5 x 105 CM2 syy_mid σ11 stresses for r = a + b [Pa] 4.17 x 105 CM2 sxx_mid σ22 stress for all r [Pa] 2.50 x 105 CM2 Dz δL radial deformations of the torus for r = a - b [m] 1.19 x 10-7 CM2 Dz δL radial deformations of the torus for r = a + b [m] 1.79 x 10-6
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1.42.3 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%
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1.43 Spherical dome under a uniform external pressure (01-0050SSLSB_FEM)
Test ID: 2480
Test status: Passed
1.43.1 Description A spherical dome of radius (a) is subjected to a uniform external pressure. The horizontal displacement and the external meridian stresses are verified.
1.43.2 Background
1.43.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 22/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
Spherical dome under a uniform external pressure
01-0050SSLSB_FEM
Units
I. S.
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Geometry
■ Radius: a = 2.54 m, ■ Thickness: h = 0.0127 m, ■ Angle: θ = 75°.
Materials properties
■ Longitudinal elastic modulus: E = 6.897 x 1010 Pa, ■ Poisson's ratio: ν = 0.2.
Boundary conditions
■ Outer: Fixed on the dome perimeter, ■ Inner: None.
Loading
■ External: Uniform pressure p = 0.6897 x 106 Pa, ■ Internal: None.
1.43.2.2 Horizontal displacement and exterior meridian stress
Reference solution
The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution.
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 401 nodes, ■ 400 planar elements.
Deformed shape
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1.43.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DX Horizontal displacements in ψ = 15° 1.73 x 10-3 CM2 DX Horizontal displacements in ψ = 45° -1.02 x 10-3 CM2 syy_mid σyy external meridian stresses in ψ = 15° -74 CM2 sxx_mid σXX external meridian stresses in ψ = 45° -68
1.43.3 Calculated results
Result name
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%
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1.44 Pinch cylindrical shell (01-0048SSLSB_FEM)
Test ID: 2478
Test status: Passed
1.44.1 Description A cylinder of length L is pinched by 2 diametrically opposite forces (F). The vertical displacement is verified.
1.44.2 Background ■ Reference: Structure Calculation Software Validation Guide, test SSLS 20/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
1.44.2.1 Model description A cylinder of length L is pinched by 2 diametrically opposite forces (F).
Pinch cylindrical shell 01-0048SSLSB_FEM
Units
I. S.
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Geometry
■ Length: L = 10.35 m (total length), ■ Radius: R = 4.953 m, ■ Thickness: h = 0.094 m.
Materials properties
■ Longitudinal elastic modulus: E = 10.5 x 106 Pa, ■ Poisson's ratio: ν = 0.3125.
Boundary conditions
■ Outer: For the modeling, we consider only half of the cylinder, so we impose symmetry conditions (nodes in the horizontal xz plane are restrained in translation along y and in rotation along x and z),
■ Inner: None.
Loading
■ External: 2 punctual loads F = 100 N, ■ Internal: None.
1.44.2.2 Vertical displacement at point A
Reference solution
The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution.
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 777 nodes, ■ 720 surface quadrangles.
1.44.2.3 Theoretical result
Solver Result name Result description Reference value CM2 DZ Vertical displacement in point A [m] -113.9 x 10-3
1.44.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement in point A [mm] -113.3 mm 0.53%
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1.45 Simply supported rectangular plate under a uniform load (01-0052SSLSB_FEM)
Test ID: 2482
Test status: Passed
1.45.1 Description A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.
1.45.2 Background
1.45.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 24/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
Simply supported rectangular plate under a uniform load Scale = 1/11 01-0052SSLSB_FEM
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.
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Boundary conditions
■ Outer: Simple support on the plate perimeter (null displacement along z-axis), ■ Inner: None.
Loading
■ External: Normal pressure of plate p = pZ = -1.0 Pa, ■ Internal: None.
1.45.2.2 Vertical displacement and bending moment at the center of the plate
Reference solution
Love-Kirchhoff thin plates theory.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 435 nodes, ■ 392 surface quadrangles.
1.45.2.3 Theoretical background
Solver Result name Result description Reference value CM2 DZ Vertical displacement at plate center [m] -1.1060 x 10-2 CM2 Mxx MX bending moment at plate center [Nm] -0.1017 CM2 Myy MY bending moment at plate center [Nm] -0.0464
1.45.3 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%
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1.46 Spherical shell under internal pressure (01-0046SSLSB_FEM)
Test ID: 2477
Test status: Passed
1.46.1 Description A spherical shell is subjected to a uniform internal pressure. The stress and the radial deformation are verified.
1.46.2 Background
1.46.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 14/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
Spherical shell under internal pressure 01-0046SSLSB_FEM
Units
I. S.
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Geometry
■ Thickness: h = 0.02 m, ■ Radius: R2 = 1 m, ■ θ = 90° (hemisphere).
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: Simple support (null displacement along vertical displacement) on the shell perimeter.
For modeling, we consider only half of the hemisphere, so we impose symmetry conditions (DOF restrains placed in the vertical plane xy in translation along z and in rotation along x and y). In addition, the node at the top of the shell is restrained in translation along x to assure the stability of the structure during calculation).
■ Inner: None.
Loading
■ External: Uniform internal pressure p = 10000 Pa ■ Internal: None.
1.46.2.2 Stresses
Reference solution
(See stresses description on the first scheme of the overview)
If 0° ≤ θ ≤ 90°
σ11 = σ22 = pR2
2
2h
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 343 nodes, ■ 324 planar elements.
1.46.2.3 Cylinder deformation
Reference solution
■ δR radial deformation of the calotte:
δR = pR22 (1 - ν) sin θ
2Eh
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 343 nodes, ■ 324 planar elements.
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Deformed shape
Spherical shell under internal pressure Scale = 1/11 Deformed
1.46.2.4 Theoretical results
Solver Result name Result description Reference value CM2 sxx_mid σ11 stress for all θ [Pa] 2.50 x 105 CM2 syy_mid σ22 stress for all θ [Pa] 2.50 x 105 CM2 Dz δR radial deformations for θ = 90° [m] 8.33 x 10-7
1.46.3 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%
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1.47 Simply supported square plate under a uniform load (01-0051SSLSB_FEM)
Test ID: 2481
Test status: Passed
1.47.1 Description A square plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.
1.47.2 Background
1.47.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 24/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
Simply supported square plate under a uniform load Scale = 1/9 01-0051SSLSB_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.
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Boundary conditions
■ Outer: Simple support on the plate perimeter (null displacement along z-axis), ■ Inner: None
Loading
■ External: Normal pressure of plate p pZ = -1.0 Pa, ■ Internal: None.
1.47.2.2 Vertical displacement and bending moment at the center of the plate
Reference solution
Love-Kirchhoff thin plates theory.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 361 nodes, ■ 324 planar elements.
1.47.2.3 Theoretical result
Solver Result name Result description Reference value CM2 DZ Vertical displacement at plate center [m] -4.43 x 10-3 CM2 Mxx MX bending moment at plate center [Nm] 0.0479 CM2 Myy MY bending moment at plate center [Nm] 0.0479
1.47.3 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%
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1.48 Simply supported rectangular plate loaded with punctual force and moments (01-0054SSLSB_FEM)
Test ID: 2484
Test status: Passed
1.48.1 Description A rectangular plate simply supported is subjected to a punctual force and moments. The vertical displacement is verified.
1.48.2 Background
1.48.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 26/89; ■ Analysis type: static linear; ■ Element type: planar.
Simply supported rectangular plate loaded with punctual force and moments 01-0054SSLSB_FEM
Units
I. S.
Geometry
■ Width: DA = CB = 20 m, ■ Length: AB = DC = 5 m, ■ Thickness: h = 1 m,
Materials properties
■ Longitudinal elastic modulus: E =1000 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: Punctual support at A, B and D (null displacement along z-axis), ■ Inner: None.
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Loading
■ External: ► In A: MX = 20 Nm, MY = -10 Nm, ► In B: MX = 20 Nm, MY = 10 Nm, ► In C: FZ = -2 N, MX = -20 Nm, MY = 10 Nm, ► In D: MX = -20 Nm, MY = -10 Nm,
■ Internal: None.
1.48.2.2 Vertical displacement at C
Reference solution
Love-Kirchhoff thin plates theory.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 867 nodes, ■ 800 surface quadrangles.
Deformed shape
Simply supported rectangular plate loaded with punctual force and moments Deformed
1.48.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -12.480
1.48.3 Calculated results
Result name Result description Value Error Dz Vertical displacement in point C [m] -12.6677 m -1.50%
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1.49 Triangulated system with hinged bars (01-0056SSLLB_FEM)
Test ID: 2486
Test status: Passed
1.49.1 Description A truss with hinged bars is placed on three punctual supports (subjected to imposed displacements) and is loaded with two punctual forces. A thermal load is applied to all the bars. The traction force and the vertical displacement are verified.
1.49.2 Background
1.49.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 12/89; ■ Analysis type: static (plane problem); ■ Element type: linear.
Units
I. S.
Geometry
■ θ = 30°, ■ Section A1 = 1.41 x 10-3 m2, ■ Section A2 = 2.82 x 10-3 m2.
Materials properties
■ Longitudinal elastic modulus: E =2.1 x 1011 Pa, ■ Coefficient of linear expansion: α = 10-5 °C-1.
Boundary conditions
■ Outer: ► Hinge at A (uA = vA = 0), ► Roller supports at B and C ( uB = v’C = 0),
■ Inner: None.
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Loading
■ External: ► Support displacement: vA = -0.02 m ; vB = -0.03 m ; v’C = -0.015 m , ► Punctual loads: FE = -150 KN ; FF = -100 KN, ► Expansion effect on all bars for a temperature variation of 150° in relation with the assembly
temperature (specified geometry), ■ Internal: None.
1.49.2.2 Tension force in BD bar
Reference solution
Determining the hyperstatic unknown with the section cut method.
Finite elements modeling
■ Linear element: S beam, automatic mesh, ■ 11 nodes, ■ 17 S beams + 1 rigid S beam for the modeling of the simple support at C.
1.49.2.3 Vertical displacement at D
Reference solution
vD displacement was determined by several software with implemented finite elements method.
Finite elements modeling
■ Linear element: S beam, automatic mesh, ■ 11 nodes, ■ 17 S beams + 1 rigid S beam for the modeling of simple support at C.
Deformed shape
Triangulated system with hinged bars 01-0056SSLLB_FEM
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1.49.2.4 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
1.49.3 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%
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1.50 Shear plate perpendicular to the medium surface (01-0055SSLSB_FEM)
Test ID: 2485
Test status: Passed
1.50.1 Description Verifies the vertical displacement of a rectangular shear plate fixed at one end, loaded with two forces.
1.50.2 Background
1.50.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 27/89; ■ Analysis type: static; ■ Element type: planar.
Shear plate Scale = 1/50 01-0055SSLSB_FEM
Units
I. S.
Geometry
■ Length: L = 12 m, ■ Width: l = 1 m, ■ Thickness: h = 0.05 m,
Materials properties
■ Longitudinal elastic modulus: E = 1.0 x 107 Pa, ■ Poisson's ratio: ν = 0.25.
Boundary conditions
■ Outer: Fixed AD edge, ■ Inner: None.
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Loading
■ External: ► At B: Fz = -1.0 N, ► At C: FZ = 1.0 N,
■ Internal: None.
1.50.2.2 Vertical displacement at C
Reference solution
Analytical solution.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 497 nodes, ■ 420 surface quadrangles.
Deformed shape
Shear plate Scale = 1/35 Deformed
1.50.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] 35.37 x 10-3
1.50.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement in point C [m] 35.6655 mm 0.84%
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1.51 Thin cylinder under its self weight (01-0044SSLSB_MEF)
Test ID: 2475
Test status: Passed
1.51.1 Description Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder subjected to its self weight only.
1.51.2 Background
1.51.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 09/89; ■ Analysis type: static, linear elastic; ■ Element type: planar. A cylinder of R radius and L length subject of self weight only.
Thin cylinder under its self weight Scale = 1/24 01-0044SSLSB_FEM
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, ■ Density: γ = 7.85 x 104 N/m3.
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Boundary conditions
■ Outer: ► Null axial displacement at z = 0, ► For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on
the nodes that are parallel with the cylinder’s axis. ■ Inner: None.
Loading
■ External: Cylinder self weight, ■ Internal: None.
1.51.2.2 Stresses
Reference solution
x axis of the local coordinate system of planar elements is parallel to the cylinders axis.
σxx = γz
σyy = 0
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 697 nodes, ■ 640 surface quadrangles.
1.51.2.3 Cylinder deformation
Reference solution
■ δL longitudinal deformation of the cylinder:
δL = γz2
2E
■ δR radial deformation of the cylinder:
δR = -γνRz
E
1.51.2.4 Theoretical results
Solver Result name Result description Reference value CM2 sxx_mid σxx stress for z = L [Pa] -314000.000000 CM2 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.
1.51.3 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%
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1.52 Spherical shell with holes (01-0049SSLSB_FEM)
Test ID: 2479
Test status: Passed
1.52.1 Description A spherical shell with holes is subjected to 4 forces, opposite 2 by 2. The horizontal displacement is verified.
1.52.2 Background ■ Reference: Structure Calculation Software Validation Guide, test SSLS 21/89; ■ Analysis type: static, linear elastic; ■ Element type: planar.
1.52.2.1 Model description
Spherical shell with holes 01-0049SSLSB_FEM
Units
I. S.
Geometry
■ Radius: R = 10 m ■ Thickness: h = 0.04 m, ■ Opening angle of the hole: ϕ0 = 18°.
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Materials properties
■ Longitudinal elastic modulus: E = 6.285 x 107 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: For modeling, we consider only a quarter of the shell, so we impose symmetry conditions (nodes in the vertical yz plane are restrained in translation along x and in rotation along y and z. Nodes on the vertical xy plane are restrained in translation along z and in rotation along x and y),
■ Inner: None.
Loading
■ External: Punctual loads F = 1 N, according to the diagram, ■ Internal: None.
1.52.2.2 Horizontal displacement at point A
Reference solution
The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution.
Finite elements modeling
■ Planar element: shell, imposed mesh, ■ 99 nodes, ■ 80 surface quadrangles.
Deformed shape
Spherical shell with holes Scale = 1/79 Deformed
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1.52.2.3 Theoretical background
Solver Result name Result description Reference value CM2 DX Horizontal displacement at point A(R,0,0) [mm] 94.0
1.52.3 Calculated results
Result name
Result description Value Error
DX Horizontal displacement at point A(R,0,0) [mm] 92.6205 mm -1.47%
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1.53 Simply supported rectangular plate under a uniform load (01-0053SSLSB_FEM)
Test ID: 2483
Test status: Passed
1.53.1 Description A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.
1.53.2 Background
1.53.2.1 Model description ■ 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 Scale = 1/25 01-0053SSLSB_FEM
Units
I. S.
Geometry
■ Width: a = 1 m, ■ Length: b = 5 m, ■ Thickness: h = 0.01 m,
Materials properties
■ Longitudinal elastic modulus: E = 1.0 x 107 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: Simple support on the plate perimeter (null displacement along z-axis), ■ Inner: None.
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Loading
■ External: Normal pressure of plate p = pZ = -1.0 Pa, ■ Internal: None.
1.53.2.2 Vertical displacement and bending moment at the center of the plate
Reference solution
Love-Kirchhoff thin plates theory.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 793 nodes, ■ 720 surface quadrangles.
1.53.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement at plate center [m] 1.416 x 10-2 CM2 Mxx MX bending moment at plate center [Nm] 0.1246 CM2 Myy MY bending moment at plate center [Nm] 0.0375
1.53.3 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%
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1.54 A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0058SSLSB_FEM)
Test ID: 2488
Test status: Passed
1.54.1 Description Verifies the vertical displacement for a square plate (0.01333 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.54.2 Background
1.54.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar. Square plate of side "a", for the modeling, only a quarter of the plate is considered.
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.01333 m,
■ Slenderness: λ = ah = 75.
Materials properties
■ Reinforcement, ■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: Fixed sides: AB and BD,
For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),
■ Inner: None.
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Loading
■ External: 1 MPa uniform pressure, ■ Internal: None.
1.54.2.2 Vertical displacement at C
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 289 nodes, ■ 256 surface quadrangles.
1.54.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -2.8053 x 10-2
1.54.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement in point C [cm] -2.79502 cm 0.37%
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1.55 A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0060SSLSB_FEM)
Test ID: 2490
Test status: Passed
1.55.1 Description Verifies the vertical displacement for a square plate (0.05 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.55.2 Background
1.55.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.05 m,
■ Slenderness: λ = ah = 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),
■ Inner: None.
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Loading
■ External: 1 MPa uniform pressure, ■ Internal: None.
1.55.2.2 Vertical displacement at C
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 289 nodes, ■ 256 surface quadrangles.
1.55.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -0.55474 x 10-3
1.55.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement in point C [cm] -0.0549874 cm
0.88%
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1.56 A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (01-0064SSLSB_FEM)
Test ID: 2494
Test status: Passed
1.56.1 Description Verifies the vertical displacement for a square plate (0.02 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.56.2 Background
1.56.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
0.02 m thick plate fixed on its perimeter Scale = 1/5 01-0064SSLSB_FEM
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.02 m, ■ Slenderness: λ = 50.
Materials properties
■ Reinforcement, ■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: Fixed edges, ■ Inner: None.
Loading
■ External: punctual force applied in the center of the plate: FZ = -106 N, ■ Internal: None.
1.56.2.2 Vertical displacement at point C (the center of the plate)
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 961 nodes, ■ 900 surface quadrangles.
1.56.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.037454
1.56.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement in point C [m] -0.0369818 m 1.26%
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1.57 A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (01-0062SSLSB_FEM)
Test ID: 2492
Test status: Passed
1.57.1 Description Verifies the vertical displacement for a square plate (0.01 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.57.2 Background ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
1.57.2.1 Model description Square plate of side "a".
0.01 m thick plate fixed on its perimeter Scale = 1/5 01-0062SSLSB_FEM
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.01 m,
■ Slenderness: λ = ah = 100.
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Materials properties
■ Reinforcement, ■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: Fixed edges, ■ Inner: None.
Loading
■ External: Punctual force applied on the center of the plate: FZ = -106 N, ■ Internal: None.
1.57.2.2 Vertical displacement at point C (center of the plate)
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 961 nodes, ■ 900 surface quadrangles.
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1.57.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.29579
1.57.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement in point C [m] -0.292146 m 1.23%
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1.58 A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0059SSLSB_FEM)
Test ID: 2489
Test status: Passed
1.58.1 Description Verifies the vertical displacement for a square plate (0.02 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.58.2 Background
1.58.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.02 m,
■ Slenderness: λ = ah = 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.
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Loading
■ External: 1 MPa uniform pressure, ■ Internal: None.
1.58.2.2 Vertical displacement at C
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 289 nodes, ■ 256 surface quadrangles.
1.58.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -0.83480 x 10-2
1.58.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement in point C [cm] -0.82559 cm 1.10%
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1.59 A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (01-0063SSLSB_FEM)
Test ID: 2493
Test status: Passed
1.59.1 Description Verifies the vertical displacement for a square plate (0.01333 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.59.2 Background
1.59.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
0.01333 m thick plate fixed on its perimeter Scale = 1/5 01-0063SSLSB_FEM
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.01333 m,
■ Slenderness: λ = ah = 75.
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Materials properties
■ Reinforcement, ■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: Fixed sides, ■ Inner: None.
Loading
■ External: Punctual force applied on the center of the plate: FZ = -106 N, ■ Internal: None.
1.59.2.2 Vertical displacement at point C (the center of the plate)
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 961 nodes, ■ 900 surface quadrangles.
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1.59.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.12525
1.59.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement in point C [m] -0.124583 m 0.53%
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1.60 A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (01-0066SSLSB_FEM)
Test ID: 2496
Test status: Passed
1.60.1 Description Verifies the vertical displacement for a square plate (0.1 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.60.2 Background
1.60.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
0.1 m thick plate fixed on its perimeter Scale = 1/5 01-0066SSLSB_FEM
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.1 m, ■ Slenderness: λ = 10.
Materials properties
■ Reinforcement, ■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: Fixed edges, ■ Inner: None.
Loading
■ External: punctual force applied in the center of the plate: FZ = -106 N, ■ Internal: None.
1.60.2.2 Vertical displacement at point C (center of the plate)
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 961 nodes, ■ 900 surface quadrangles.
1.60.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.42995 x 10-3
1.60.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement in point C [mm] -0.412094 mm
4.15%
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1.61 Vibration mode of a thin piping elbow in space (case 2) (01-0068SDLLB_FEM)
Test ID: 2498
Test status: Passed
1.61.1 Description Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with two straight elements (0.6 m long) and subjected to its self weight only.
1.61.2 Background
1.61.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; ■ Analysis type: modal analysis (in space); ■ Element type: linear.
Vibration mode of a thin piping elbow Scale = 1/11 01-0068SDLLB_FEM
Units
I. S.
Geometry
■ Average radius of curvature: OA = R = 1 m, ■ L = 0.6 m, ■ Straight circular hollow section: ■ Outer diameter: de = 0.020 m, ■ Inner diameter: di = 0.016 m, ■ Section: A = 1.131 x 10-4 m2, ■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4,
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■ Polar inertia: Ip = 9.274 x 10-9 m4. ■ Points coordinates (in m):
► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 ) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: ► Fixed at points C and D ► In A: translation restraint along y and z, ► In B: translation restraint along x and z,
■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.61.2.2 Eigen modes frequencies
Reference solution
The Rayleigh method applied to a thin curved beam is used to determine parameters such as:
■ transverse bending:
fj = μi
2
2π R2 GIpρA where i = 1,2.
Finite elements modeling
■ Linear element: beam, ■ 23 nodes, ■ 22 linear elements.
Eigen mode shapes
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1.61.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode Transverse 1 frequency [Hz] 33.4 CM2 Eigen mode Eigen mode Transverse 2 frequency [Hz] 100
1.61.3 Calculated results
Result name
Result description Value Error
Eigen mode Transverse 1 frequency [Hz] 33.19 Hz -0.63% Eigen mode Transverse 2 frequency [Hz] 94.62 Hz -5.38%
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1.62 Vibration mode of a thin piping elbow in space (case 1) (01-0067SDLLB_FEM)
Test ID: 2497
Test status: Passed
1.62.1 Description Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, fixed on its ends and subjected to its self weight only.
1.62.2 Background
1.62.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; ■ Analysis type: modal analysis (space problem); ■ Element type: linear.
Vibration mode of a thin piping elbow Scale = 1/7 01-0067SDLLB_FEM
Units
I. S.
Geometry
■ Average radius of curvature: OA = R = 1 m, ■ Straight circular hollow section: ■ Outer diameter: de = 0.020 m, ■ Inner diameter: di = 0.016 m, ■ Section: A = 1.131 x 10-4 m2, ■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4,
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■ Polar inertia: Ip = 9.274 x 10-9 m4. ■ Points coordinates (in m):
► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: Fixed at points A and B, ■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.62.2.2 Eigen modes frequencies
Reference solution
The Rayleigh method applied to a thin curved beam is used to determine parameters such as:
■ transverse bending:
fj = μi
2
2π R2 GIpρA where i = 1,2.
Finite elements modeling
■ Linear element: beam, ■ 11 nodes, ■ 10 linear elements.
Eigen mode shapes
1.62.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode Transverse 1 frequency [Hz] 44.23 CM2 Eigen mode Eigen mode Transverse 2 frequency [Hz] 125
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1.62.3 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%
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1.63 A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0057SSLSB_FEM)
Test ID: 2487
Test status: Passed
1.63.1 Description Verifies the vertical displacement for a square plate (0.01 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.63.2 Background
1.63.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.01 m,
■ Slenderness: λ = ah = 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.
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Loading
■ External: 1 MPa uniform pressure, ■ Internal: None.
1.63.2.2 Vertical displacement at C
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 289 nodes, ■ 256 surface quadrangles.
1.63.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -6.639 x 10-2
1.63.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement in point C [cm] -6.56563 cm 1.11%
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1.64 A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0061SSLSB_FEM)
Test ID: 2491
Test status: Passed
1.64.1 Description Verifies the vertical displacement for a square plate (0.1 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.
1.64.2 Background
1.64.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.1 m,
■ Slenderness: λ = ah = 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.
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Loading
■ External: 1 MPa uniform pressure, ■ Internal: None.
1.64.2.2 Vertical displacement at C
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 289 nodes, ■ 256 surface quadrangles.
1.64.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -0.78661 x 10-4
1.64.3 Calculated results
Result name
Result description Value Error
Dz Vertical displacement in point C [mm] -0.0781846 mm
0.61%
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1.65 A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (01-0065SSLSB_FEM)
Test ID: 2495
Test status: Passed
1.65.1 Description Verifies the vertical displacement for a square plate (0.05 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.
1.65.2 Background
1.65.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar.
0.05 m thick plate fixed on its perimeter Scale = 1/5 01-0065SSLSB_FEM
Units
I. S.
Geometry
■ Side: a = 1 m, ■ Thickness: h = 0.05 m, ■ Slenderness: λ = 20.
Materials properties
■ Reinforcement, ■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: Fixed sides, ■ Inner: None.
Loading
■ External: Punctual force applied at the center of the plate: FZ = -106 N, ■ Internal: None.
1.65.2.2 Vertical displacement at point C center of the plate)
Reference solution
This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%.
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 961 nodes, ■ 900 surface quadrangles.
1.65.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.2595 x 10-2
1.65.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement in point C [m] -0.00257232 m 0.86%
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1.66 Reactions on supports and bending moments on a 2D portal frame (Rafters) (01-0077SSLPB_FEM)
Test ID: 2500
Test status: Passed
1.66.1 Description Moments and actions on supports calculation on a 2D portal frame. The purpose of this test is to verify the results of Advance Design for the M. R. study of a 2D portal frame.
1.66.2 Background
1.66.2.1 Model description ■ Reference: Design and calculation of metal structures. ■ Analysis type: static linear; ■ Element type: linear.
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1.66.2.2 Moments and actions on supports M.R. calculation on a 2D portal frame. RDM results, for the linear load perpendicular on the rafters, are:
2qLVV EA == ( ) ( ) H
fh3f3k²hf5h8
32²qLHH EA =
++++
==
HhMM DB −== ( )fhH8
²qLMC +−=
1.66.2.3 Theoretical results
Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the chords
Solver Result name Result description Reference value CM2 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
1.66.3 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%
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1.67 Slender beam of variable rectangular section (fixed-fixed) (01-0086SDLLB_FEM)
Test ID: 2504
Test status: Passed
1.67.1 Description Verifies the eigen modes (flexion) for a slender beam with variable rectangular section (fixed-fixed).
1.67.2 BackgroundOverview
1.67.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 10/89; ■ Analysis type: modal analysis (plane problem); ■ Element type: linear.
Units
I. S.
Geometry
■ Length: L = 0.6 m, ■ Constant thickness: h = 0.01 m ■ Initial section:
► b0 = 0.03 m ► A0 = 3 x 10-4 m²
■ Section variation: ► with (α = 1) ► b = b0e-2αx ► A = A0e-2αx
Materials properties
■ E = 2 x 1011 Pa ■ ν = 0.3 ■ ρ = 7800 kg/m3
Boundary conditions
■ Outer: ► Fixed at end x = 0, ► Fixed at end x = 0.6 m.
■ Inner: None.
Loading
■ External: None, ■ Internal: None.
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1.67.2.2 Reference results
Calculation method used to obtain the reference solution
ωi pulsation is given by the roots of the equation:
( ) ( ) ( ) ( ) 0rlsinslshrs2
²r²sslchrlcos1 =−
+−
with
( ) 0²²s;²r;EIA 2
isi2
i2i
zo
2i04
i >α−λ⎯→⎯α−λ=λ+α=ωρ
=λ
Therefore, the ν translation components of φi(x) mode, are:
( ) ( ) ( ) ⎥⎦
⎤⎢⎣
⎡−
−−
+−=Φ α ))sx(rsh)rxsin(s()rlsin(s)sl(rsh
)sl(ch)rlcos(sxchrxcosex xi
Uncertainty about the reference: analytical solution:
Reference values
Eigen mode φi(x)* Eigen mode order
Frequency (Hz) x = 0 0.1 0.2 0.3 0.4 0.5 0.6
1 143.303 0 0.237 0.703 1 0.859 0.354 0 2 396.821 0 -0.504 -0.818 0 0.943 0.752 0 3 779.425 0 0.670 0.210 -0.831 0.257 1 0 4 1289.577 0 -0.670 0.486 0 -0.594 1 0
* φi(x) eigen modes* standardized to 1 at the point of maximum amplitude.
Eigen modes
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1.67.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Eigen mode Frequency of eigen mode 1 [Hz] 143.303 CM2 Eigen mode Frequency of eigen mode 2 [Hz] 396.821 CM2 Eigen mode Frequency of eigen mode 3 [Hz] 779.425 CM2 Eigen mode Frequency of eigen mode 4 [Hz] 1289.577
1.67.3 Calculated results
Result name
Result description Value Error
Frequency of eigen mode 1 [Hz] 145.88 Hz 1.80% Frequency of eigen mode 2 [Hz] 400.26 Hz 0.87% Frequency of eigen mode 3 [Hz] 783.15 Hz 0.48% Frequency of eigen mode 4 [Hz] 1293.42 Hz 0.30%
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1.68 Short beam on two hinged supports (01-0084SSLLB_FEM)
Test ID: 2502
Test status: Passed
1.68.1 Description Verifies the deflection magnitude on a non-slender beam with two hinged supports.
1.68.2 Background
1.68.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 02/89 ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Units
I. S.
Geometry
■ Length: L = 1.44 m, ■ Area: A = 31 x 10-4 m² ■ Inertia: I = 2810 x 10-8 m4 ■ Shearing coefficient: az = 2.42 = A/Ar
Materials properties
■ E = 2 x 1011 Pa ■ ν = 0.3
Boundary conditions
■ Hinge at end x = 0, ■ Hinge at end x = 1.44 m.
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Loading
Uniformly distributed force of p = -1. X 105 N/m on beam AB.
1.68.2.2 Reference results Calculation method used to obtain the reference solution
The deflection on the middle of a non-slender beam considering the shear force deformations given by the Timoshenko function:
GA8pl
EIpl
3845v
r
24
+=
where ( )υ+=
12EG and
zr a
AA =
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 Magnitudes and units Value C V, deflection (m) -1.25926 x 10-3
1.68.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Dz Deflection magnitude in point C [m] -1.25926
1.68.3 Calculated results
Result name
Result description Value Error
Dz Deflection magnitude in node C [m] -0.00125926 m
0.00%
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1.69 Double fixed beam in Eulerian buckling with a thermal load (01-0091HFLLB_FEM)
Test ID: 2506
Test status: Passed
1.69.1 Description Verifies the normal force on the nodes of a double fixed beam in Eulerian buckling with a thermal load.
1.69.2 Background
1.69.2.1 Model description ■ Reference: Euler theory; ■ Analysis type: Eulerian buckling; ■ Element type: linear.
Units
I. S.
Geometry
L = 10 m
Cross Section Sx m² Sy m² Sz m² Ix m4 Iy m4 Iz m4 Vx m3 V1y m3 V1z m3 V2y m3 V2z m3
IPE200 0.002850 0.001400 0.001799 0.0000000646 0.0000014200 0.0000194300 0.00000000 0.00002850 0.00019400 0.00002850 0.00019400
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.
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Loading
■ External: Punctual load FZ = 1 N at = L/2 (load that initializes the deformed shape), ■ Internal: ΔT = 5°C corresponding to a compression force of:
kN 925.29500001.000285.011E1.2TESN =×××=Δα=
1.69.2.2 Displacement of the model in the linear elastic range
Reference solution
The reference critical load established by Euler is:
93.3724.117925.29k 724.117
2
P 2
2
critical ==⇒=
⎟⎠⎞
⎜⎝⎛
= λπ NLEI
Observation: in this case, the thermal load has no effect over the critical coefficient
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 11 nodes, ■ 10 elements.
Deformed shape of mode 1
1.69.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Fx Normal Force on Node 6 - Case 101 [kN] -29.925 CM2 Fx Normal Force on Node 6 - Case 102 [kN] -117.724
1.69.3 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%
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1.70 Reactions on supports and bending moments on a 2D portal frame (Columns) (01-0078SSLPB_FEM)
Test ID: 2501
Test status: Passed
1.70.1 Description Moments and actions on supports calculation on a 2D portal frame. The purpose of this test is to verify the results of Advance Design for the M. R. study of a 2D portal frame.
1.70.2 Background
1.70.2.1 Model description ■ Reference: Design and calculation of metal structures. ■ Analysis type: static linear; ■ Element type: linear.
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1.70.2.2 Moments and reactions on supports M.R. calculation on a 2D portal frame. RDM results, for the linear load perpendicular on the column, are:
L2²qhVV EA −=−=
( )( ) ( )fh3f3k²h
fh26kh516
²qhHE +++++
= qhHH EA −=
hHqhM EB −=2
² ( )fhH
4²qhM EC +−= hHM ED −=
1.70.2.3 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
1.70.3 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%
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1.71 Plane portal frame with hinged supports (01-0089SSLLB_FEM)
Test ID: 2505
Test status: Passed
1.71.1 Description Calculation of support reactions of a 2D portal frame with hinged supports.
1.71.2 Background
1.71.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLL 14/89; ■ Analysis type: static linear; ■ Element type: linear.
Units
I. S.
Geometry
■ Length: L = 20 m, ■ I1 = 5.0 x 10-4 m4 ■ a = 4 m ■ h = 8 m ■ b = 10.77 m ■ I2 = 2.5 x 10-4 m4
Materials properties
■ Isotropic linear elastic material. ■ E = 2.1 x 1011 Pa
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Boundary conditions
Hinged base plates A and B (uA = vA = 0 ; uB = vB = 0).
Loading
■ p = -3 000 N/m ■ F1 = -20 000 N ■ F2 = -10 000 N ■ M = -100 000 Nm
1.71.2.2 Calculation method used to obtain the reference solution ■ K = (I2/b)(h/I1) ■ p = a/h ■ m = 1 + p ■ B = 2(K + 1) + m ■ C = 1 + 2m ■ N = B + mC ■ VA = 3pl/8 + F1/2 – M/l + F2h/l ■ HA = pl²(3 + 5m)/(32Nh) + (F1l/(4h))(C/N) + F2(1-(B + C)/(2N)) + (3M/h)((1 + m)/(2N))
1.71.2.3 Reference values Point Magnitudes and units Value
A V, vertical reaction (N) 31 500.0 A H, horizontal reaction (N) 20 239.4 C vc (m) -0.03072
1.71.2.4 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
1.71.3 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%
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1.72 A 3D bar structure with elastic support (01-0094SSLLB_FEM)
Test ID: 2508
Test status: Passed
1.72.1 Description A 3D bar structure with elastic support is subjected to a vertical load of -100 kN. The V2 magnitude on node 5, the normal force magnitude, the reaction magnitude on supports and the action magnitude are verified.
1.72.1.1 Model description ■ Reference: Internal GRAITEC; ■ Analysis type: static linear; ■ Element type: linear.
Units
I. S.
Geometry
For all bars:
■ H = 3 m ■ B = 3 m ■ S = 0.02 m2
Element Node i Node j 1 (bar) 1 5 2 (bar) 2 5 3 (bar) 3 5 4 (bar) 4 5
5 (spring) 5 6
Materials properties
■ Isotropic linear elastic materials ■ Longitudinal elastic modulus: E = 2.1 E8 N/m2,
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Boundary conditions
■ Outer: At node 5: K = 50000 kN/m ; ■ Inner: None.
Loading
■ External: Vertical load at node: P = -100 kN, ■ Internal: None.
1.72.1.2 Theoretical results
System solution
2
22 BHL += . Also, U1 = V1 = U5 = U6 = V6 = 0
■ Stiffness matrix of bar 1
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( ) ( )
1L2x= where
.121.1
21)(.).1()(
−
++−=⇔+−=
ξ
ξξξ jiji uuuuLxu
Lxxu
in the local coordinate system:
[ ] [ ] [ ][ ] [ ] [ ]
)()()()(
0000010100000101
)()(
1111
41
41
41
41
2=
221
21
2121
1
1
1
101
j
j
i
i
j
i
L Te
T
v
vuvu
LES
uu
LESd
LES
dL
ESdxBBESdVBHBke
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
=⎥⎦
⎤⎢⎣
⎡−
−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
⎭⎬⎫
⎩⎨⎧−
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−===
∫
∫∫∫
−
−
ξ
ξ
where [ ]
)v()u()v()u(
0000010100000101
LESk
5
5
1
1
1
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
=
The elementary matrix [ ]ek expressed in the global coordinate system XY is the following: (θ angle allowing the transition from the global base to the local base):
[ ] [ ] [ ][ ] [ ]
[ ]
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
θθθθ−θθ−θθθθθ−θ−
θ−θθ−θθθθθ−θ−θθθ
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
θθ−θθ
θθ−θθ
==−
22
22
22
22
e
eeeT
ee
sinsincossinsincossincoscossincoscos
sinsincossinsincossincoscossincoscos
LESK
cossin00sincos00
00cossin00sincos
Ravec RkRK
Knowing that LHsin and
2cos == θθ
LB
, then:
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=θθ
=θ
=θ
2
2
22
2
22
L2HBcossin
LHsin
L2Bcos
[ ]
)()()()(
22
2222
22
2222
:)DH(arctan =5,1 nodes 1element for
5
5
1
1
22
22
22
22
31
VUVU
HHBHHB
HBBHBB
HHBHHB
HBBHBB
LESK
⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−
−−
−−
−−
=→ θ
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■ Stiffness matrix of spring support 5
[ ]
)()()()(
0000010100000101
)()(
1111
:system coordinate local in the
4KKsay We
5
j
j
i
i
j
i
vuvu
Kuu
Kk
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−
′=⎥⎦
⎤⎢⎣
⎡−
−′=
=′
[ ]
)()()()(
101000001010
0000
':90=6,5 nodes 5element for
6
6
5
5
5
VUVU
KK
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−=°→ θ
■ System [ ]{ } { }FQK =
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
−=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
′′−
′−′+−−
−−
−−
−−
6Y
6X
5X
1Y
1X
6
6
5
5
1
1
233
233
3
2
33
2
3
233
233
3
2
33
2
3
RR
PRRR
VUVUVU
K0K000000000
K0KHLES
2HB
LESH
LES
2HB
LES
002
HBLES
2B
LES
2HB
LES
2B
LES
00HLES
2HB
LESH
LES
2HB
LES
002
HBLES
2B
LES
2HB
LES
2B
LES
If U1 = V1 = U5 = U6 = V6 = 0, then:
m 001885.0
4KH
LES
4P
KHLES
4P
V2
32
3
5 −=+
−=
′+
−=
And N 23563V
4KRN 1436VH
LESR
0RN 1015V2
HBLESRN 1015V
2HB
LESR
56Y52
31Y
6X535X531X
=−==−=
=−===−=
Note:
■ The values on supports specified by Advance Design correspond to the actions, ■ RY6 calculated value must be multiplied by 4 in relation to the double symmetry,
■ x1 value is similar to the one found by Advance Design by dividing this by 2
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Effort in bar 1:
⎭⎬⎫
⎩⎨⎧
−=
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡−
−
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
17591759
1111
and
200
200
002
002
5
1
5
1
5
5
1
1
5
5
1
1
NN
uu
LES
VUVU
LB
LH
LH
LB
LB
LH
LH
LB
vuvu
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡−
−
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
5
1
5
1
5
5
1
1
5
5
1
1
1111
and
cossin00sincos00
00cossin00sincos
NN
uu
LES
VUVU
vuvu
θθθθ
θθθθ
Reference values
Point Magnitude Units Value 5 V2 m -1.885 10-3
All bars Normal force N -1759 Supports 1, 3, 4 and 5 Fz action N -1436 Supports 1, 3, 4 and 5 Action Fx=Fy N 7182/1015 ±=
Support 6 Fz action N 23563 x 4=94253
Finite elements modeling
■ Linear element: beam, automatic mesh, ■ 5 nodes, ■ 4 linear elements.
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Deformed shape
Normal forces diagram
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1.72.1.3 Reference values
Solver Result name Result description Reference value CM2 D V2 magnitude on node 5 [m] -1.885 10-3 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
1.72.2 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%
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1.73 Fixed/free slender beam with eccentric mass or inertia (01-0096SDLLB_FEM)
Test ID: 2510
Test status: Passed
1.73.1 Description Fixed/free slender beam with eccentric mass or inertia.
Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending, transverse bending, punctual mass.
1.73.2 Background
1.73.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 15/89; ■ Analysis type: modal analysis; ■ Element type: linear. ■ Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending,
transverse bending, punctual mass..
1.73.2.2 Problem data
Units
I. S.
Geometry
■ Outer diameter: de= 0.35 m, ■ Inner diameter: di = 0.32 m, ■ Beam length: l = 10 m, ■ Distance BC: lBC = 1 m ■ Area: A =1.57865 x 10-2 m2 ■ Inertia: Iy = Iz = 2.21899 x 10-4m4 ■ Polar inertia: Ip = 4.43798 x 10-4m4 ■ Punctual mass: mc = 1000 kg
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Materials properties
■ Longitudinal elasticity modulus of AB element: E = 2.1 x 1011 Pa, ■ Density of the linear element AB: ρ = 7800 kg/m3 ■ Poisson's ratio ν=0.3(this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the more
appropriate to obtain the correct frequency value of modes No. 4 and 5 with NE/NASTRAN: ■ Elastic modulus of BC element: E = 1021 Pa ■ Density of the linear element BC: ρ = 0 kg/m3
Boundary conditions
Fixed at point A, x = 0,
Loading
None for the modal analysis
1.73.2.3 Reference frequencies
Reference solutions
The different eigen frequencies are determined using a finite elements model of Euler beam (slender beam).
fz + t0 = flexion x,z + torsion
fy + tr = flexion x,y + traction
Mode Units Reference 1 (fz + t0) Hz 1.636 2 (fy + tr) Hz 1.642 3 (fy + tr) Hz 13.460 4 (fz + t0) Hz 13.590 5 (fz + t0) Hz 28.900 6 (fy + tr) Hz 31.960 7 (fz + t0) Hz 61.610 1 (fz + t0) Hz 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
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1.73.2.4 Theoretical results
Solver Result name Result description Reference value CM2 Frequency Eigen mode 1 frequency (fz + t0) [Hz] 1.636 CM2 Frequency Eigen mode 2 frequency (fy + tr) [Hz] 1.642 CM2 Frequency Eigen mode 3 frequency (fy + tr) [Hz] 13.46 CM2 Frequency Eigen mode 4 frequency (fz + t0) [Hz] 13.59 CM2 Frequency Eigen mode 5 frequency (fz + t0) [Hz] 28.90 CM2 Frequency Eigen mode 6 frequency (fy + tr) [Hz] 31.96 CM2 Frequency Eigen mode 7 frequency (fz + t0) [Hz] 61.61 CM2 Frequency Eigen mode 8 frequency (fy + tr) [Hz] 63.93
Note:
fz + t0 = flexion x,z + torsion
fy + tr = flexion x,y + traction
Observation: because the mass matrix of Advance Design is condensed and not consistent, the torsion modes obtained are not taking into account the self rotation mass inertia of the beam.
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1.73.3 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 100%
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1.74 Fixed/free slender beam with centered mass (01-0095SDLLB_FEM)
Test ID: 2509
Test status: Passed
1.74.1 Description Fixed/free slender beam with centered mass.
Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending, transverse bending, punctual mass.
1.74.2 Background ■ Reference: Structure Calculation Software Validation Guide, test SDLL 15/89; ■ Analysis type: modal analysis; ■ Element type: linear. ■ Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending,
transverse bending, punctual mass.
1.74.2.1 Test data
Units
I. S.
Geometry
■ Outer diameter de = 0.35 m, ■ Inner diameter: di = 0.32 m, ■ Beam length: l = 10 m, ■ Area: A = 1.57865 x 10-2 m2 ■ Polar inertia: IP = 4.43798 x 10-4m4 ■ Inertia: Iy = Iz = 2.21899 x 10-4m4 ■ Punctual mass: mc = 1000 kg ■ Beam self-weight: M
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Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Density: ρ = 7800 kg/m3 ■ Poisson's ratio: ν=0.3 (this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the
more appropriate to obtain the correct frequency value of mode No. 8 with NE/NASTRAN)
Boundary conditions
■ Outer: Fixed at point A, x = 0, ■ Inner: none
Loading
None for the modal analysis
1.74.2.2 Reference results
Reference frequency
For the first mode, the Rayleigh method gives the approximation formula
)M24.0m(IEI3
x2/1fc
3z
1 +Π=
Mode Shape Units Reference 1 Flexion Hz 1.65 2 Flexion Hz 1.65 3 Flexion Hz 16.07 4 Flexion Hz 16.07 5 Flexion Hz 50.02 6 Flexion Hz 50.02 7 Traction Hz 76.47 8 Torsion Hz 80.47 9 Flexion Hz 103.2 10 Flexion Hz 103.2
Comment: The mass matrix associated with the beam torsion on two nodes, is expressed as:
⎥⎦
⎤⎢⎣
⎡×
××ρ12/12/11
3Il P
And to the extent that Advance Design uses a condensed mass matrix, the value of the torsion mass inertia
introduced in the model is set to: 3
Il p××ρ
Uncertainty about the reference frequencies
■ Analytical solution mode 1 ■ Other modes: ± 1%
Finite elements modeling
■ Linear element AB: Beam ■ Beam meshing: 20 elements.
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Modal deformations
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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.
Eigen modes vector 8
Node DX DY DZ RX RY RZ 1 -3.336e-033 6.479e-031 -6.316e-031 1.055e-022 5.770e-028 5.980e-028 2 -5.030e-013 1.575e-008 -1.520e-008 1.472e-002 6.022e-008 6.243e-008 3 -1.005e-012 6.185e-008 -5.966e-008 2.944e-002 1.171e-007 1.214e-007 4 -1.505e-012 1.365e-007 -1.317e-007 4.416e-002 1.705e-007 1.769e-007 5 -2.002e-012 2.381e-007 -2.296e-007 5.887e-002 2.206e-007 2.289e-007 6 -2.495e-012 3.648e-007 -3.517e-007 7.359e-002 2.673e-007 2.774e-007 7 -2.983e-012 5.149e-007 -4.963e-007 8.831e-002 3.106e-007 3.225e-007 8 -3.464e-012 6.867e-007 -6.618e-007 1.030e-001 3.506e-007 3.641e-007 9 -3.939e-012 8.785e-007 -8.464e-007 1.177e-001 3.873e-007 4.023e-007 10 -4.406e-012 1.088e-006 -1.049e-006 1.325e-001 4.207e-007 4.371e-007 11 -4.863e-012 1.315e-006 -1.267e-006 1.472e-001 4.508e-007 4.684e-007 12 -5.310e-012 1.556e-006 -1.499e-006 1.619e-001 4.777e-007 4.964e-007 13 -5.746e-012 1.811e-006 -1.744e-006 1.766e-001 5.015e-007 5.210e-007 14 -6.169e-012 2.077e-006 -2.000e-006 1.913e-001 5.221e-007 5.423e-007 15 -6.580e-012 2.353e-006 -2.265e-006 2.061e-001 5.396e-007 5.605e-007 16 -6.976e-012 2.637e-006 -2.539e-006 2.208e-001 5.541e-007 5.755e-007 17 -7.357e-012 2.928e-006 -2.819e-006 2.355e-001 5.658e-007 5.874e-007 18 -7.723e-012 3.224e-006 -3.104e-006 2.502e-001 5.746e-007 5.965e-007 19 -8.072e-012 3.524e-006 -3.393e-006 2.649e-001 5.808e-007 6.028e-007 20 -8.403e-012 3.826e-006 -3.685e-006 2.797e-001 5.844e-007 6.065e-007 21 -8.717e-012 4.130e-006 -3.977e-006 2.944e-001 5.856e-007 6.077e-007
With NE/NASTRAN, the results associated with mode No. 8, are:
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1.74.2.3 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).
1.74.3 Calculated results
Result name
Result description Value Error
Eigen mode 1 frequency [Hz] 1.65 Hz 0.00% Eigen mode 2 frequency [Hz] 1.65 Hz 0.00% Eigen mode 3 frequency [Hz] 16.06 Hz -0.06% Eigen mode 4 frequency [Hz] 16.06 Hz -0.06% Eigen mode 5 frequency [Hz] 50 Hz -0.04% Eigen mode 6 frequency [Hz] 50 Hz -0.04% Eigen mode 7 frequency [Hz] 76.46 Hz -0.01% Eigen mode 9 frequency [Hz] 103.14 Hz -0.06% Eigen mode 10 frequency [Hz] 103.14 Hz 100%
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1.75 Vibration mode of a thin piping elbow in space (case 3) (01-0069SDLLB_FEM)
Test ID: 2499
Test status: Passed
1.75.1 Description Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with two straight elements (2 m long) and subjected to its self weight only.
1.75.1.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; ■ Analysis type: modal analysis (space problem); ■ Element type: linear.
Vibration mode of a thin piping elbow Scale = 1/12 01-0069SDLLB_FEM
Units
I. S.
Geometry
■ Average radius of curvature: OA = R = 1 m, ■ L = 2 m, ■ Straight circular hollow section: ■ Outer diameter: de = 0.020 m, ■ Inner diameter: di = 0.016 m, ■ Section: A = 1.131 x 10-4 m2, ■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, ■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4, ■ Polar inertia: Ip = 9.274 x 10-9 m4.
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■ Points coordinates (in m): ► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 ) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.
Boundary conditions
■ Outer: ► Fixed at points C and D ► At A: translation restraint along y and z, ► At B: translation restraint along x and z,
■ Inner: None.
Loading
■ External: None, ■ Internal: None.
1.75.1.2 Eigen modes frequencies
Reference solution
The Rayleigh method applied to a thin curved beam is used to determine parameters such as:
■ transverse bending:
fj = μi
2
2π R2 GIpρA where i = 1,2 with i = 1,2:
Finite elements modeling
■ Linear element: beam, ■ 41 nodes, ■ 40 linear elements.
Eigen mode shapes
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1.75.1.3 Theoretical results
Reference
Solver Result name Result description Reference value CM2 Eigen mode Eigen mode Transverse 1 frequency [Hz] 17.900 CM2 Eigen mode Eigen mode Transverse 2 frequency [Hz] 24.800
1.75.2 Calculated results
Result name
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%
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1.76 Slender beam of variable rectangular section with fixed-free ends (ß=5) (01-0085SDLLB_FEM)
Test ID: 2503
Test status: Passed
1.76.1 Description Verifies the eigen modes (bending) for a slender beam with variable rectangular section (fixed-free).
1.76.2 Background
1.76.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SDLL 09/89; ■ Analysis type: modal analysis (plane problem); ■ Element type: linear.
Units
I. S.
Geometry
■ Length: L = 1 m, ■ Straight initial section:
► h0 = 0.04 m ► b0 = 0.05 m ► A0 = 2 x 10-3 m²
■ Straight final section ► h1 = 0.01 m ► b1 = 0.01 m ► A1 = 10-4 m²
Materials properties
■ E = 2 x 1011 Pa ■ ρ = 7800 kg/m3
Boundary conditions
■ Outer: ► Fixed at end x = 0, ► Free at end x = 1
■ Inner: None.
Loading
■ External: None, ■ Internal: None.
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1.76.2.2 Reference results Calculation method used to obtain the reference solution
Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories):
²t²A
²x²EIz
x2
2
δνδ
ρ−=⎟⎠
⎞⎜⎝
⎛δ
νδδδ
where Iz and A vary with the abscissa.
The result is:
( ) ρβαλπ
= 12E
²l1h,i
21fi with
⎪⎪⎩
⎪⎪⎨
⎧
=ο
=β
=ο
=α
51b
b
41h
h
λ1 λ2 λ3 λ4 λ5 β = 5 24.308 75.56 167.21 301.9 480.4
Uncertainty about the reference: analytical solution:
Reference values
Eigen mode type Frequency (Hz) 1 56.55 2 175.79 3 389.01 4 702.36
Flexion
5 1117.63
MODE 1 Scale = 1/4
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MODE 2 Scale = 1/4
MODE 3 Scale = 1/4
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MODE 4 Scale = 1/4
MODE 5 Scale = 1/4
1.76.2.3 Theoretical results
Theoretical Frequency
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
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1.76.3 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%
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1.77 Cantilever beam in Eulerian buckling with thermal load (01-0092HFLLB_FEM)
Test ID: 2507
Test status: Passed
1.77.1 Description Verifies the vertical displacement and the normal force on a cantilever beam in Eulerian buckling with thermal load.
1.77.2 Background
1.77.2.1 Model description ■ Reference: Euler theory; ■ Analysis type: Eulerian buckling; ■ Element type: linear.
Units
I. S.
Geometry
■ L = 10 m ■ S=0.01 m2 ■ I = 0.0002 m4
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)
( 5000001.0T0005.001.010.2
100000ESN
100 −×=Δα==×
−==ε )
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1.77.2.2 Displacement of the model in the linear elastic range
Reference solution
The reference critical load established by Euler is:
98696.010000098696 98696
4P 2
2
critical ==⇒== λπ NLEI
Observation: in this case, the thermal load has no effect over the critical coefficient
Finite elements modeling
■ Linear element: beam, imposed mesh, ■ 5 nodes, ■ 4 elements.
Deformed shape
1.77.2.3 Theoretical results
Solver Result name Result description Reference value CM2 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
1.77.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement on Node 5 [cm] -1 cm 0.00% Fx Normal Force - Case 101 [N] -100000 N 0.00% Fx Normal Force - Case 102 [N] -98699.3 N 0.00%
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1.78 Simple supported beam in free vibration (01-0098SDLLB_FEM)
Test ID: 2512
Test status: Passed
1.78.1 Description Simple supported beam in free vibration.
Tested functions: Shear force, eigen frequencies.
1.78.2 Background ■ Reference: NAFEMS, FV5 ■ Analysis type: modal analysis; ■ Tested functions: Shear force, eigen frequencies.
1.78.2.1 Problem data
Units
I. S.
Geometry
Full square section:
■ Dimensions: a x b = 2m x 2 m ■ Area: A = 4 m2 ■ Inertia: IP = 2.25 m4
Iy = Iz = 1.333 m4
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Materials properties
■ Longitudinal elastic modulus: E = 2 x 1011 Pa, ■ Poisson's ratio: ν = 0.3. ■ Density: ρ = 8000 kg/m3
Boundary conditions
■ Outer: ► x = y = z = Rx = 0 at A ; ► y = z =0 at B ;
■ Inner: None.
Loading
None for the modal analysis
1.78.2.2 Reference frequencies Mode Shape Units Reference
1 Flexion Hz 42.649 2 Flexion Hz 42.649 3 Torsion Hz 77.542 4 Traction Hz 125.00 5 Flexion Hz 148.31 6 Flexion Hz 148.31 7 Torsion Hz 233.10 8 Flexion Hz 284.55 9 Flexion Hz 284.55
Comment: Due to the condensed (lumped) nature of the mass matrix of Advance Design, the frequencies values of 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
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Modal deformations
1.78.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Frequency of eigen mode 1 [Hz] 42.649 CM2 Frequency of eigen mode 2 [Hz] 42.649 CM2 Frequency of eigen mode 3 [Hz] 77.542 CM2 Frequency of eigen mode 4 [Hz] 125.00 CM2 Frequency of eigen mode 5 [Hz] 148.31 CM2 Frequency of eigen mode 6 [Hz] 148.31 CM2 Frequency of eigen mode 7 [Hz] 233.10
Comment: The torsion modes No. 3 and 7 that are calculated with NASTRAN cannot be calculated with Advance Design CM2 solver and therefore the mode No. 3 of the Advance Design analysis corresponds to mode No. 4 of the reference. The same problem in the case of No. 7 - Advance Design, that corresponds to mode No. 8 of the reference.
1.78.3 Calculated results
Result name
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%
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1.79 Membrane with hot point (01-0099HSLSB_FEM)
Test ID: 2513
Test status: Passed
1.79.1 Description Membrane with hot point.
Tested functions: Stresses.
1.79.2 Background ■ Reference: NAFEMS, Test T1 ■ Analysis type: static, thermo-elastic; ■ Tested functions: Stresses.
1.79.2.1 Problem data
Observation: the units system of the initial NAFEMS test, defined in mm, was transposed in m for practical reasons. However, this has no influence on the results values.
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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 ΔT = 100°C;
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1.79.3 σyy stress at point A:
Reference solution:
Reference value: σyy = 50 MPa in A
Finite elements modeling
■ Planar elements: membranes, ■ 28 planar elements, ■ 39 nodes.
1.79.3.1 Theoretical results
Solver Result name Result description Reference value CM2 syy_mid σyy in A [MPa] 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.
1.79.4 Calculated results
Result name
Result description Value Error
syy_mid Sigma yy in A [MPa] 50.8666 MPa 1.73%
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1.80 Beam on 3 supports with T/C (k = -10000 N/m) (01-0102SSNLB_FEM)
Test ID: 2516
Test status: Passed
1.80.1 Description Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k = -10000 N/m).
1.80.2 Background
1.80.2.1 Model description ■ Reference: internal GRAITEC test; ■ Analysis type: static non linear; ■ Element type: linear, T/C.
Units
I. S.
Geometry
■ L = 10 m ■ Section: IPE 200, Iz = 0.00001943 m4
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 N/m2, ■ Poisson's ratio: ν = 0.3.
Boundary conditions
■ Outer: ► Support at node 1 restrained along x and y (x = 0), ► Support at node 2 restrained along y (x = 10 m), ► T/C ky Rigidity = -10000 N/m (the – sign corresponds to an upwards restraint),
■ Inner: None.
Loading
■ External: Vertical punctual load P = -100 N at x = 5 m, ■ Internal: None.
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1.80.2.2 References solutions
Displacements
( )( )
( )( )
( )( )
( ) rad 000034.0Lk2EI3EI32
LkEI6PL
m 00058.0Lk2EI316
PL3v
rad 000106.0Lk2EI3EI16
LkEI3PL
rad 000129.0Lk2EI3EI32
LkEI2PL3
3yzz
3yz
2
3
3yz
3
3
3yzz
3yz
2
2
3yzz
3yz
2
1
=+
+−=β
=+
−=
=+
+−=β
−=+
+=β
Mz Moments
( )( )
N.m 9.2202
MM4
PL)m5x(M
N.m 15.58Lk2EI316
PLk3M
0M
1z2zz
3yz
4y
2z
1z
−=−
+==
−=+
=
=
Finite elements modeling
■ Linear element: S beam, automatic mesh, ■ 3 nodes, ■ 2 linear elements + 1 T/C.
Deformed shape
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Moment diagram
1.80.2.3 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
1.80.3 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] -3.44295e-005 Rad -1.26% My M moment - node 1 [Nm] 1.77636e-015 N*m 0.00% My M moment - node 2 [Nm] 58.1169 N*m -0.06% My M moment - middle span 1 [Nm] -220.942 N*m -0.02%
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1.81 Beam on 3 supports with T/C (k = 0) (01-0100SSNLB_FEM)
Test ID: 2514
Test status: Passed
1.81.1 Description Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k = 0).
1.81.2 Background
1.81.2.1 Model description ■ Reference: internal GRAITEC test; ■ Analysis type: static non linear; ■ Element type: linear, T/C.
Units
I. S.
Geometry
■ L = 10 m ■ 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 stiffness ky = 0,
■ Inner: None.
Loading
■ External: Vertical punctual load P = -100 N at x = 5 m, ■ Internal: None.
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1.81.2.2 References solutions ky being null, the non linear model behaves the same way as the structure without support 3.
Displacements
( )( )
( )( )
( )( )
( ) rad 000153.0Lk2EI3EI32
LkEI6PL
m 00153.0Lk2EI316
PL3v
rad 000153.0Lk2EI3EI16
LkEI3PL
rad 000153.0Lk2EI3EI32
LkEI2PL3
3yzz
3yz
2
3
3yz
3
3
3yzz
3yz
2
2
3yzz
3yz
2
1
=+
+−=β
=+
−=
=+
+−=β
−=+
+=β
Mz Moments
( )( )
N.m 2502
MM4
PL)m5x(M
0Lk2EI316
PLk3M
0M
1z2zz
3yz
4y
2z
1z
−=−
+==
=+
=
=
Finite elements modeling
■ Linear element: S beam, automatic mesh, ■ 3 nodes, ■ 2 linear elements + 1 T/C.
Deformed shape
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Moment diagrams
1.81.2.3 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
1.81.3 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] 5.05771e-014 N*m 0.00% My Moment M - middle span 1 [Nm] -250 N*m 0.00%
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1.82 Non linear system of truss beams (01-0104SSNLB_FEM)
Test ID: 2518
Test status: Passed
1.82.1 Description Verifies the displacement and the normal force for a bar system containing 4 elements of the same length and 2 diagonals.
1.82.2 Background
1.82.2.1 Model description ■ Reference: internal GRAITEC test; ■ Analysis type: static non linear; ■ Element type: linear, bar, tie.
Units
I. S.
Geometry
■ L = 5 m ■ Section S = 0.005 m2
Materials properties
Longitudinal elastic modulus: E = 2.1 x 1011 N/m2.
Boundary conditions
■ Outer: ► Support at node 1 restrained along x and y, ► Support at node 2 restrained along x and y,
■ Inner: None.
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Loading
■ External: Horizontal punctual load P = 50000 N at node 3, ■ Internal: None.
1.82.2.2 References solutions In non linear analysis without large displacement, the introduction of ties for the diagonal bars removes bar 5 (test No. 0103SSLLB_FEM allows finding an compression force in this bar at the linear calculation).
Displacements
0v
m 000238.0ESPLv
m 001195.0ES11PL5uu
4
3
43
=
−=−=
===
N normal forces
0N 0NN 70711P2N N 50000PN
0N 0N
4243
1323
1412
==
==−=−=
==
Finite elements modeling
■ Linear element: bar, without meshing, ■ 4 nodes, ■ 6 linear elements.
Deformed shape
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Normal forces
1.82.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DX u3 displacement on Node 3 [m] 0.001195 CM2 DZ v3 displacement on Node 3 [m] -0.000238 CM2 DX u4 displacement on Node 4 [m] 0.001195 CM2 DZ v4 displacement on Node 4 [m] 0 CM2 Fx N12 normal force on Element 1 [N] 0 CM2 Fx N23 normal force on Element 2 [N] -50000 CM2 Fx N34 normal force on Element 3 [N] 0 CM2 Fx N14 normal force on Element 4 [N] 0 CM2 Fx N13 normal effort on Element 5 [N] 70711 CM2 Fx N24 normal force on Element 6 [N] 0
1.82.3 Calculated results
Result name Result description Value Error DX u3 displacement on Node 3 [m] 0.00119035 m -0.39% DZ v3 displacement on Node 3 [m] -0.000238095 m -0.04% DX u4 displacement on Node 4 [m] 0.00119035 m -0.39% DZ v4 displacement on Node 4 [m] 9.94686e-316 m 0.00% Fx N12 normal force on Element 1 [N] 0 N 0.00% Fx N23 normal force on Element 2 [N] -50000 N 0.00% Fx N34 normal force on Element 3 [N] 0 N 0.00% Fx N14 normal force on Element 4 [N] 2.08884e-307 N 0.00% Fx N13 normal effort on Element 5 [N] 70710.7 N 0.00% Fx N24 normal force on Element 6 [N] 0 N 0.00%
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1.83 Linear system of truss beams (01-0103SSLLB_FEM)
Test ID: 2517
Test status: Passed
1.83.1 Description Verifies the displacement and the normal force for a bar system containing 4 elements of the same length and 2 diagonals.
1.83.2 Background
1.83.2.1 Model description ■ Reference: internal GRAITEC test; ■ Analysis type: static linear; ■ Element type: linear, bar.
Units
I. S.
Geometry
■ L = 5 m ■ Section S = 0.005 m2
Materials properties
Longitudinal elastic modulus: E = 2.1 x 1011 N/m2.
Boundary conditions
■ Outer: ► Support at node 1 restrained along x and y, ► Support at node 2 restrained along x and y,
■ Inner: None.
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Loading
■ External: Horizontal punctual load P = 50000 N at node 3, ■ Internal: None.
1.83.2.2 References solutions
Displacements
m 000108.0ES11PL5v
m 000541.0ES11PL25u
m 000129.0ES11PL6v
m 000649.0ES11PL30u
4
4
3
3
==
==
−=−
=
==
N normal forces
N 32141P11
25N N 22727P115N
N 38569P11
26N N 27272P116N
N 22727P115N 0N
4243
1323
1412
−=−===
==−=−=
===
Finite elements modeling
■ Linear element: bar, without meshing, ■ 4 nodes, ■ 6 linear elements.
Deformed shape
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Normal forces
1.83.2.3 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
1.83.3 Calculated results
Result name 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] 0 N 0.00% Fx N23 normal force on Element 2 [N] -27272.9 N 0.00% Fx N43 normal force on Element 3 [N] 22727.1 N 0.00% Fx N14 normal force on Element 4 [N] 22727.1 N 0.00% Fx N13 normal effort on Element 5 [N] 38569.8 N 0.00% Fx N42 normal force on Element 6 [N] -32140.9 N 0.00%
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1.84 Linear element in combined bending/tension - without compressed reinforcements - Partially tensioned section (02-0158SSLLB_B91)
Test ID: 2520
Test status: Passed
1.84.1 Description Verifies the reinforcement results for a concrete beam with 8 isostatic spans subjects to uniform loads and compression normal forces.
1.84.1.1 Model description ■ Reference: J. Perchat (CHEC) reinforced concrete course ■ Analysis type: static linear; ■ Element type: planar.
Units
■ Forces: kN ■ Moment: kN.m ■ Stresses: MPa ■ Reinforcement density: cm²
Geometry
■ Beam dimensions: 0.2 x 0.5 ht ■ Length: l = 48 m in 8 spans of 6m,
Materials properties
■ Longitudinal elastic modulus: E = 20000 MPa, ■ Poisson's ratio: ν = 0.
Boundary conditions
■ Outer: ► Hinged at end x = 0, ► Vertical support at the same level with all other supports
■ Inner: Hinged at each beam end (isostatic)
Loading
■ External: ► Case 1 (DL):uniform linear load g= -5kN/m (on all spans except 8)
Fx = 10 kN at x = 42m: Ng = -10 kN for spans from 6 to 7
Fx = 140 kN at x = 32m: Ng = -150 kN for span 5
Fx = -50 kN at x = 24m: Ng = -100 kN for span 4
Fx = 50 kN at x = 18m: Ng = -50 kN for span 3
Fx = 50 kN at x = 12m: Ng = -100 kN for span 2
Fx = -70 kN at x = 6m: Ng = -30 kN for span 1
► Case 10 (DL):uniform linear load g = -5 kN/m (span 8)
Fx = 10 kN at x = 48m: Ng = -10 kN
Fx = -10 kN at x = 42m
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► Case 2 to 8 (LL):uniform linear load q = -9 kN/m (on spans 1, 3 to 7) uniform linear load q = -15 kN/m (on span 2)
Fx = 30 kN at x = 6m (case 2 span 1)
Fx = -50 kN at x = 6m (case 3 span 2)
Fx = 50 kN at x = 12m (case 3 span 2)
Fx = -40 kN at x = 12m (case 4 span 3)
Fx = 40 kN at x = 18m (case 4 span 3)
Fx = -100 kN at x = 18m (case 5 span 4)
Fx = 100 kN at x = 24m (case 5 span 4)
Fx = -150 kN at x = 24m (case 6 span 5)
Fx = 150 kN at x = 30m (case 6 span 5)
Fx = -8 kN at x = 30m (case 7 span 6)
Fx = 8 kN at x = 36m (case 7 span 6)
Fx = -8 kN at x = 36m (case 8 span 7
Fx = 8 kN at x = 42m (case 8 span 7)
► Case 9 (ACC):uniform linear load a = -25 kN/m (on 8th span)
Fx = 8 kN at x = 36m (case 9 span 8)
Fx = -8 kN at x = 42m (case 9 span 8)
Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107)
Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102)
Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103)
Comb BAELS: 1xDL + 1*LL (comb 108 to 114)
Comb BAELA: 1xDL + 1xACC with duration of less than 1h (comb 115)
■ Internal: None.
Reinforced concrete calculation hypothesis:
All concrete covers are set to 5 cm
BAEL 91 calculation (according to 99 revised version)
Span Concrete Reinforcement Application Concrete Cracking 1 B20 HA fe500 D>24h No Non
prejudicial 2 B35 Adx fe235 1h<D<24h No Non
prejudicial 3 B50 HA fe 400 D<1h Yes Non
prejudicial 4 B25 HA fe500 D>24h Yes Prejudicial 5 B25 HA fe500 D>24h No Very
prejudicial 6 B30 Adx fe235 D>24h Yes Prejudicial 7 B40 HA fe500 D>24h Yes 160 MPa 8 B45 HA fe500 D<1h Yes Non
prejudicial
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1.84.1.2 Reinforcement calculation Reference solution
Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 fc28 20 35 50 25 25 30 40 45 ft28 1.8 2.7 3.6 2.1 2.1 2.4 3 3.3 fe 500 235 400 500 500 235 500 500
teta 1 0.9 0.85 1 1 1 1 0.85 gamb 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.15 gams 1.15 1.15 1.15 1.15 1.15 1.15 1.15 1
h 1.6 1 1.6 1.6 1.6 1 1.6 1.6
fbu 11.33 22.04 33.33 14.17 14.17 17.00 22.67 39.13 fed 434.78 204.35 347.83 434.78 434.78 204.35 434.78 500.00
sigpreju 250.00 156.67 264.00 250.00 250.00 156.67 160.00 252.76 sigtpreju 200.00 125.33 211.20 200.00 200.00 125.33 160.00 202.21
g 5.00 5.00 5.00 5.00 5.00 5.00 5.00 5.00 q 9.00 15.00 9.00 9.00 9.00 9.00 9.00 25.00 pu 20.25 29.25 20.25 20.25 20.25 20.25 20.25 30.00
pser 14.00 20.00 14.00 14.00 14.00 14.00 14.00 G -30.00 -100.00 -50.00 -100.00 -150.00 -10.00 -10.00 -10.00 Q -30.00 -50.00 -40.00 -100.00 -100.00 -8.00 -8.00 -8.00 l 6.00 6.00 6.00 6.00 6.00 6.00 6.00 6.00
Mu 91.13 131.63 91.13 91.13 91.13 91.13 91.13 135.00 Nu -85.50 -210.00 -127.50 -285.00 -352.50 -25.50 -25.50 -18.00
Mser 63.00 90.00 63.00 63.00 63.00 63.00 63.00 Nser -60.00 -150.00 -90.00 -200.00 -250.00 -18.00 -18.00 Vu 60.75 87.75 60.75 60.75 60.75 60.75 60.75 90.00
Main reinforcement calculation according to ULS Mu/A 74.03 89.63 65.63 34.13 20.63 86.03 86.03 131.40 ubu 0.161 0.100 0.049 0.059 0.036 0.125 0.094 0.083 a 0.221 0.133 0.062 0.077 0.046 0.167 0.123 0.108 z 0.410 0.426 0.439 0.436 0.442 0.420 0.428 0.430
Au 6.12 20.57 7.97 8.35 9.18 11.27 5.21 6.46 Main reinforcement calculation with prejudicial cracking according to SLS
Mser/A 51.000 60.000 45.000 23.000 13.000 59.400 59.400 0.000 a 0.4186 0.6678 0.6303 0.4737 0.4737 0.6328 0.6923 0.6157
Mrb 87.53 220.78 302.44 121.16 121.16 182.01 258.82 267.55 A 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 B -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 C -0.4533 -0.8511 -0.3788 -0.2044 -0.1156 -0.8426 -0.8250 0.0000 D 0.4533 0.8511 0.3788 0.2044 0.1156 0.8426 0.8250 0.0000
alpha1 0.238 0.432 0.428 z 0.414 0.385 0.386
Aserp 10.22 10.99 10.75 Main reinforcement calculation with very prejudicial cracking according to SLS
Mser/A 51.00 60.00 45.00 23.00 13.00 59.40 59.40 0.00 a 0.47 0.72 0.68 0.53 0.53 0.68 0.69 0.67
Mrb 96.93 231.67 319.66 132.43 132.43 192.27 258.82 283.60 A 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 B -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 C -0.5667 -1.0638 -0.4735 -0.2556 -0.1444 -1.0532 -0.8250 0.0000
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Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 D 0.5667 1.0638 0.4735 0.2556 0.1444 1.0532 0.8250 0.0000
alpha1 0.203 z 0.420
Asertp 14.049 Transverse reinforcement calculation
tu 0.68 0.98 0.68 0.68 0.68 0.68 0.68 1.00 k 0.57 0.40 0.00 -0.14 -0.41 0.00 0.00 0.00
At/st 1.87 7.08 4.31 3.90 4.77 7.34 3.45 4.44 Recapitulation
Aflex 6.12 20.57 7.97 10.22 14.05 11.27 10.75 6.46 e0 -0.95 -1.67 -1.43 -3.17 -3.97 -0.29 -0.29 -0.13
Aminfsimp 0.75 2.38 1.86 0.87 0.87 2.11 1.24 1.37 Aminfcomp 0.83 2.54 2.01 0.90 0.90 2.80 1.65 0.30
At 1.87 7.08 4.31 3.90 4.77 7.34 3.45 4.44 Atmin 1.60 3.40 2.00 1.60 1.60 3.40 1.60 1.60
Finite elements modeling
■ Linear elements: beams with imposed mesh ■ 29 nodes, ■ 28 linear elements.
1.84.1.3 Theoretical results
Solver Result name Result description Reference value CM2 Az Inf. main reinf. T1 [cm2] 6.12 CM2 Amin Min. main reinf. T1 [cm2] 0.75 CM2 Atz Trans. reinf. T1 [cm2] 1.87 CM2 Az Inf. main reinf. T2 [cm2] 20.57 CM2 Amin Min. main reinf. T2 [cm2] 2.38 CM2 Atz Trans. reinf. T2 [cm2] 7.08 CM2 Az Inf. main reinf. T3 [cm2] 7.97 CM2 Amin Min. main reinf. T3 [cm2] 1.86 CM2 Atz Trans. reinf. T3 [cm2] 4.31 CM2 Az Inf. main reinf. T4 [cm2] 10.22 CM2 Amin Min. main reinf. T4 [cm2] 0.87 CM2 Atz Trans. reinf. T4 [cm2] 3.90 CM2 Az Inf. main reinf. T5 [cm2] 14.05 CM2 Amin Min. main reinf. T5 [cm2] 4.20 CM2 Atz Trans. reinf. T5 [cm2] 4.77 CM2 Az Inf. main reinf. T6 [cm2] 11.27 CM2 Amin Min. main reinf. T6 [cm2] 2.11 CM2 Atz Trans. reinf. T6 [cm2] 7.34 CM2 Az Inf. main reinf. T7 [cm2] 10.75 CM2 Amin Min. main. reinf. T7 [cm2] 1.24 CM2 Atz Trans. reinf. T7 [cm2] 3.45 CM2 Az Inf. main reinf. T8 [cm2] 6.46 CM2 Amin Min. main reinf. T8 [cm2] 1.37 CM2 Atz Trans. reinf. T8 [cm2] 4.44
The "Mu limit" method must be applied in order to achieve the same results.
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1.84.2 Calculated results
Result name
Result description Value Error
Az Inf. main reinf. T1 [cm²] -6.11718 cm² 0.05% Amin Min. main reinf. T1 [cm²] 0.7452 cm² -0.64% Atz Trans. reinf. T1 [cm²] 1.8699 cm² -0.01% Az Inf. main reinf. T2 [cm²] -20.5688 cm² 0.01% Amin Min. main reinf. T2 [cm²] 2.3783 cm² -0.07% Atz Trans. reinf. T2 [cm²] 7.07943 cm² -0.01% Az Inf. main reinf. T3 [cm²] -7.96552 cm² 0.06% Amin Min. main reinf. T3 [cm²] 1.863 cm² 0.16% Atz Trans. reinf. T3 [cm²] 4.3125 cm² 0.06% Az Inf. main reinf. T4 [cm²] -10.2301 cm² -0.10% Amin Min. main reinf. T4 [cm²] 0.8694 cm² -0.07% Atz Trans. reinf. T4 [cm²] 3.9008 cm² 0.02% Az Inf. main reinf. T5 [cm²] -14.0512 cm² -0.01% Amin Min. main reinf. T5 [cm²] 4.2 cm² 0.00% Atz Trans. reinf. T5 [cm²] 4.7702 cm² 0.00% Az Inf. main reinf. T6 [cm²] -11.2742 cm² -0.04% Amin Min. main reinf. T6 [cm²] 2.11404 cm² 0.19% Atz Trans. reinf. T6 [cm²] 7.34043 cm² 0.01% Az Inf. main reinf. T7 [cm²] -10.7634 cm² -0.12% Amin Min. main. reinf. T7 [cm²] 1.242 cm² 0.16% Atz Trans. reinf. T7 [cm²] 3.45 cm² 0.00% Az Inf. main reinf. T8 [cm²] -6.47718 cm² -0.27% Amin Min. main reinf. T8 [cm²] 1.3662 cm² -0.28% Atz Trans. reinf. T8 [cm²] 4.44444 cm² 0.10%
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1.85 Design of a Steel Structure according to CM66 (03-0206SSLLG_CM66)
Test ID: 2522
Test status: Passed
1.85.1 Description Verifies the steel calculation results (maximum displacement, normal force, bending moment, deflections, buckling lengths, lateral-torsional buckling and cross section optimization) for a simple metallic framework with a concrete floor, according to CM66.
1.85.2 Background
1.85.2.1 Model description ■ Calculation model: Simple metallic framework with a concrete floor. ■ Load case:
► Permanent loads: 150 kg/m² for the floor and 25kg/m² for the roof. ► Overloads: 250 kg/m² on the floor. ► Wind loads on region II for a normal location ► Snow loads on region 2B at an altitude of 750m.
■ CM66 Combinations
Model preview
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Structure’s load case Code No. Type Title CMP 1 Static SW + Dead loads CMS 2 Static Overloads for usage CMV 3 Static Wind overloads along +X in overpressure CMV 4 Static Wind overloads along +X in depression CMV 5 Static Wind overloads along -X in overpressure CMV 6 Static Wind overloads along -X in depression CMV 7 Static Wind overloads along +Z in overpressure CMV 8 Static Wind overloads along +Z in depression CMV 9 Static Wind overloads along -Z in overpressure CMV 10 Static Wind overloads along -Z in depression CMN 11 Static Normal snow overloads
1.85.2.2 Effel Structure results Displacement Envelope (“CMCD" load combinations)
Envelope of linear element forces D DX DY DZ Env. Case No. Max.
location (cm) (cm) (cm) (cm) Max(D) 213 148 CENTER 12.115 0.037 12.035 -1.393 Min(D) 188 1.1 START 0.000 0.000 0.000 0.000
Max(DX) 204 72.1 START 3.138 3.099 0.434 0.244 Min(DX) 204 313 END 2.872 -1.872 -0.129 -2.174 Max(DY) 213 148 CENTER 12.115 0.037 12.035 -1.393 Min(DY) 213 61.5 END 9.986 -0.118 -9.985 0.046 Max(DZ) 201 371 CENTER 4.149 -0.006 -0.188 4.145 Min(DZ) 203 370 CENTER 4.124 -0.006 -0.240 -4.118
Envelope of forces on linear elements (“CMCFN” load combinations)
Envelope of linear element forces Fx Fy Fz Mx My Mz Env. Case No. MaxSite (T) (T) (T) (T*m) (T*m) (T*m)
Max (Fx) 120 4.1 START 19.423 -4.108 -1.384 -0.003 1.505 7.551 Min (Fx) 138 98 START -41.618 -0.962 -0.192 0.000 0.000 0.000 Max(Fy) 120 57 END -13.473 16.349 -0.016 -0.003 0.002 55.744 Min(Fy) 120 60 START -15.994 -16.112 -0.006 -3E-004 6E-006 53.096 Max(Fz) 177 371 START -3.486 -0.118 2.655 0.000 0.000 0.000 Min(Fz) 187 370 START -3.666 -0.147 -2.658 0.000 0.000 0.000 Max(Mx) 120 111 END 3.933 4.840 0.278 0.028 -4E-005 11.531 Min(Mx) 120 21 END -22.324 13.785 -0.191 -0.028 -0.004 42.562 Max(My) 177 371 CENTER -3.099 -0.118 -0.323 0.000 4.403 -0.500 Min(My) 179 370 CENTER -3.283 -0.155 0.321 0.000 -4.373 -0.660 Max (Mz) 120 57 END -13.473 16.349 -0.016 -0.003 0.002 55.744 Min (Mz) 120 59.2 END -19.455 -8.969 -0.702 -0.003 -0.001 -57.105 Envelope of linear element stresses (“CMCFN” load combinations)
Envelope of linear element stresses sxxMax sxyMax sxzMax sFxx sMxxMax Env. Case No. MaxSite (MPa) (MPa) (MPa) (MPa) (MPa)
Max(sxxMax) 120 59.2 END 273.860 -14.696 -1.024 -16.453 290.312 Min(sxxMax) 120 292 START -150.743 0.000 0.000 -150.743 0.000 Max(sxyMax) 120 57 START 262.954 37.139 -0.030 -15.609 278.562 Min(sxyMax) 120 60 END 241.643 -36.595 -0.011 -18.536 260.179 Max(sxzMax) 185 371 START -2.949 -0.183 3.876 -2.949 0.000 Min(sxzMax) 179 370 START -3.104 -0.255 -3.882 -3.104 0.000 Max(sFxx) 120 293 END 161.095 9E-005 -0.002 161.095 0.000 Min(sFxx) 120 292 START -150.743 0.000 0.000 -150.743 0.000
Max(sMxxMax) 120 59.2 END 273.860 -14.696 -1.024 -16.453 290.312 Min(sMxxMax) 1 1.1 START -4.511 3.155 -0.646 -4.511 0.000
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1.85.2.3 CM66 Effel Expertise results
Hypotheses
For columns
■ Deflections: 1/150 Envelopes deflections calculation.
■ Buckling XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on fixed nodes
■ Lateral-torsional buckling: Ldi automatic calculation: hinged restraint Lds automatic calculation: hinged restraint
For rafters
■ Deflections: 1/200 Envelopes deflections calculation.
■ Buckling: XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on fixed nodes
■ Lateral-torsional buckling: Ldi automatic calculation: no restraint Lds automatic calculation: hinged restraint
For columns
■ Deflections: 1/150 Envelopes deflections calculation.
■ Buckling: XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on displaceable nodes
■ Lateral-torsional buckling: Ldi automatic calculation: hinged restraint Lds automatic calculation: hinged restraint
Optimization parameters
■ Work ratio optimization between 90 and 100% ■ All the sections from the library are available. ■ Labels optimization.
The results of the optimization given below correspond to an iteration of the finite elements calculation.
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Deflection verification
Ratio
Max values on the element
■ Columns: L / 168 ■ Rafter: L / 96 ■ Column: L / 924
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CM Stress diagrams
Work ratio
Stresses
Max values on the element
■ Columns: 375.16 MPa ■ Rafter: 339.79 MPa ■ Column: 180.98 MPa
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Buckling lengths
Lfy
Lfz
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Lateral-torsional buckling lengths
Ldi
Lds
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Optimization
1.85.2.4 Theoretical results
Solver Result name Result description Reference value CM2 D Maximum displacement (CMCD) [cm] 12.115 CM2 Fx Envelope normal force (CMCFN) Min (Fx) [T] -41.618 CM2 Fx Envelope normal force (CMCFN) Max (Fx) [T] 19.423 CM2 My Envelope bending moment (CMCFN) Min (Mz) [Tm] -57.105 CM2 My Envelope bending moment (CMCFN) Max (Mz) [Tm] 55.744
Warning, the Mz bending moment of Effel Structure corresponds to the My bending moment of Advance Design.
Solver Result name Result description Reference value CM2 Deflection CM deflections on Columns [adm] L / 168 (89%) CM2 Deflection CM deflections on Rafters [adm] L / 96 (208%) CM2 Deflection CM deflections on Columns [adm] L / 924 (16%) CM2 Stress CM stresses on Columns [MPa] 374.67 CM2 Stress CM stresses on Rafters [MPa] 339.74 CM2 Stress CM stresses on Columns [MPa] 180.98 CM2 Lfy Buckling lengths on Columns Lfy [m] 8.02 CM2 Lfz Buckling lengths on Columns Lfz [m] 24.07 CM2 Lfy Buckling lengths on Rafters Lfy [m] 1.72 CM2 Lfz Buckling lengths on Rafters Lfz [m] 20.25 CM2C Lfy Buckling lengths on Columns Lfy [m] 4.20 CM2 Lfz Buckling lengths on Columns Lfz [m] 5.67
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Warning, the local axes in Effel Structure are opposite to those in Advance Design.
Solver Result name Result description Reference value CM2 Ldi Lateral-torsional buckling lengths on Columns Ldi [m] 8.5 CM2 Lds Lateral-torsional buckling lengths on Columns Lds [m] 8.5 CM2 Ldi Lateral-torsional buckling lengths on Rafters Ldi [m] 8.61 CM2 Lds Lateral-torsional buckling lengths on Rafters Lds [m] 1.72 CM2 Ldi Lateral-torsional buckling lengths on Columns Ldi [m] 2 CM2 Lds Lateral-torsional buckling lengths on Columns Lds [m] 2
Solver Result name Result description Rate (%) Final section CM2 Work ratio IPE500 columns - section optimization [adm] 1.59 IPE600 CM2 Work ratio IPE400 rafters - section optimization [adm] 1.45 IPE500 CM2 Work ratio IPE400 columns - section optimization [adm] 0.77 IPE360
1.85.3 Calculated results
Result name
Result description Value Error
D Maximum displacement (CMCD) [cm] 12.143 cm 0.23% Fx Envelope normal force (CMCFN) Min (Fx) [T] -41.6548 T -0.09% Fx Envelope normal force (CMCFN) Max (Fx) [T] 19.4828 T 0.31% My Envelope bending moment (CMCFN) Min (Mz)[Tm] -57.1283 T*m -0.04% My Envelope bending moment (CMCFN) Max (Mz) [Tm] 55.7785 T*m 0.06% Deflection CM deflections on Columns [adm] 167.736 Adim. -0.16% Deflection CM deflections on Rafters [adm] 96.1185 Adim. 0.12% Deflection CM deflections on Columns [adm] 924.996 Adim. 0.11% Stress CM stresses on Columns [MPa] 374.568 MPa -0.03% Stress CM stresses on Rafters [MPa] 352.227 MPa 3.68% Stress CM stresses on Columns [MPa] 180.7 MPa -0.15% Lfy Buckling lengths on Columns Lfy [m] 7.96718 m -0.66% Lfz Buckling lengths on Columns Lfz [m] 24.0693 m 0.00% Lfy Buckling lengths on Rafters Lfy [m] 6.63414 m 0.06% Lfz Buckling lengths on Rafters Lfz [m] 20.2452 m -0.02% Lfy Buckling lengths on Columns Lfy [m] 4.19567 m -0.10% Lfz Buckling lengths on Columns Lfz [m] 5.67211 m 0.04% Ldi Lateral-torsional buckling lengths on Columns Ldi [m] 8.5 m 0.00% Lds Lateral-torsional buckling lengths on Columns Lds [m] 8.5 m 0.00% Ldi Lateral-torsional buckling lengths on Rafters Ldi [m] 8.61187 m 0.02% Lds Lateral-torsional buckling lengths on Rafters Lds [m] 8.61187 m 0.02% Ldi Lateral-torsional buckling lengths on Columns Ldi [m] 2 m 0.00% Lds Lateral-torsional buckling lengths on Columns Lds [m] 2 m 0.00% Work ratio IPE500 columns - section optimization [adm] 1.59391 Adim. 0.25% Work ratio IPE400 rafters - section optimization [adm] 1.49884 Adim. 3.37% Work ratio IPE400 columns - section optimization [adm] 0.768938 Adim. -0.14%
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1.86 Linear element in simple bending - without compressed reinforcement (02-0162SSLLB_B91)
Test ID: 2521
Test status: Passed
1.86.1 Description Verifies the reinforcement results for a concrete beam with 8 isostatic spans subjected to uniform loads.
1.86.1.1 Model description ■ Reference: J. Perchat (CHEC) reinforced concrete course ■ Analysis type: static linear; ■ Element type: planar.
Units
■ Forces: kN ■ Moment: kN.m ■ Stresses: MPa ■ Reinforcement density: cm2
Geometry
■ Beam dimensions: 0.2 x 0.5 ht ■ Length: l = 42 m in 7 spans of 6m,
Materials properties
■ Longitudinal elastic modulus: E = 20000 MPa, ■ Poisson's ratio: ν = 0.
Boundary conditions
■ Outer: ► Hinged at end x = 0, ► Vertical support at the same level with all other supports
■ Inner: Hinge z at each beam end (isostatic)
Loading
■ External: ► Case 1 (DL):uniform linear load g = -5 kN/m (on all spans except 8)
► Case 2 to 8 (LL):uniform linear load q = -9 kN/m (on spans 1, 3 to 7)
uniform linear load q = -15 kN/m (on span 2)
► Case 9 (ACC): uniform linear load a = -25 kN/m (on 8th span)
► Case 10 (DL):uniform linear load g = -5 kN/m (on 8th span)
Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107)
Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102)
Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103)
Comb BAELS: 1xDL + 1*LL (comb 108 to 114)
Comb BAELUA: 1xDL + 1xACC (comb 115)
■ Internal: None.
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Reinforced concrete calculation hypothesis:
■ All concrete covers are set to 5 cm ■ BAEL 91 calculation with the revised version 99
Span Concrete Reinforcement Application Concrete Cracking 1 B20 HA fe500 D>24h No Non
prejudicial 2 B35 Adx fe235 1h<D<24h No Non
prejudicial 3 B50 HA fe 400 D<1h Yes Non
prejudicial 4 B25 HA fe500 D>24h Yes Prejudicial 5 B60 HA fe500 D>24h No Very
prejudicial 6 B30 Adx fe235 D>24h Yes Prejudicial 7 B40 HA fe500 D>24h Yes 160 MPa 8 B45 HA fe500 D<1h Yes Non
prejudicial
1.86.1.2 Reinforcement calculation
Reference solution
Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 fc28 20 35 50 25 60 30 40 45 ft28 1.8 2.7 3.6 2.1 4.2 2.4 3 3.3 fe 500 235 400 500 500 235 500 500
teta 1 0.9 0.85 1 1 1 1 0.85 gamb 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.15 gams 1.15 1.15 1.15 1.15 1.15 1.15 1.15 1
h 1.6 1 1.6 1.6 1.6 1 1.6 1.6
fbu 11.33 22.04 33.33 14.17 34.00 17.00 22.67 39.13 fed 434.78 204.35 347.83 434.78 434.78 204.35 434.78 500.00
sigpreju 250.00 156.67 264.00 250.00 285.15 156.67 160.00 252.76 sigtpreju 200.00 125.33 211.20 200.00 228.12 125.33 160.00 202.21
g 5.00 5.00 5.00 5.00 5.00 5.00 5.00 5.00 q 9.00 15.00 9.00 9.00 9.00 9.00 9.00 25.00
pu 20.25 29.25 20.25 20.25 20.25 20.25 20.25 30.00 pser 14.00 20.00 14.00 14.00 14.00 14.00 14.00
l 6.00 6.00 6.00 6.00 6.00 6.00 6.00 6.00 Mu 91.13 131.63 91.13 91.13 91.13 91.13 91.13 135.00
Mser 63.00 90.00 63.00 63.00 63.00 63.00 63.00 Vu 60.75 87.75 60.75 60.75 60.75 60.75 60.75 90.00
Longitudinal reinforcement calculation according to ELU ubu 0.199 0.147 0.068 0.159 0.066 0.132 0.099 0.085 a 0.279 0.200 0.087 0.217 0.086 0.178 0.131 0.111 z 0.400 0.414 0.434 0.411 0.435 0.418 0.426 0.430
Au 5.24 15.56 6.03 5.10 4.82 10.67 4.91 6.28 Main reinforcement calculation with prejudicial cracking according to SLS
A 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 B -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 C -0.56000 -0.89362 -0.87500 D 0.56000 0.89362 0.87500
alpha1 0.367 0.442 0.438 z 0.395 0.384 0.384
Aserp 6.38 10.48 10.25 Main reinforcement calculation with very prejudicial cracking according to SLS
A 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 B -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 C -0.70 -1.60 -0.66 -0.70 -0.61371 -1.12 -0.88 0.00 D 0.70 1.60 0.66 0.70 0.61371 1.12 0.88 0.00
alpha1 0.381 z 0.393
Asertp 7.030
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Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8 Transversal reinforcement calculation
tu 0.68 0.98 0.68 0.68 0.68 0.68 0.68 1.00 k 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00
At/st 0.69 1.79 4.31 3.45 3.45 7.34 3.45 4.44 Recapitulation
Aflex 5.24 15.56 6.03 6.38 7.03 10.67 10.25 6.28 Aminflex 0.75 2.38 1.86 0.87 1.74 2.11 1.24 1.37
At 0.69 1.79 4.31 3.45 3.45 7.34 3.45 4.44 Atmin 1.60 3.40 2.00 1.60 1.60 3.40 1.60 1.60
Finite elements modeling
■ Linear elements: beams with imposed mesh ■ 29 nodes, ■ 28 linear elements.
1.86.1.3 Theoretical results
Reference
Solver Result name Result description Reference value CM2 Az Inf. main reinf. T1 [cm2] 5.24 CM2 Amin Min. main reinf. T1 [cm2] 0.75 CM2 Atz Trans. reinf. T1 [cm2] 0.69 CM2 Az Inf. main reinf. T2 [cm2] 15.56 CM2 Amin Min. main reinf. T2 [cm2] 2.38 CM2 Atz Trans. reinf. T2 [cm2] 1.79 CM2 Az Inf. main reinf. T3 [cm2] 6.03 CM2 Amin Min. main reinf. T3 [cm2] 1.86 CM2 Atz Trans. reinf. T3 [cm2] 4.31 CM2 Az Inf. main reinf. T4 [cm2] 6.38 CM2 Amin Min. main reinf. T4 [cm2] 0.87 CM2 Atz Trans. reinf. T4 [cm2] 3.45 CM2 Az Inf. main reinf. T5 [cm2] 7.03 CM2 Amin Min. main reinf. T5 [cm2] 1.74 CM2 Atz Trans. reinf. T5 [cm2] 3.45 CM2 Az Inf. main reinf. T6 [cm2] 10.67 CM2 Amin Min. main reinf. T6 [cm2] 2.11 CM2 Atz Trans. reinf. T6 [cm2] 7.34 CM2 Az Inf. main reinf. T7 [cm2] 10.25 CM2 Amin Min. main. reinf. T7 [cm2] 1.24 CM2 Atz Trans. reinf. T7 [cm2] 3.45 CM2 Az Inf. main reinf. T8 [cm2] 6.28 CM2 Amin Min. main reinf. T8 [cm2] 1.37 CM2 Atz Trans. reinf. T8 [cm2] 4.44
The "Mu limit" method must be applied to attain the same results.
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1.86.2 Calculated results
Result name
Result description Value Error
Az Inf. main reinf. T1 [cm²] -5.24348 cm² -0.07% Amin Min. main reinf. T1 [cm²] 0.7452 cm² -0.64% Atz Trans. reinf. T1 [cm²] 0.69 cm² 0.00% Az Inf. main reinf. T2 [cm²] -15.5613 cm² -0.01% Amin Min. main reinf. T2 [cm²] 2.3783 cm² -0.07% Atz Trans. reinf. T2 [cm²] 1.79433 cm² 0.24% Az Inf. main reinf. T3 [cm²] -6.03286 cm² -0.05% Amin Min. main reinf. T3 [cm²] 1.863 cm² 0.16% Atz Trans. reinf. T3 [cm²] 4.3125 cm² 0.06% Az Inf. main reinf. T4 [cm²] -6.38336 cm² -0.05% Amin Min. main reinf. T4 [cm²] 0.8694 cm² -0.07% Atz Trans. reinf. T4 [cm²] 3.45 cm² 0.00% Az Inf. main reinf. T5 [cm²] -7.03527 cm² -0.07% Amin Min. main reinf. T5 [cm²] 1.7388 cm² -0.07% Atz Trans. reinf. T5 [cm²] 3.45 cm² 0.00% Az Inf. main reinf. T6 [cm²] -10.6698 cm² 0.00% Amin Min. main reinf. T6 [cm²] 2.11404 cm² 0.19% Atz Trans. reinf. T6 [cm²] 7.34043 cm² 0.01% Az Inf. main reinf. T7 [cm²] -10.2733 cm² -0.23% Amin Min. main. reinf. T7 [cm²] 1.242 cm² 0.16% Atz Trans. reinf. T7 [cm²] 3.45 cm² 0.00% Az Inf. main reinf. T8 [cm²] -6.29338 cm² -0.21% Amin Min. main reinf. T8 [cm²] 1.3662 cm² -0.28% Atz Trans. reinf. T8 [cm²] 4.44444 cm² 0.10%
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1.87 Double cross with hinged ends (01-0097SDLLB_FEM)
Test ID: 2511
Test status: Passed
1.87.1 Description Double cross with hinged ends.
Tested functions: Eigen frequencies, crossed beams, in plane bending.
1.87.2 Background ■ Reference: NAFEMS, FV2 test ■ Analysis type: modal analysis; ■ Tested functions: Eigen frequencies, Crossed beams, In plane bending.
1.87.2.1 Problem data
Units
I. S.
Geometry
Full square section:
■ Arm length: L = 5 m ■ Dimensions: a x b = 0.125 x 0.125 ■ Area: A = 1.563 10-2 m2 ■ Inertia: IP = 3.433 x 10-5m4
Iy = Iz = 2.035 x 10-5m4
Materials properties
■ Longitudinal elastic modulus: E = 2 x 1011 Pa, ■ Density: ρ = 8000 kg/m3
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Boundary conditions
■ Outer: A, B, C, D, E, F, G, H points restraint along x and y; ■ Inner: None.
Loading
None for the modal analysis
1.87.2.2 Reference frequencies Mode Units Reference
1 Hz 11.336 2,3 Hz 17.709
4 to 8 Hz 17.709 9 Hz 45.345
10,11 Hz 57.390 12 to 16 Hz 57.390
Finite elements modeling
■ Linear elements type: Beam ■ Imposed mesh: 4 Elements / Arms
Modal deformations
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1.87.2.3 Theoretical results
Solver Result name Result description Reference value CM2 Frequency Frequency of Eigen Mode 1 [Hz] 11.336 CM2 Frequency Frequency of Eigen Mode 2 [Hz] 17.709 CM2 Frequency Frequency of Eigen Mode 3 [Hz] 17.709 CM2 Frequency Frequency of Eigen Mode 4 [Hz] 17.709 CM2 Frequency Frequency of Eigen Mode 5 [Hz] 17.709 CM2 Frequency Frequency of Eigen Mode 6 [Hz] 17.709 CM2 Frequency Frequency of Eigen Mode 7 [Hz] 17.709 CM2 Frequency Frequency of Eigen Mode 8 [Hz] 17.709 CM2 Frequency Frequency of Eigen Mode 9 [Hz] 45.345 CM2 Frequency Frequency of Eigen Mode 10 [Hz] 57.390 CM2 Frequency Frequency of Eigen Mode 11 [Hz] 57.390 CM2 Frequency Frequency of Eigen Mode 12 [Hz] 57.390 CM2 Frequency Frequency of Eigen Mode 13 [Hz] 57.390 CM2 Frequency Frequency of Eigen Mode 14 [Hz] 57.390 CM2 Frequency Frequency of Eigen Mode 15 [Hz] 57.390 CM2 Frequency Frequency of Eigen Mode 16 [Hz] 57.390
1.87.3 Calculated results
Result name
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%
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1.88 Beam on 3 supports with T/C (k -> infinite) (01-0101SSNLB_FEM)
Test ID: 2515
Test status: Passed
1.88.1 Description Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k -> infinite).
1.88.2 Background
1.88.2.1 Model description ■ Reference: internal GRAITEC test; ■ Analysis type: static non linear; ■ Element type: linear, T/C.
Units
I. S.
Geometry
■ L = 10 m ■ 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 stiffness ky → ∞ (1.1030N/m),
■ Inner: None.
Loading
■ External: Vertical punctual load P = -100 N at x = 5 m, ■ Internal: None.
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1.88.2.2 References solutions ky being infinite, the non linear model behaves the same way as a beam on 3 supports.
Displacements
( )( )
( )( )
( )( )
( ) rad 000038.0Lk2EI3EI32
LkEI6PL
0Lk2EI316
PL3v
rad 000077.0Lk2EI3EI16
LkEI3PL
rad 000115.0Lk2EI3EI32
LkEI2PL3
3yzz
3yz
2
3
3yz
3
3
3yzz
3yz
2
2
3yzz
3yz
2
1
−=+
+−=β
=+
−=
=+
+−=β
−=+
+=β
Mz Moments
( )( )
N.m 13.2032
MM4
PL)m5x(M
N.m 75.93Lk2EI316
PLk3M
0M
1z2zz
3yz
4y
2z
1z
−=−
+==
−=+
=
=
Finite elements modeling
■ Linear element: S beam, automatic mesh, ■ 3 nodes, ■ 2 linear elements + 1 T/C.
Deformed shape
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Moment diagram
1.88.2.3 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
1.88.3 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] -7.65875e-005 Rad 0.54% DZ Displacement - node 3 [m] 9.36486e-030 m 0.00% RY Rotation Ry in node 3 [rad] 3.81695e-005 Rad 0.45% My Moment M - node 1 [Nm] 1.3145e-013 N*m 0.00% My Moment M - node 2 [Nm] 93.6486 N*m -0.11% My Moment M - middle span 1 [Nm] -203.176 N*m -0.02%
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1.89 Study of a mast subjected to an earthquake (02-0112SMLLB_P92)
Test ID: 2519
Test status: Passed
1.89.1 Description A structure consisting of 2 beams and 2 punctual masses, subjected to a lateral earthquake along X. The frequency modes, the eigen vectors, the participation factors, the displacement at the top of the mast and the forces at the top of the mast are verified.
1.89.1.1 Model description ■ Reference: internal GRAITEC test; ■ Analysis type: modal and spectral analyses; ■ Element type: linear, mass.
1.89.1.2 Material strength model
Units
I. S.
Geometry
■ Length: L = 35 m, ■ Outer radius: Rext = 3.00 m ■ Inner radius: Rint = 2.80 m
■ Axial section: S= 3.644 m2 ■ Polar inertia: Ip = 30.68 m4 ■ Bending inertias: Ix =15.34 m4
Iy = 15.34 m4
Masses
■ M1 =203873.6 kg ■ M2 =101936.8 kg
Materials properties
■ Longitudinal elastic modulus: E = 1.962 x 1010 N/m2, ■ Poisson's ratio: ν = 0.1, ■ Density: ρ = 25 kN/m3
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Boundary conditions
■ Outer: Fixed in X = 0, Y = 0 m,
Loading
■ External: Seismic excitation on X direction
Finite elements modeling
Linear element: beam, automatic mesh,
1.89.1.3 Seismic hypothesis in conformity with PS92 regulation ■ Zone: Nice Sophia Antipolis (Zone II). ■ Site: S1 (Medium soil, 10m thickness). ■ Construction type class: B ■ Behavior coefficient: 3 ■ Material damping: 4% (Reinforced concrete).
1.89.1.4 Modal analysis
Eigen periods reference solution
Substract the value of structure’s specific horizontal periods by solving the following equation:
( ) 0MKdet 2 =ω−
⎟⎟⎠
⎞⎜⎜⎝
⎛≡
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−≡
2
1
3
M00M
M
25516
L7EI48K
Eigen modes Units Reference 1 Hz 2.085 2 Hz 10.742
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Modal vectors
For ω1:
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛χ⇒=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
ωω
−⎟⎟⎠
⎞⎜⎜⎝
⎛−
−055.31
UU
0UU
M00M
25516
L7EI48
2
11
2
1
12
2
211
3
For ω2:
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=⎟⎟⎠
⎞⎜⎜⎝
⎛χ
655.01
UU
2
12
Normalizing relative to the mass
⎟⎟⎠
⎞⎜⎜⎝
⎛
××
χ−
−
3
4
1 10842.210305.9
; ⎟⎟⎠
⎞⎜⎜⎝
⎛
×−×
χ−
−
3
3
2 10316.11001.2
Modal deformations
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1.89.1.5 Spectral study
Design spectrum
Nominal acceleration:
2n1 sm5411.5a 2.085Hzf =⇒=
2n2 sm25.6aHz742.01f =⇒=
Observation: the gap between pulses is greater than 10%, so the modal responses can be regarded as independent.
Reference participation factors
Δχ=γ Mii
earthquake direction vector
Eigen modes Reference 1 479.427 2 275.609
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Pseudo-acceleration
iiii a χ×γ×ξ×=Γ in (m/s2)
4.0%5
⎟⎟⎠
⎞⎜⎜⎝
⎛η
=ξ : Damping correction factor.
η: Structure damping.
⎟⎟⎠
⎞⎜⎜⎝
⎛=Γ
8.25562.7026
1 ⎟⎟⎠
⎞⎜⎜⎝
⎛=Γ
2.4783-3.7852
1
Reference modal displacement
⎟⎟⎠
⎞⎜⎜⎝
⎛−−
=ψ024.814E021.576E
1 ⎟⎟⎠
⎞⎜⎜⎝
⎛−
−=ψ
045.446E-048.318E
2
Equivalent static forces
⎟⎟⎠
⎞⎜⎜⎝
⎛++
=058.415E055.510E
F1 ⎟⎟⎠
⎞⎜⎜⎝
⎛+
+=
052.526E-057.717E
F2
Displacement at the top of the mast
( ) ( )( )221 04E446.502E81.4U −−+−=
Units Reference m 4.814 E-02
Shear force at the top of the mast
( ) ( )( )3
05E526.205E415.8T
22
1+−++
=
3: Being the behavior coefficient of forces
Units Reference N 2.929 E+05
Moment at the base
Units Reference N.m 1.578 E+07
1.89.1.6 Theoretical results
Reference
Solver Result name Result description Reference value CM2 Frequency Frequency Mode 1 [Hz] 2.085 CM2 Frequency Frequency Mode 2 [Hz] 10.742 CM2 D Displacement at the top of the mast [cm] 4.814 CM2 Fz Forces at the top of the mast [N] 2.929E+05
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1.89.2 Calculated results
Result name Result description Value Error Frequency Mode 1 [Hz] 2.08 Hz -0.24% Frequency Mode 2 [Hz] 10.74 Hz -0.02% D Displacement at the top of the mast [cm] 4.81159 cm -0.05% Fz Forces at the top of the mast [N] 292677 N -0.08%
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1.90 Design of a concrete floor with an opening (03-0208SSLLG_BAEL91)
Test ID: 2524
Test status: Passed
1.90.1 Description Verifies the displacements, bending moments and reinforcement results for a 2D concrete slab with supports and punctual loads.
1.90.2 Background
1.90.2.1 Model description ■ Calculation model: 2D concrete slab.
► Slab thickness: 20 cm ► Slab length: 20m ► Slab width: 10m ► The supports (punctual and linear) are considered as hinged. ► Supports positioning (see scheme below) ► 1,50m*2,50m opening => see positioning on the following scheme
■ Materials: ► Concrete B25 ► Young module: E= 36000 MPa
■ Load case: ► Permanent loads: 100 kg/m2 ► Permanent loads: 200 kg/ml around the opening ► Punctual loads of 2T in permanent loads (see the following definition) ► Usage overloads: 250 kg/m2
■ Mesh density: 0.5 m
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Slab geometry
Support positions
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Positions of punctual loads
Global loading overview
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Load Combinations
Code Numbers Type Title BAGMAX 1 Static Permanent loads + self weight
BAQ 2 Static Usage overloads BAELS 101 Comb_Lin Gmax+Q BAELU 102 Comb_Lin 1.35Gmax+1.5Q
1.90.2.2 Effel Structure Results
SLS max displacements (load combination 101)
Mx bending moment for ULS load combination
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My bending moment for ULS load combination
Mxy bending moment for ULS load combination
1.90.2.3 Effel RC Expert Results
Main hypothesis
■ Top and bottom concrete covers: 3 cm ■ Slightly dangerous cracking ■ Concrete B25 => Fc28= 25 MPa ■ Reinforcement calculation according to Wood method. ■ Calculation starting from non averaged forces.
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Axi reinforcements
Ayi reinforcements
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Axs reinforcements
Ays reinforcements
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1.90.2.4 Theoretical results
Solver
Result name
Result description Reference value
CM2 D Max displacement for SLS (load combination 101) [cm] 0.176 CM2 Myy Mx and My bending moments for ULS (load combination 102) Max(Mx) [kN.m] 25.20 CM2 Myy Mx and My bending moments for ULS (load combination 102) Min(Mx) [kN.m] -15.71 CM2 Mxx Mx and My bending moments for ULS (load combination 102) Max(My) [kN.m] 31.17 CM2 Mxx Mx and My bending moments for ULS (load combination 102) Min(My) [kN.m] -18.79 CM2 Mxy Mx and My bending moments for ULS (load combination 102) Max (Mxy) [kN.m] 10.26 CM2 Mxy Mx and My bending moments for ULS (load combination 102) Min (Mxy) [kN.m] -10.14 CM2 Axi Theoretic reinforcements Axi [cm2] 3.84 CM2 Axs Theoretic reinforcements Axs [cm2] 3.55 CM2 Ayi Theoretic reinforcements Ayi [cm2] 3.75 CM2 Ays Theoretic reinforcements Ays [cm2] 4.53
These values are obtained from the maximum values from the mesh.
1.90.3 Calculated results
Result name
Result description Value Error
D Max displacement for SLS (load combination 101) [cm] 0.174641 cm -0.77% Myy Mx and My bending moments for ULS (load combination
102) Max(Mx) [kNm] 25.2594 kN*m
0.24%
Myy Mx and My bending moments for ULS (load combination 102) Min(Mx) [kNm]
-15.6835 kN*m
0.17%
Mxx Mx and My bending moments for ULS (load combination 102) Max(My) [kNm]
31.2449 kN*m
0.24%
Mxx Mx and My bending moments for ULS (load combination 102) Min(My) [kNm]
-18.7726 kN*m
0.09%
Mxy Mx and My bending moments for ULS (load combination 102) Max (Mxy) [kNm]
10.1558 kN*m
-1.02%
Mxy Mx and My bending moments for ULS (load combination 102) Min (Mxy) [kNm]
-10.2508 kN*m
-1.09%
Axi Theoretic reinforcements Axi [cm2] 3.83063 cm² -0.24% Axs Theoretic reinforcementsAxs [cm2] 3.629 cm² 2.23% Ayi Theoretic reinforcements Ayi [cm2] 3.72879 cm² -0.57% Ays Theoretic reinforcements Ays [cm2] 4.61909 cm² 1.97%
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1.91 Design of a 2D portal frame (03-0207SSLLG_CM66)
Test ID: 2523
Test status: Passed
1.91.1 Description Verifies the steel calculation results (displacement at ridge, normal forces, bending moments, deflections, stresses, buckling lengths, lateral torsional buckling lengths and cross section optimization) for a 2D metallic portal frame, according to CM66.
1.91.2 Background
1.91.2.1 Model description ■ Calculation model: 2D metallic portal frame.
► Column section: IPE500 ► Rafter section: IPE400 ► Base plates: hinged. ► Portal frame width: 20m ► Columns height: 6m ► Portal frame height at the ridge: 7.5m
■ Load case: ► Permanent loads: 150 kg/m on the roof + elements self weight. ► Usage overloads: 800 kg/ml on the roof
■ Mesh density: 1m
Model preview
Combinations
Code Numbers Type Title CMP 1 Static Permanent load + self weight CMS 2 Static Usage overloads
CMCFN 101 Comb_Lin 1.333P CMCFN 102 Comb_Lin 1.333P+1.5S CMCFN 103 Comb_Lin P+1.5S CMCD 104 Comb_Lin P+S
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1.91.2.2 Effel Structure Results
Ridge displacements (combination 104)
Diagram of normal force envelope
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Envelope of bending moments diagram
1.91.2.3 Effel Expert CM results
Main hypotheses
For columns
■ Deflections: 1/150 Envelopes deflections calculation.
■ Buckling: XY plane: Automatic calculation of the structure on fixed nodes XZ plane: Automatic calculation of the structure on fixed nodes
Ka-Kb Method
■ Lateral-torsional buckling: Ldi automatic calculation: no restraints Lds imposed value: 2 m
For the rafters
■ Deflections: 1/200 Envelopes deflections calculation.
■ Buckling: XY plane: Automatic calculation of the structure on fixed nodes XZ plane: Automatic calculation of the structure on fixed nodes
Ka-Kb Method
■ Lateral-torsional buckling: Ldi automatic calculation: No restraints Lds imposed value: 1.5m
Optimization criteria
■ Work ratio optimization between 90 and 100% ■ Labels optimization (on Advance Design templates)
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Deflection verification
Ratio
CM Stress diagrams
Work ratio
Stresses
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Buckling lengths
Lfy
Lfz
Lateral-torsional buckling lengths
Ldi
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Lds
Optimization
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1.91.2.4 Theoretical results
Solver Result name Result description Reference value CM2 D Displacement at the ridge [cm] 9.36 CM2 Fx Envelope normal forces on Columns (min) [T] -15.77 CM2 Fx Envelope normal forces on Rafters (max) [T] -1.02 CM2 My Envelope bending moments on Columns (min) [T.m] -42.41 CM2 My Envelope bending moments on Rafters (max) [T.m] 42.41 CM2 Deflection CM deflections on Columns [%] L / 438 (34%) CM2 Deflection CM deflections on Rafters [%] L / 111 (180%) CM2 Stress CM stresses on Columns [MPa] 230.34 CM2 Stress CM stresses on Rafters [MPa] 458.38 CM2 Lfy Buckling lengths on Columns - Lfy [m] 5.84 CM2 Lfz Buckling lengths on Columns - Lfz [m] 6 CM2 Lfy Buckling lengths on Rafters - Lfy [m] 7.08 CM2 Lfz Buckling lengths on Rafters - Lfz [m] 10.11
Warning, the local axes in Effel Structure have different orientation in Advance Design.
Solver Result name Result description Reference value CM2 Ldi Lateral-torsional buckling lengths on Columns Ldi [m] 6 CM2 Lds Lateral-torsional buckling lengths on Columns Lds [m] 2 CM2 Ldi Lateral-torsional buckling lengths on Rafters Ldi [m] 10.11 CM2 Lds Lateral-torsional buckling lengths on Rafters Lds [m] 1.5
Solver Result name Result description Rate (%) Final section CM2 Work ratio IPE500 columns - section optimization 98 IPE500 CM2 Work ratio IPE400 rafters - section optimization 195 IPE550
1.91.3 Calculated results
Result name Result description Value Error D Displacement at the ridge [cm] 9.36473 cm 0.05% Fx Envelope normal forces on Columns (min) [T] -15.7798 T -0.06% Fx Envelope normal forces on Rafters (max) [T] -1.0161 T 0.38% My Envelope bending moments on Columns (min) [Tm] 42.4226 T*m 0.03% My Envelope bending moments on Rafters (max) [Tm] 42.4226 T*m 0.03% Deflection CM deflections on Columns [adm] 438.077 Adim. 0.02% Deflection CM deflections on Rafters [adm] 111.325 Adim. 0.29% Stress CM stresses on Columns [adm] 230.331 MPa 0.00% Stress CM stresses on Rafters [adm] 458.367 MPa 0.00% Lfy Buckling lengths on Columns - Lfy [m] 6 m 0.00% Lfz Buckling lengths on Columns - Lfz [m] 5.84401 m 0.07% Lfy Buckling lengths on Rafters - Lfy [m] 10.1119 m 0.02% Lfz Buckling lengths on Rafters - Lfz [m] 7.07904 m -0.01% Ldi Lateral-torsional buckling lengths on Columns Ldi [m] 6 m 0.00% Lds Lateral-torsional buckling lengths on Columns Lds [m] 2 m 0.00% Ldi Lateral-torsional buckling lengths on Rafters Ldi [m] 10.1119 m 0.02% Lds Lateral-torsional buckling lengths on Rafters Lds [m] 1.5 m 0.00% Work ratio IPE500 columns - section optimization [adm] 0.980131 Adim. 0.01% Work ratio IPE400 rafters - section optimization [adm] 1.9505 Adim. 0.03%
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1.92 Cantilever rectangular plate (01-0001SSLSB_FEM)
Test ID: 2433
Test status: Passed
1.92.1 Description Verifies the vertical displacement on the free extremity of a cantilever rectangular plate fixed on one side. The plate is 1 m long, subjected to a uniform planar load.
1.92.2 Background
1.92.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 01/89. ■ Analysis type: linear static. ■ Element type: planar.
Cantilever rectangular plate Scale =1/4 01-0001SSLSB_FEM
Units
S.I.
Geometry
■ Thickness: e = 0.005 m, ■ Length: l = 1 m, ■ Width: b = 0.1 m.
Materials properties
■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.
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Boundary conditions
■ Outer: Fixed at end x = 0, ■ Inner: None.
Loadings
■ External: Uniform load p = -1700 Pa on the upper surface, ■ Internal: None.
1.92.2.2 Displacement of the model in the linear elastic range
Reference solution
The reference displacement is calculated for the unsupported end located at x = 1m.
u = bl4p8EIz =
0.1 x 14 x 1700
8 x 2.1 x 1011 x 0.1 x 0.0053
12 = -9.71 cm
Finite elements modeling
■ Planar element: plate, imposed mesh, ■ 1100 nodes, ■ 990 surface quadrangles.
Deformed shape
Deformed cantilever rectangular plate Scale =1/4 01-0001SSLSB_FEM
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1.92.2.3 Theoretical results
Solver Result name Result description Reference value CM2 DZ Vertical displacement on the free extremity [cm] -9.71
1.92.3 Calculated results
Result name
Result description Value Error
DZ Vertical displacement on the free extremity [cm] -9.58696 cm 1.27%
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1.93 Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontal force (TTAD #13175)
Test ID: 5088
Test status: Passed
1.93.1 Description Calculates torsors using different mesh sizes for a concrete wall subjected to a horizontal force.
1.94 Verifying torsors on a single story coupled walls subjected to horizontal forces
Test ID: 4804
Test status: Passed
1.94.1 Description Verifies torsors on a single story coupled walls subjected to horizontal forces
1.95 Verifying diagrams for Mf Torsors on divided walls (TTAD #11557)
Test ID: 3610
Test status: Passed
1.95.1 Description Performs the finite elements calculation and verifies the results diagrams for Mf torsors on a high wall divided in 6 walls (by height).
The loads applied on the model: self weight, two live load cases and seism loads according to Eurocodes 8.
1.96 Verifying the level mass center (TTAD #11573, TTAD #12315)
Test ID: 3609
Test status: Passed
1.96.1 Description Performs the finite elements calculation on a model with two planar concrete elements with a linear support. Verifies the level mass center and generates the "Excited total masses" and "Level modal mass and rigidity centers" reports.
The model consists of two planar concrete elements with a linear fixed support. The loads applied on the model: self weight, a planar live load of -1 kN and seism loads according to French standards of Eurocodes 8.
1.97 Generating results for Torsors NZ/Group (TTAD #11633)
Test ID: 3594
Test status: Passed
1.97.1 Description Performs the finite elements calculation on a complex concrete structure with four levels. Generates results for Torsors NZ/Group. Verifies the legend results.
The structure has 88 linear elements, 30 planar elements, 48 windwalls, etc.
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1.98 Verifying Sxx results on beams (TTAD #11599)
Test ID: 3595
Test status: Passed
1.98.1 Description Performs the finite elements calculation on a complex model with concrete, steel and timber elements. Verifies the Sxx results on beams. Generates the maximum stresses report.
The structure has 40 timber linear elements, 24 concrete linear elements, 143 steel elements. The loads applied on the structure: dead loads, live loads, snow loads, wind loads and temperature loads (according to Eurocodes).
1.99 Verifying forces results on concrete linear elements (TTAD #11647)
Test ID: 3551
Test status: Passed
1.99.1 Description Verifies forces results on concrete beams consisting of a linear element and on beams consisting of two linear elements. Generates the linear elements forces by load case report.
1.100 Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD #11854)
Test ID: 3539
Test status: Passed
1.100.1Description Verifies the results diagrams display after changing the view from standard (top, left,...) to user view.
1.101 Verifying stresses in beam with "extend into wall" property (TTAD #11680)
Test ID: 3491
Test status: Passed
1.101.1Description Verifies the results on two concrete beams which have the "Extend into the wall" option enabled. One of the beams is connected to 2 walls on both sides and one with a wall and a pole. Generates the linear elements forces by elements report.
1.102 Verifying constraints for triangular mesh on planar elements (TTAD #11447)
Test ID: 3460
Test status: Passed
1.102.1Description Performs the finite elements calculation, verifies the stresses for triangular mesh on a planar element and generates a report for planar elements stresses in neutral fiber.
The planar element is 20 cm thick, C20/25 material with a linear rigid support. A linear load of 30.00 kN is applied on FX direction.
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1.103 Verifying the displacement results on linear elements for vertical seism (TTAD #11756)
Test ID: 3442
Test status: Passed
1.103.1Description Verifies the displacements results on an inclined steel bar for vertical seism according to Eurocodes 8 localization and generates the corresponding report.
The steel bar has a rigid support and IPE100 cross section and is subjected to self weight and seism load on Z direction (vertical).
1.104 Verifying forces for triangular meshing on planar element (TTAD #11723)
Test ID: 3463
Test status: Passed
1.104.1Description Performs the finite elements calculation, verifies the forces for triangular meshing on a planar element and generates a report for planar elements forces by load case.
The planar element is a square shell (5 m) with a thickness of 20 cm, C20/25 material with a linear rigid support. A linear load of -10.00 kN is applied on FZ direction.
1.105 Generating planar efforts before and after selecting a saved view (TTAD #11849)
Test ID: 3454
Test status: Passed
1.105.1Description Generates efforts for all planar elements before and after selecting the third saved view.
1.106 Verifying tension/compression supports on nonlinear analysis (TTAD #11518)
Test ID: 4198
Test status: Passed
1.106.1Description Verifies the behavior of supports with several rigidities fields defined.
Performs the finite elements calculation and generates the "Displacements of linear elements by element" report.
The model consists of a vertical linear element (concrete B20, R20*30 cross section) with a rigid punctual support at the base and a T/C punctual support at the top. A value of 15000.00 kN/m is defined for the KTX and KTZ stiffeners of the T/C support. Two loads of 500.00 kN are applied.
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1.107 Verifying tension/compression supports on nonlinear analysis (TTAD #11518)
Test ID: 4197
Test status: Passed
1.107.1Description Verifies the supports behavior when the rigidity has a high value.
Performs the finite elements calculation and generates the "Displacements of linear elements by element" report.
The model consists of a vertical linear element (concrete B20, R20*30 cross section) with a rigid punctual support at the base and a T/C punctual support at the top. A large value of the KTX stiffener of the T/C support is defined. Two loads of 500.00 kN are applied.
1.108 Verifying the display of the forces results on planar supports (TTAD #11728)
Test ID: 4375
Test status: Passed
1.108.1Description Performs the finite elements calculation and verifies the display of the forces results on a planar support. The model consists of a concrete vertical element with a planar support.
1.109 Verifying results on punctual supports (TTAD #11489)
Test ID: 3693
Test status: Passed
1.109.1Description Performs the finite elements calculation and generates the punctual supports report, containing the following tables: "Displacements of point supports by load case", "Displacements of point supports by element", "Point support actions by load case", "Point support actions by element" and "Sum of actions on supports and nodes restraints".
The structure consists of concrete, steel and timber linear elements with punctual supports.
1.110 Generating a report with torsors per level (TTAD #11421)
Test ID: 3774
Test status: Passed
1.110.1Description Generates a report with the torsors per level results.
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1.111 Verifying nonlinear analysis results for frames with semi-rigid joints and rigid joints (TTAD #11495)
Test ID: 3795
Test status: Passed
1.111.1Description Verifies the nonlinear analysis results for two frames with one level. One of the frames has semi-rigid joints and the other has rigid joints.
1.112 Verifying forces on a linear elastic support which is defined in a user workplane (TTAD #11929)
Test ID: 4553
Test status: Passed
1.112.1Description Verifies forces on a linear elastic support, which is defined in a user workplane, and generates a report with forces for linear support in global and local workplane.
1.113 Verifying the internal forces results for a simple supported steel beam
Test ID: 4533
Test status: Passed
1.113.1Description Performs the finite elements calculation for a horizontal element (S235 material and IPE180 cross section) with two hinge rigid supports at each end. One of the supports has translation restraints on X, Y and Z, the other support has restraints on Y and Z.
Verifies the internal forces My, Fz.
Validated according to:
Example: 3.1 - Simple beam bending without the stability loss
Publication: Steel structures members - Examples according to Eurocodes
By: F. Wald a kol.
1.114 Verifying the main axes results on a planar element (TTAD #11725)
Test ID: 4310
Test status: Passed
1.114.1Description Verifies the main axes results on a planar element.
Performs the finite elements calculation for a concrete wall (20 cm thick) with a linear support. Displays the forces results on the planar element main axes.
2 CAD, Rendering and Visualization
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2.1 Verifying annotation on selection (TTAD #12700)
Test ID: 4575
Test status: Passed
2.1.1 Description Verifying annotation on selection (TTAD #12700)
2.2 Verifying rotation for steel beam with joint (TTAD #12592)
Test ID: 4560
Test status: Passed
2.2.1 Description Verifying rotation for steel beam with joint at one end (TTAD #12592)
2.3 Verifying hide/show elements command (TTAD #11753)
Test ID: 3443
Test status: Passed
2.3.1 Description Verifies the hide/show elements command for the whole structure using the right-click option.
2.4 System stability during section cut results verification (TTAD #11752)
Test ID: 3457
Test status: Passed
2.4.1 Description Performs the finite elements calculation and verifies the section cut results on a concrete planar element with an opening.
2.5 Verifying the grid text position (TTAD #11704)
Test ID: 3464
Test status: Passed
2.5.1 Description Verifies the grid text position from different views.
2.6 Verifying the grid text position (TTAD #11657)
Test ID: 3465
Test status: Passed
2.6.1 Description Verifies the grid text position from different views.
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2.7 Generating combinations (TTAD #11721)
Test ID: 3468
Test status: Passed
2.7.1 Description Generates combinations for three types of loads: live loads, dead loads and snow (with an altitude > 1000 m and base effect); generates the combinations description report.
2.8 Verifying the coordinates system symbol (TTAD #11611)
Test ID: 3550
Test status: Passed
2.8.1 Description Verifies the coordinates system symbol display from different views.
2.9 Verifying descriptive actors after creating analysis (TTAD #11589)
Test ID: 3579
Test status: Passed
2.9.1 Description Generates the finite elements calculation on a complex concrete structure (C35/45 material). Verifies the descriptive actors after creating the analysis.
The structure consists of 42 linear elements, 303 planar elements, 202 supports, etc. 370 planar loads are applied: live loads, dead loads and temperature.
2.10 Creating a circle (TTAD #11525)
Test ID: 3607
Test status: Passed
2.10.1 Description Creates a circle.
2.11 Creating a camera (TTAD #11526)
Test ID: 3608
Test status: Passed
2.11.1 Description Verifies the camera creation and visibility.
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2.12 Verifying the local axes of a section cut (TTAD #11681)
Test ID: 3637
Test status: Passed
2.12.1 Description Changes the local axes of a section cut in the descriptive model and verifies if the local axes are kept in analysis model.
2.13 Verifying the snap points behavior during modeling (TTAD #11458)
Test ID: 3644
Test status: Passed
2.13.1 Description Verifies the snap points behavior when the "Allowed deformation" function is enabled (stretch points) and when it is disabled (grip points).
2.14 Verifying the representation of elements with HEA cross section (TTAD #11328)
Test ID: 3701
Test status: Passed
2.14.1 Description Verifies the representation of elements with HEA340 cross section.
2.15 Verifying the descriptive model display after post processing results in analysis mode (TTAD #11475)
Test ID: 3733
Test status: Passed
2.15.1 Description Performs the finite elements calculation and displays the forces results on linear elements. Returns to the model mode to verify the descriptive model display.
2.16 Verifying holes in horizontal planar elements after changing the level height (TTAD #11490)
Test ID: 3740
Test status: Passed
2.16.1 Description Verifies holes in horizontal planar elements after changing the level height.
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2.17 Verifying the display of elements with compound cross sections (TTAD #11486)
Test ID: 3742
Test status: Passed
2.17.1 Description Creates an element with compound cross section (CS1 IPE400 IPE240) and verifies the cross section display.
2.18 Modeling using the tracking snap mode (TTAD #10979)
Test ID: 3745
Test status: Passed
2.18.1 Description Enables the "tracking" snap mode to model structure elements.
2.19 Turning on/off the "ghost" rendering mode (TTAD #11999)
Test ID: 4304
Test status: Passed
2.19.1 Description Verifies the on/off function for the "ghost" rendering mode when the workplane display is disabled.
2.20 Moving a linear element along with the support (TTAD #12110)
Test ID: 4302
Test status: Passed
2.20.1 Description Moves a linear element along with the element support, after selecting both elements.
2.21 Verifying the "ghost" display after changing the display colors (TTAD #12064)
Test ID: 4349
Test status: Passed
2.21.1 Description Verifies the "ghost" display on selected elements after changing the element display color.
2.22 Verifying the "ghost display on selection" function for saved views (TTAD #12054)
Test ID: 4347
Test status: Passed
2.22.1 Description Verifies the display of saved views which contain elements with the "ghost on selection" function enabled.
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2.23 Verifying the steel connections modeling (TTAD #11698)
Test ID: 4440
Test status: Passed
2.23.1 Description Verifies the modeling of steel connections.
2.24 Verifying the fixed load scale function (TTAD #12183).
Test ID: 4429
Test status: Passed
2.24.1 Description Verifies the "fixed load scale" function.
2.25 Verifying the dividing of planar elements which contain openings (TTAD #12229)
Test ID: 4483
Test status: Passed
2.25.1 Description Verifies the dividing of planar elements which contain openings.
2.26 Verifying the program behavior when trying to create lintel (TTAD #12062)
Test ID: 4507
Test status: Passed
2.26.1 Description Verifies the program behavior when trying to create lintel on a planar element with an inappropriate opening.
2.27 Verifying the program behavior when launching the analysis on a model with overlapped loads (TTAD #11837)
Test ID: 4511
Test status: Passed
2.27.1 Description Verifies the program behavior when launching the analysis on a model that had overlapped loads.
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2.28 Verifying the display of punctual loads after changing the load case number (TTAD #11958)
Test ID: 4508
Test status: Passed
2.28.1 Description Creates a punctual load and verifies the display of the load after placing it in another load case using the load case number from the properties window.
2.29 Verifying the display of a beam with haunches (TTAD #12299)
Test ID: 4513
Test status: Passed
2.29.1 Description Verifies the display of a beam with haunches, in the "Linear contour" rendering mode.
2.30 Creating base plate connections for non-vertical columns (TTAD #12170)
Test ID: 4534
Test status: Passed
2.30.1 Description Creates a base plate connection on a non-vertical column.
2.31 Verifying drawing of joints in y-z plan (TTAD #12453)
Test ID: 4551
Test status: Passed
2.31.1 Description Verifying drawing of joints in y-z plan (TTAD #12453)
3 Climatic Generator
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3.1 EC1: generating snow loads on a 3 slopes 3D portal frame with parapets (NF EN 1991-1-3/NA) (TTAD #11111)
Test ID: 4546
Test status: Passed
3.1.1 Description Generates snow loads on a 3 slopes 3D portal frame with parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA)
3.2 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.3 - Wind - Example C)
Test ID: 4523
Test status: Passed
3.2.1 Description Generates wind loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).
3.3 EC1: generating wind loads on a 3D portal frame with one slope roof (NF EN 1991-1-4/NA) (VT : 3.2 - Wind - Example B)
Test ID: 4521
Test status: Passed
3.3.1 Description Generates wind loads on a 3D portal frame with one slope roof, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).
3.4 EC1: generating wind loads on a triangular based lattice structure with compound profiles and automatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12276)
Test ID: 4525
Test status: Passed
3.4.1 Description Generates the wind loads on a triangular based lattice structure with compound profiles, using automatic calculation of "n" - eigen mode frequency (NF EN 1991-1-4/NA). The wind loads are generated according to Eurocodes 1 - French standards.
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3.5 EC1: generating snow loads on a 2 slopes 3D portal frame (NF EN 1991-1-3/NA) (VT : 3.4 - Snow - Example A)
Test ID: 4518
Test status: Passed
3.5.1 Description Generates snow loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA)
3.6 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.1 - Wind - Example A)
Test ID: 4520
Test status: Passed
3.6.1 Description Generates wind loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).
3.7 EC1: wind loads on a triangular based lattice structure with compound profiles and user defined "n" (NF EN 1991-1-4/NA) (TTAD #12276)
Test ID: 4526
Test status: Passed
3.7.1 Description Generates wind loads on a triangular based lattice structure with compound profiles using user defined "n" - eigen mode frequency - (NF EN 1991-1-4/NA) (TTAD #12276).
3.8 EC1: Verifying the geometry of wind loads on an irregular shed. (TTAD #12233)
Test ID: 4478
Test status: Passed
3.8.1 Description Verifies the geometry of wind loads on an irregular shed. The wind loads are generated according to Eurocodes 1 - French standard.
3.9 EC1: Verifying the wind loads generated on a building with protruding roof (TTAD #12071, #12278)
Test ID: 4510
Test status: Passed
3.9.1 Description Generates wind loads on the windwalls of a concrete structure with protruding roof, according to the Eurocodes 1 - French standard. Verifies the wind loads from both directions and generates the "Description of climatic loads" report.
The structure has concrete columns and beams (R2*3 cross section and B20 material) and rigid supports.
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3.10 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11531)
Test ID: 4090
Test status: Passed
3.10.1 Description Generates wind loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 French standards - Martinique wind speed.
The structure consists of concrete (C20/25) beams and columns with rigid fixed supports, with rectangular cross section (R20*30).
3.11 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11569)
Test ID: 4087
Test status: Passed
3.11.1 Description Generates snow loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanian standards.
The structure consists of concrete (C20/25) beams and columns with rigid fixed supports, with rectangular cross section (R20*30).
3.12 Generating the description of climatic loads report according to EC1 Romanian standards (TTAD #11688)
Test ID: 4104
Test status: Passed
3.12.1 Description Generates the "Description of climatic loads" report according to EC1 Romanian standards.
The model consists of a steel portal frame with rigid fixed supports. Haunches are defined at both ends of the beams. Dead loads and SR EN 1991-1-4/NB wind loads are generated.
3.13 EC1: generating snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height (TTAD #11943)
Test ID: 3706
Test status: Passed
3.13.1 Description Generates snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height, according to Eurocodes 1 French standard.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
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3.14 EC1: generating wind loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11687)
Test ID: 4055
Test status: Passed
3.14.1 Description Generates wind loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanian standards.
The structure consists of concrete (C20/25) beams and columns with rigid fixed supports.
3.15 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11699)
Test ID: 4085
Test status: Passed
3.15.1 Description Generates wind loads on a 2 slopes 3D portal frame according to Eurocodes 1 French standards, using the "Case 1" formula for calculating the turbulence factor.
The structure consists of steel elements with hinge rigid supports.
3.16 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11570)
Test ID: 4086
Test status: Passed
3.16.1 Description Generates snow loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanian standards.
The structure consists of concrete (C20/25) beams and columns with rigid fixed supports, with rectangular cross section (R20*30).
3.17 NV2009: generating wind loads and snow loads on a simple structure with planar support (TTAD #11380)
Test ID: 4091
Test status: Passed
3.17.1 Description Generates wind loads and snow loads on the windwalls of a concrete structure with a planar support, according to NV2009 French standards.
The structure has concrete columns and beams (R20*30 cross section and C20/25 material).
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3.18 EC1: verifying the snow loads generated on a monopitch frame (TTAD #11302)
Test ID: 3713
Test status: Passed
3.18.1 Description Generates wind loads on the windwalls of a monopitch frame, according to the Eurocodes 1 French standard.
The structure has concrete beams and columns (R20*30 cross section and B20 material) with rigid fixed supports.
3.19 EC1: generating wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls (TTAD #11937)
Test ID: 3705
Test status: Passed
3.19.1 Description Generates wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls, according to the Eurocodes 1 French standards.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with rigid fixed supports.
3.20 EC1: generating snow loads on two side by side roofs with different heights, according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)
Test ID: 3615
Test status: Passed
3.20.1 Description Generates snow loads on two side by side roofs with different heights, according to Eurocodes 1 German standards.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
3.21 EC1: generating wind loads on a 55m high structure according to German standards (DIN EN 1991-1-4/NA) (DEV2012 #3.12)
Test ID: 3618
Test status: Passed
3.21.1 Description Generates wind loads on the windwalls of a 55m high structure, according to Eurocodes 1 German standards.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
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3.22 EC1: generating wind loads on double slope 3D portal frame according to Czech standards (CSN EN 1991-1-4) (DEV2012 #3.18)
Test ID: 3621
Test status: Passed
3.22.1 Description Generates wind loads on the windwalls of a double slope 3D portal frame, according to the Eurocodes 1 Czech standard (CSN EN 1991-1-4).
The structure has concrete columns and beams (R20*30 cross section and C20/25 material).
3.23 EC1: generating snow loads on duopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)
Test ID: 3613
Test status: Passed
3.23.1 Description Generates snow loads on the windwalls of duopitch multispan roofs structure, according to Eurocodes 1 German standards.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
3.24 EC1: generating snow loads on two close roofs with different heights according to Czech standards (CSN EN 1991-1-3) (DEV2012 #3.18)
Test ID: 3623
Test status: Passed
3.24.1 Description Generates snow loads on two close roofs with different heights, according to Eurocodes 1 Czech standards.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
3.25 EC1: generating snow loads on monopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)
Test ID: 3614
Test status: Passed
3.25.1 Description Generates snow loads on the windwalls of a monopitch multispan roofs structure, according to Eurocodes 1 German standards.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
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3.26 EC1: snow on a 3D portal frame with horizontal roof and parapet with height reduction (TTAD #11191)
Test ID: 3605
Test status: Passed
3.26.1 Description Generates snow loads on the windwalls of a 3D portal frame with horizontal roof and 2 parapets, according to Eurocodes 1. The height of one parapet is reduced.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
3.27 EC1: generating snow loads on a 2 slopes 3D portal frame with gutter (TTAD #11113)
Test ID: 3603
Test status: Passed
3.27.1 Description Generates snow loads on the windwalls of a 2 slopes 3D portal frame with gutter, according to Eurocodes 1. The structure consists of concrete (C20/25) beams and columns with rigid fixed supports.
3.28 EC1: generating snow loads on a 3D portal frame with horizontal roof and gutter (TTAD #11113)
Test ID: 3604
Test status: Passed
3.28.1 Description Generates snow loads on the windwalls of a 3D portal frame with horizontal roof and gutter, according to Eurocodes 1.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
3.29 EC1: generating snow loads on a 3D portal frame with a roof which has a small span (< 5m) and a parapet (TTAD #11735)
Test ID: 3606
Test status: Passed
3.29.1 Description Generates snow loads on the windwalls of a 3D portal frame with a roof which has a small span (< 5m) and a parapet, according to Eurocodes 1.
The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.
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3.30 EC1: generating wind loads on double slope 3D portal frame with a fully opened face (DEV2012 #1.6)
Test ID: 3535
Test status: Passed
3.30.1 Description Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 French standard.
The structure has concrete columns and beams (R20*30 cross section and C20/25 material), a double slope roof and a fully opened face.
3.31 EC1: generating wind loads on an isolated roof with two slopes (TTAD #11695)
Test ID: 3529
Test status: Passed
3.31.1 Description Generates wind loads on a concrete structure, according to the Eurocodes 1 French standard.
The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and an isolated roof with two slopes.
3.32 EC1: generating wind loads on duopitch multispan roofs with pitch < 5 degrees (TTAD #11852)
Test ID: 3530
Test status: Passed
3.32.1 Description Generates wind loads on the windwalls of a steel structure, according to the Eurocodes 1 French standard.
The structure has steel columns and beams (I cross section and S275 material), rigid hinge supports and multispan roofs with pitch < 5 degrees.
3.33 EC1: wind load generation on a simple 3D structure with horizontal roof
Test ID: 3099
Test status: Passed
3.33.1 Description Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 standard.
The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and horizontal roof.
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3.34 EC1 NF: generating wind loads on a 3D portal frame with 2 slopes roof (TTAD #11932)
Test ID: 3004
Test status: Passed
3.34.1 Description Generates the wind loads on a concrete structure according to the French Eurocodes 1 standard.
The structure has a roof with two slopes, concrete columns and beams (R20*30 cross section and C20/25 material). The columns have rigid supports.
3.35 EC1: wind load generation on simple 3D portal frame with 4 slopes roof (TTAD #11604)
Test ID: 3104
Test status: Passed
3.35.1 Description Generates wind loads on the windwalls of a concrete structure with 4 slopes roof, according to the Eurocodes 1 standard.
The structure has 6 concrete columns (R20*30 cross section and C20/25 material) with rigid supports and C20/25 concrete walls.
3.36 EC1: wind load generation on a signboard
Test ID: 3107
Test status: Passed
3.36.1 Description Generates wind loads on the windwall of a concrete signboard, according to the Eurocodes 1 standard.
The signboard has concrete elements (R20*30 cross section and C20/25 material) and rigid supports.
3.37 EC1: wind load generation on a simple 3D portal frame with 2 slopes roof (TTAD #11602)
Test ID: 3103
Test status: Passed
3.37.1 Description Generates wind loads on the windwalls of a concrete structure with 2 slopes roof, according to the Eurocodes 1 standard.
The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and rigid supports.
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3.38 EC1: wind load generation on a building with multispan roofs
Test ID: 3106
Test status: Passed
3.38.1 Description Generates wind loads on the windwalls of a concrete structure with multispan roofs, according to the Eurocodes 1 standard.
The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and rigid supports.
3.39 EC1: wind load generation on a high building with horizontal roof
Test ID: 3101
Test status: Passed
3.39.1 Description Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 standard.
The structure is 63m high, has 4 columns and 4 beams (R20*30 cross section and C20/25 material), rigid supports and horizontal roof.
3.40 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528)
Test ID: 4569
Test status: Passed
3.40.1 Description Generates snow loads on a 4 slopes shed with gutters on each slope and middle parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
3.41 EC1: Generating snow loads on a single slope with lateral parapets (TTAD #12606)
Test ID: 4570
Test status: Passed
3.41.1 Description Generates snow loads on a single slope with lateral parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
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3.42 EC1: generating wind loads on a square based lattice structure with compound profiles and automatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12744)
Test ID: 4580
Test status: Passed
3.42.1 Description Generates the wind loads on a square based lattice structure with compound profiles, using automatic calculation of "n" - eigen mode frequency. The wind loads are generated according to Eurocodes 1 - French standards (NF EN 1991-1-4/NA).
3.43 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528)
Test ID: 4568
Test status: Passed
3.43.1 Description Generates snow loads on a 4 slopes shed with gutters on each slope and lateral parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
3.44 EC1: Generating snow loads on two side by side buildings with gutters (TTAD #12806)
Test ID: 4848
Test status: Passed
3.44.1 Description Generates snow loads on two side by side buildings with gutters, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
3.45 EC1: Generating wind loads on a square based structure according to UK standards (BS EN 1991-1-4:2005) (TTAD #12608)
Test ID: 4845
Test status: Passed
3.45.1 Description Generates the wind loads on a square based structure. The wind loads are generated according to Eurocodes 1 - UK standards (BS EN 1991-1-4:2005).
3.46 EC1: Generating snow loads on a 4 slopes with gutters building. (TTAD #12719)
Test ID: 4847
Test status: Passed
3.46.1 Description Generates snow loads on a model from CTCIM which contains 4 slopes with gutters, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
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3.47 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12808)
Test ID: 4849
Test status: Passed
3.47.1 Description Generates snow loads on 2 closed building with gutters, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
3.48 EC1: Generating snow loads on a 4 slopes with gutters building (TTAD #12716)
Test ID: 4846
Test status: Passed
3.48.1 Description Generates snow loads on a model from CTCIM which contains 4 slopes with gutters and lateral parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
3.49 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12835)
Test ID: 4850
Test status: Passed
3.49.1 Description Generates snow loads on 2 closed building with gutters. The lower building is longer. The wind loads are generated according to Eurocodes 1 - French standards (NF EN 1991-1-3/NA).
3.50 EC1: Generating snow loads on a 2 slope building with gutters and parapets. (TTAD #12878)
Test ID: 4852
Test status: Passed
3.50.1 Description Generates snow loads on a 2 slope building with gutters and lateral parapets on all sides, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
3.51 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12841)
Test ID: 4851
Test status: Passed
3.51.1 Description Generates snow loads on 2 closed building with gutters. The lower building is longer and has a 4 slope shed and the higher building has a 2 slope roof. The snow loads are generated according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).
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3.52 NV2009: Verifying wind and snow reports for a protruding roof (TTAD #11318)
Test ID: 4536
Test status: Passed
3.52.1 Description Generates wind loads and snow loads according to NV2009 - French climatic standards. Verifies wind and snow reports for a protruding roof.
3.53 NV2009: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604)
Test ID: 4567
Test status: Passed
3.53.1 Description Generates wind loads on a 2 slopes 3D portal frame at 15m height, according to the French standard (NV2009).
4 Combinations
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4.1 Verifying combinations for CZ localization (TTAD #12542)
Test ID: 4550
Test status: Passed
4.1.1 Description Verifies simplified combinations for CZ localization.
4.2 Generating combinations (TTAD #11673)
Test ID: 3471
Test status: Passed
4.2.1 Description Generates concomitance between three types of loads applied on a structure (live loads, dead loads and seismic loads - EN 1998-1), using the quadratic combination function. Generates the combinations description report and the point support actions by element report.
4.3 Defining concomitance rules for two case families (TTAD #11355)
Test ID: 3749
Test status: Passed
4.3.1 Description Generates live loads and dead loads on a steel structure. Defines the concomitance rules between the two load case families and generates the concomitance matrix.
4.4 Generating load combinations with unfavorable and favorable/unfavorable predominant action (TTAD #11357)
Test ID: 3751
Test status: Passed
4.4.1 Description Generates load combinations with unfavorable and favorable/unfavorable predominant action. Predominant action is a case family with 2 static load cases.
4.5 Generating combinations for NEWEC8.cbn (TTAD #11431)
Test ID: 3746
Test status: Passed
4.5.1 Description Generates combinations for NEWEC8.cbn.
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4.6 Generating load combinations after changing the load case number (TTAD #11359)
Test ID: 3756
Test status: Passed
4.6.1 Description Generates load combinations with concomitance matrix after changing the load case number. A report with the combinations description is generated.
4.7 Generating the concomitance matrix after adding a new dead load case (TTAD #11361)
Test ID: 3766
Test status: Passed
4.7.1 Description Creates a new dead load case, after two case families were created, and generates the concomitance matrix. A report with the combinations description is generated.
4.8 Generating a set of combinations with seismic group of loads (TTAD #11889)
Test ID: 4350
Test status: Passed
4.8.1 Description Generates a set of combinations with seismic group of loads.
4.9 Generating the concomitance matrix after switching back the effect for live load (TTAD #11806)
Test ID: 4219
Test status: Passed
4.9.1 Description Generates the concomitance matrix and the combinations description reports after switching back the effect for live load.
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4.10 Performing the combinations concomitance standard test no. 5 (DEV2012 #1.7)
Test ID: 4394
Test status: Passed
4.10.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - E (Base) - only one seismic load !
4 - Q (Base or Acco)
Set value "0" (exclusive) between seismic and all Q loads (seism only combination).
Generates combinations. Generates the combinations report.
4.11 Performing the combinations concomitance standard test no.9 (DEV2012 #1.7)
Test ID: 4408
Test status: Passed
4.11.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - E Group (Base)
3 - EX
4 - EY
5 - EZ
Un-group 3 - E Group to independent loads : 3 - EX 4 - EY 5 - EZ
Generates combinations. Generates the combinations report.
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4.12 Performing the combinations concomitance standard test no.4 (DEV2012 #1.7)
Test ID: 4391
Test status: Passed
4.12.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base)
3 - G (Favorable or Unfavorable)
4 - Q (Base)
Generates combinations. Generates the combinations report.
4.13 Performing the combinations concomitance standard test no.6 (DEV2012 #1.7)
Test ID: 4397
Test status: Passed
4.13.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - E (Base) - only one seismic load
Generates combinations. Generates the combinations report.
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4.14 Performing the combinations concomitance standard test no.8 (DEV2012 #1.7)
Test ID: 4407
Test status: Passed
4.14.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q Group(Base or Acco)
2 - Q
3 - Q
4 - Q
5 - Q (Base or Acco)
6 - Snow (Base or Acco)
Generates combinations. Generates the combinations report.
Un-group 2-Q Group to independent 2-Q, 3-Q, 4-Q loads.
Generates combinations. Generates the combinations report.
4.15 Performing the combinations concomitance standard test no.10 (DEV2012 #1.7)
Test ID: 4409
Test status: Passed
4.15.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - E Group (Base)
3 - EX
4 - EY
5 - EZ
Defines the seismic group type as "Quadratic".
Generates the corresponding combinations and the combinations report.
Resets the combination set. Defines the seismic group type as "Newmark".
Generates the corresponding combinations and the combinations report.
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4.16 Performing the combinations concomitance standard test no.7 (DEV2012 #1.7)
Test ID: 4405
Test status: Passed
4.16.1 Description Creates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - G (Favorable or Unfavorable)
4 - Q (Acco)
5 - Acc (Base)
Generates combinations. Generates the combinations report.
4.17 Generating a set of combinations with Q group of loads (TTAD #11960)
Test ID: 4353
Test status: Passed
4.17.1 Description Generates a set of combinations with Q group of loads.
4.18 Performing the combinations concomitance standard test no.1 (DEV2010#1.7)
Test ID: 4382
Test status: Passed
4.18.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - Q (Acco)
Generates combinations. Generates the combinations report.
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4.19 Generating a set of combinations with different Q "Base" types (TTAD #11806)
Test ID: 4357
Test status: Passed
4.19.1 Description Generates a set of combinations with different Q "Base" types
1 - G
2 - Q - Base
3 - G
4 - Q -Base or acco
Generates the first set of combinations and the first combinations report.
1 - G
2 - Q - Base
3 - G
4 - Q -Base
Generates the second set of combinations and the second combinations report.
4.20 Performing the combinations concomitance standard test no.2 (DEV2012 #1.7)
Test ID: 4384
Test status: Passed
4.20.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - G (Favorable or Unfavorable)
3 - Q (Base)
Generates combinations. Generates the combinations report.
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4.21 Performing the combinations concomitance standard test no.3 (DEV2012 #1.7)
Test ID: 4386
Test status: Passed
4.21.1 Description Generates loads:
1 - G (Favorable or Unfavorable)
2 - Q (Base or Acco)
3 - G (Favorable or Unfavorable)
4 - Q (Base or Acco)
Generates combinations. Generates the combinations report.
5 Concrete Design
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5.1 EC2: Verifying the minimum reinforcement area for a simply supported beam
Test ID: 4517
Test status: Passed
5.1.1 Description Verifies the minimum reinforcement area for a simply supported concrete beam subjected to self weight. The verification is made with Eurocodes 2 - French annex.
5.2 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load
Test ID: 4519
Test status: Passed
5.2.1 Description Verifies the longitudinal reinforcement area of a beam under a linear load (horizontal level behavior law).
Verification is done with Eurocodes 2 norm French Annex.
5.3 EC2: Verifying the longitudinal reinforcement area for a beam subjected to point loads
Test ID: 4527
Test status: Passed
5.3.1 Description Verifies the longitudinal reinforcement area for a beam subjected to point loads (applied at the middle of the beam).
The verification is performed according to EC2 norm with French Annex.
5.4 Modifying the "Design experts" properties for concrete linear elements (TTAD #12498)
Test ID: 4542
Test status: Passed
5.4.1 Description Defines the "Design experts" properties for a concrete (EC2) linear element in analysis model and verifies properties from descriptive model.
5.5 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - horizontal level behavior law
Test ID: 4557
Test status: Passed
5.5.1 Description Verifies the longitudinal reinforcement area of a beam under self-weight and linear loads - horizontal level behavior law. The verification is made according to EC2 norm with French Annex.
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5.6 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - bilinear stress-strain diagram
Test ID: 4541
Test status: Passed
5.6.1 Description Verifies the longitudinal reinforcement area of a simply supported beam under a linear load - bilinear stress-strain diagram.
Verification is done according to Eurocodes 2 norm with French Annex.
5.7 Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular cross section
Test ID: 4179
Test status: Passed
5.7.1 Description Performs the finite elements calculation and the reinforced concrete calculation according to the Eurocodes 2 - French DAN. Verifies the longitudinal reinforcement and generates the corresponding report: "Longitudinal reinforcement linear elements".
The model consists of a concrete linear element with rectangular cross section (R18*60) with rigid hinge supports at both ends and two linear vertical loads: -15.40 kN and -9.00 kN.
5.8 Verifying the reinforced concrete results on a structure with 375 load cases combinations (TTAD #11683)
Test ID: 3475
Test status: Passed
5.8.1 Description Verifies the reinforced concrete results for a model with more than 100 load cases combinations.
On a concrete structure there are applied: dead loads, self weight, live loads, wind loads (according to NV2009) and accidental loads. A number of 375 combinations are obtained.
Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforcement areas planar elements report.
5.9 Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678)
Test ID: 3543
Test status: Passed
5.9.1 Description Verifies the longitudinal reinforcement for a vertical concrete bar.
Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforcement report.
The bar has a circular cross section with a radius of 40.00 cm and a rigid hinge support. A vertical punctual load of -5000.00 kN is applied.
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5.10 Verifying the reinforced concrete results on a fixed beam (TTAD #11836)
Test ID: 3542
Test status: Passed
5.10.1 Description Verifies the concrete results on a fixed horizontal beam.
Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforced concrete calculation results report.
The beam has a R25*60 cross section, C25/30 material and has a rigid hinge support at one end and a rigid support with translation restraints on Y and Z at the other end. A linear dead load (-28.75 kN) and a live load (-50.00 kN) are applied.
5.11 Verifying the longitudinal reinforcement for a fixed linear element (TTAD #11700)
Test ID: 3547
Test status: Passed
5.11.1 Description Verifies the longitudinal reinforcement for a horizontal concrete bar.
Performs the finite elements calculation and the reinforced concrete calculation. Verifies the longitudinal reinforcement and generates the reinforcement report.
The bar has a rectangular cross section R40*80, has a rigid hinge support at one end and a rigid support with translation restraints on Y and Z. Four loads are applied: a linear dead load of -50.00 kN on FZ, a punctual dead load of -30.00 kN on FZ, a linear live load of -60.00 kN on FZ and a punctual live load of -25.00 kN on FZ.
5.12 Verifying the longitudinal reinforcement for linear elements (TTAD #11636)
Test ID: 3545
Test status: Passed
5.12.1 Description Verifies the longitudinal reinforcement for a vertical concrete bar.
Performs the finite elements calculation and the reinforced concrete calculation. Verifies the longitudinal reinforcement and generates the reinforcement report.
The bar has a square cross section of 30.00 cm, a rigid fixed support at the base and a support with translation restraints on X and Y. A vertical punctual load of -1260.00 kN is applied.
5.13 EC2 : calculation of a square column in traction (TTAD #11892)
Test ID: 3509
Test status: Passed
5.13.1 Description The test is performed on a single column in tension, according to Eurocodes 2.
The column has a section of 20 cm square and a rigid support. A permanent load (traction of 100 kN) and a live load (40 kN) are applied.
Performs the finite elements calculation and the reinforced concrete calculation. Generates the longitudinal reinforcement report.
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5.14 Verifying Aty and Atz for a fixed concrete beam (TTAD #11812)
Test ID: 3528
Test status: Passed
5.14.1 Description Performs the finite elements calculation and the reinforced concrete calculation of a model with a horizontal concrete beam.
The beam has a R20*50 cross section and two hinge rigid supports.
Verifies Aty and Atz for the fixed concrete beam.
5.15 Verifying concrete results for planar elements (TTAD #11583)
Test ID: 3548
Test status: Passed
5.15.1 Description Verifies the reinforcement results on planar elements.
Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforced concrete analysis report: data and results.
The model consists of two planar elements (C20/25 material) with rigid fixed linear supports. On each element, a punctual load of 50.00 kN on FX is applied.
5.16 Verifying concrete results for linear elements (TTAD #11556)
Test ID: 3549
Test status: Passed
5.16.1 Description Verifies the reinforcement results for a horizontal concrete bar.
Performs the finite elements calculation and the reinforced concrete calculation. Verifies the reinforcement and generates the reinforced concrete analysis report: data and results.
The bar has a rectangular cross section R20*50, a rigid hinge support at one end, a rigid support with translation restraints on X, Y and Z and rotation restraint on X.
5.17 Verifying the reinforcement of concrete columns (TTAD #11635)
Test ID: 3564
Test status: Passed
5.17.1 Description Verifies the reinforcement of a concrete column.
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5.18 Verifying the minimum transverse reinforcement area results for an articulated beam (TTAD #11342)
Test ID: 3638
Test status: Passed
5.18.1 Description Verifies the minimum transverse reinforcement area for an articulated horizontal beam.
Performs the finite elements calculation and the reinforced concrete calculation and generates the "Transverse reinforcement linear elements" report.
The beam has a rectangular cross section (R20*50), B25 material and two hinge rigid supports at both ends.
5.19 Verifying the minimum transverse reinforcement area results for articulated beams (TTAD #11342)
Test ID: 3639
Test status: Passed
5.19.1 Description Verifies the minimum transverse reinforcement area for two articulated horizontal beams.
Performs the finite elements calculation and the reinforced concrete calculation and generates the "Transverse reinforcement linear elements" report.
Each beam has rectangular cross section (R30*70), B25 material and two hinge rigid supports at both ends.
On each beam there are applied:
- Dead loads: a linear load of -25.00 kN and two punctual loads of -55.00 kN and -65.00 kN
- Live loads: a linear load of -20.00 kN and two punctual loads of -40.00 kN and -35.00 kN.
5.20 EC2: column design with “Nominal Stiffness method” square section (TTAD #11625)
Test ID: 3001
Test status: Passed
5.20.1 Description Verifies and generates the corresponding report for the longitudinal reinforcement bars of a column. The column is designed with "Nominal stiffness method", with a square cross section (C40).
5.21 EC2: Verifying the transverse reinforcement area for a beam subjected to linear loads
Test ID: 4555
Test status: Passed
5.21.1 Description Verifies the transverse reinforcement area for a beam subjected to linear loads. The verification is made according to EC2 norm with French Annex.
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5.22 EC2 Test 1: Verifying a rectangular cross section beam made from concrete C25/30 to resist simple bending - Bilinear stress-strain diagram
Test ID: 4969
Test status: Passed
5.22.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending.
During this test, the determination of stresses will be made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
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5.23 EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinear stress-strain diagram
Test ID: 4978
Test status: Passed
5.23.1 Description Verifies a T concrete section, without compressed reinforcement - Bilinear stress-strain diagram.
The purpose of this test is to verify the My resulted stresses for the USL load combination and the results of the theoretical reinforcement area Az.
The objective is to verify:
- The stresses results
- The theoretical reinforcement area results
5.23.2 Background This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.
5.23.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 52.3 kN/m ■ Exploitation loadings (category A): Q = 13kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram The objective is to verify:
■ The stresses results ■ The theoretical reinforcement area results
Simply supported beam
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Units
Metric System
Geometry
Beam cross section characteristics:
■ Beam length: 7m ■ Concrete cover: c=3.50 cm ■ Effective height: d=h-(0.6*h+ebz)=0.595 m; d’=ebz=0.035m
Materials properties
Rectangular solid concrete C16/20 and S400B reinforcement steel is used. The following characteristics are used in relation to this material:
■ Exposure class XC1 ■ Concrete density: 16kN/m3 ■ Reinforcement steel ductility: Class A ■ The calculation is made considering bilinear stress-strain diagram
■ Concrete C16/20: MPaffc
ckcd 67.10
5,116
===γ
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N
■ MPa.*.f*.f //ckctm 90116300300 3232 ===
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1
■ Steel S400B : MPaf
fs
ykyd 8.347
15,1400
===γ
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X.
■ Inner: None.
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Loading
The beam is subjected to the following load combinations:
■ Load combinations: The ultimate limit state (ULS) combination is:
Cmax = 1.35 x G + 1.5 x Q=1.35*352.3+1.5*13=90.105kN/ml
Characteristic combination of actions:
CCQ = 1.0 x G + 1.0 x Q=52.3+13=65.3kN/ml
Load calculations:
kNmMEd 89.5518
²7*105.70==
kNmMEcq 96.3998
²7*3.65==
5.23.2.2 Reference results in calculating the concrete beam moment At first, it will be determined the moment resistance of the concrete section only:
kNm,*,,*,*,f*hd*h*bM cdf
feffbtu 52367102100595010090
2=⎟
⎠
⎞⎜⎝
⎛ −=⎟⎠
⎞⎜⎝
⎛ −=
Comparing Mbtu with MEd:
Therefore the concrete section is not entirely compressed;
Therefore, calculations considering the T section are required.
5.23.2.3 Reference reinforcement calculation:
Theoretical section 2:
The moment corresponding to this section is:
MNm
hdfhbbM f
cdfweffEd
418.021.00595*67.10*1.0*)18.09.0(
2***)(2
=
=⎟⎠⎞
⎜⎝⎛ −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
According to this value, the steel section is:
²08.228.347*
21.0595.0
418.0
*2
22 cm
fh
d
MA
ydf
Ed =⎟⎠⎞
⎜⎝⎛ −
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=
kNmMkNmM Edbtu 89.551523 =<=
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Theoretical section 1:
The theoretical section 1 corresponds to a calculation for a rectangular shape beam section
kNmMMM EdEdEd 133418.0552.021 =−=−=
197.067.10*²595.0*18.0
133.0*²*1 ===
cdw
Edcu Fdb
Mμ
For a S400B reinforcement and for a XC1 exposure class, there will be a: 3720μμ .lucu =< , therefore there will be no compressed reinforcement.
There will be a calculation without compressed reinforcement:
[ ] ( )[ ] 276.0197.0*211*25.1)*21(1*25.1 =−−=−−= cuu μα
mdz uc 529.0)276.0*40.01(*595.0)*4.01(*1 =−=−= α
²25.78.347*529.0
133.0*1
11 cm
fzMA
ydc
Ed ===
Theoretical section 1:
In conclusion the entire reinforcement steel area is A=A1+A2=4.25+22.08=29.33cm2
Finite elements modeling
■ Linear element: S beam, ■ 8 nodes, ■ 1 linear element.
ULS and SLS load combinations(kNm)
Simply supported beam subjected to bending
ULS (reference value: 552kNm)
Theoretical reinforcement area(cm2)
For Class B reinforcement steel ductility (reference value: A=29.33cm2)
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5.23.2.4 Reference results
Result name Result description Reference value My,ULS My corresponding to the 101 combination (ULS) [kNm] 552 kNm Az (Class B) Theoretical reinforcement area [cm2] 29.33 cm2
5.23.3 Calculated results
Result name
Result description Value Error
My My USL -540.63 kN*m 2.0408 % Az Az -28.6439 cm² 2.3507 %
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5.24 EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement – Inclined stress-strain diagram
Test ID: 4982
Test status: Passed
5.24.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
For these tests, the constitutive law for reinforcement steel, on the inclined stress-strain diagram is applied.
This test performs the verification of the theoretical reinforcement area for a rectangular concrete beam subjected to the defined loads. The test confirms the presence of the compressed reinforcement for this model.
5.24.2 Background Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses, the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage are performed.
For these tests, the constitutive law for reinforcement steel, on the inclined stress-strain diagram is applied.
This test performs the verification of the theoretical reinforcement area for a rectangular concrete beam subjected to the defined loads. The test confirms the presence of the compressed reinforcement for this model.
5.24.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Linear loadings : Loadings from the structure: G = 55 kN/m+ dead load,
Live loads: Q=60kN/m
■ Point loads: Loadings from the structure: G = 35kN;
Live loads: Q=25kN
■ 8,02 =ψ
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering inclined stress-strain diagram The objective is to verify:
■ The stresses results ■ The longitudinal reinforcement ■ The minimum reinforcement percentage
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Simply supported beam
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.80 m, ■ Width: b = 0.40 m, ■ Length: L = 6.30 m, ■ Section area: A = 0.32 m2 , ■ Concrete cover: c=4.50 cm ■ Effective height: d=h-(0.6*h+ebz)=0.707 m; d’=ebz=0.045m
Materials properties
Rectangular solid concrete C25/30 and S500A reinforcement steel is used. The following characteristics are used in relation to this material:
■ Exposure class XD1 ■ Concrete density: 25kN/m3 ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering inclined stress-strain diagram ■ Cracking calculation required ■ Concrete C25/30:
MPa,,
ff
c
ckcd 6716
5125
γ===
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1
MPafE ckcm 31476
1082522000
10822000
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
■ Steel S500 :
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2
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Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 6.3) restrained in translation along Y and Z, and restrained rotation along X.
■ Inner: None.
Loading
The beam is subjected to the following load combinations:
■ Dead load: G’=0.4*0.8*2.5=8.00 kN/ml
■ Linear load combinations: The ultimate limit state (ULS) combination is:
Cmax = 1.35 x G + 1.5 x Q=1.35*(55+8)+1.5*60=175.05 kN/ml
Characteristic combination of actions:
CCQ = 1.0 x G + 1.0 x Q=55+8+60=123 kN/ml
Quasi-permanent combination of actions:
CQP = 1.0 x G + 0.6 x Q=55+8+0.8*60=111 kN/m
■ Point load combinations: The ultimate limit state (ULS) combination is:
Cmax = 1.35 x G + 1.5 x Q=1.35*35+1.5*25=84.75 kN
Characteristic combination of actions:
CCQ = 1.0 x G + 1.0 x Q=35+25=60 kN
Quasi-permanent combination of actions:
CQP = 1.0 x G + 0.6 x Q=35+0.8*25=55 kN
■ Load calculations:
kNmMEd 10024
30.6*75.1848
²30.6*05.175=+=
kNmMEcq 7054
30.6*608
²30.6*123=+=
kNmMEqp 6374
30.6*558
²30.6*111=+=
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5.24.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity:
)(*)(*),( 00 tft cmRH ββϕϕ =∞
According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.2
925.2825
8.168.16)( =+
==cm
cm ffβ
According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.4
at t0 = 28 days
According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.5
The value of the φRH coefficient depends of the concrete quality:
2130
***1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.3a
121 == αα if MPafCM 35≤
If not:
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.8c
In this case,
MpaMpaff ckcm 338 =+=
121 == αα
78.167.266*1.0
100501
167.266)800400(*2
800*400*2*230 =
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=⇒=
+== RHmm
uAch ϕ
According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.6
54.2488.0*92.2*78.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ
The coefficient of equivalence is determined by the following formula:
94.20
705637*54..21
31476200000
*),(1 0
=
+
=
∞+
=
Ecar
Eqp
cm
se
MM
t
EE
ϕ
α
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5.24.2.3 Reference results in calculating the concrete beam reduced moment limit Due to exposure class XD3, we will verify the section having non-compressed steel reinforcement, determining the reduced moment limit, using the formula defined by Jeans Roux in his book “Practice of EC2”:
This value can be determined by the next formula if MPafck 50< valid for a constitutive law to horizontal plateau:
)*62.7969.165(*)*66.162.4(*)(
γγαμ
−+−=
ck
ckeluc f
fK
( ) ( )24 ***10 eee cbaK ααα ++= −
Where:
7.16928,18925*3,75 =−=a
5.7345,87425*6,5 =+−=b
121325*04,0 −=−=c
( ) ( ) 181.1²94.20*1294.20*5.7347.1692*10 4 =−+= −eK α
Then:
422,1705
1002==γ
271.0)*62.7969.165(*)*66.162.4(
*)( =−+−
=γγ
αμck
ckeluc f
fK
Calculation of reduced moment:
301.067.16*²707.0*40.0
002.1*²*
===cdw
Edcu fdb
Mμ
271.0301.0 =>= luccu μμ 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 luμ :
NmfdbM cdwluEd6
1 10*902.067.16*²707.0*40.0*271.0*²** === μ
■ The α value:
[ ] [ ] 404.0)271.0*21(1*25.1)*21(1*25.1 =−−=−−= lulu μα
■ Calculation of the lever arm zc:
mdz lulu 593.0)404.0*4.01(*707.0)*4.01(* =−=−= α
■ Tensioned reinforcement elongation calculation:
17.55.3*404.0
404.01*12 =
−=
−= cu
u
usu ε
ααε
‰
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■ Tensioned reinforcement efforts calculation(S500A):
MPasusu 454*38,95271,432 ≤+= εσ
MpaMPasu 45463.43700517.0*38,95271,432 ≤=+=σ
■ Calculation of the reinforcement area:
²76.3463.437*593.0
10*902.0.
61
1 cmMPamNm
fzMA
ydlu
Ed ===−
Compressed steel reinforcement reduction (Section As2):
Reduction coefficient:
( ) 9572045070704040707040401000
53αα100053ε ...*.
.*.*.)'dd*(
d**,
lulu
sc =−=−= ‰
MPascydsc 52.43500295.0*38.95271.43200217.000295.0 =+=⇒=>= σεε
Compressed reinforcement calculation:
²47.352.435*)045.0707.0(
902.0002.1)'(
12 cm
ddMMA
sc
EdEds =
−−
=−
−=
σ
The steel reinforcement condition:
²48.364.43752.435*47.3.22 cm
fAA
yd
scs ===
σ
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: kNmMEcq 705=
■ The equivalence coefficient: 9420α .e =
■ The stress on the concrete will be limited at 0.6*fck=15Mpa and the stress on steel at 0.8*fyk, or 400MPa
Calculation of the resistance moment MRd for detecting the presence of the compressed reinforcement:
mdxsce
ce 311.0707.0*40015*94.20
15*94.20**
*1 =
+=
+=
σσασα
NxbF cwc6
1 10*933.015311.0*40.0*21***
21
=×== σ
mxdzc 603.03311.0707.0
31 =−=−=
NmzFM ccrb66 10*563.0603.0*10*933.0* ===
Therefore the compressed reinforced established earlier was correct.
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Theoretical section 1 (tensioned reinforcement only)
NmMM rb6
1 10*563.0==
440.0707.0311.01
1 ===dxα
mxdzc 603.03311.0707.0
31 =−=−=
²34.23400603.0
563.011 cm
zMA
sc
=×
=×
=σ
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
NmMMM rbcqser66
,2 10*142.010*)591.0705.0( =−=−=
Compressed reinforcement stresses:
ddd
cesc *'***
1
1
αασασ −
=
MPasc 66.268707.0*440.0
045.0707.0*440.0*15*94.20 =−
=σ
Compressed reinforcement area:
( ) ²98.766.268*045.0707.0
142.0*)'(
' 2 cmddMA
sc
=−
=−
=σ
Complementary tensioned reinforcement area:
²36.5400
66.268*98.7*' cmAAs
scs ===
σσ
Section area:
Tensioned reinforcement: 23.34+5.36=28.7 cm2
Compressed reinforcement: 7.98 cm2
Considering an envelope calculation of ULS and SLS, it will be obtained:
Tensioned reinforcement ULS: A=38.24cm2
Compressed reinforcement SLS: A=7.98cm2
To optimize the reinforcement area, it is preferable a third iteration by recalculating with SLS as a baseline amount of tensioned reinforcement (after ULS: Au=38.24cm2)
Reference reinforcement third calculation at SLS:
For this third iteration, the calculation will begin considering the section of the tensile reinforcement found when calculating for ULS: Au=38.24cm2
From this value, it will be calculated the stress obtained in the tensioned reinforcement:
MPaAA
sELU
ELSs 21.300400*
24.3870.28* === σσ
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Calculating the moment resistance Mrb for detecting the presence of compressed steel reinforcement:
mdxsce
ce 361.0707.0*21.30015*94.20
15*94.20**
*1 =
+=
+=
σσασα
NxbF cwc6
1 10*083.115361.0*40.0*21***
21
=×== σ
mxdzc 587.03361.0707.0
31 =−=−=
NmzFM ccrb66 10*636.0587.0*10*083.1* ===
Therefore the compressed reinforced established earlier was correct.
Theoretical section 1 (tensioned reinforcement only)
NmMM rb6
1 10*636.0==
511.0707.0361.01
1 ===dxα
mxdzc 587.03361.0707.0
31 =−=−=
²09.36400*578.0
36.0*
11 cmzMA
sc
===σ
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
NmMMM rbcqser66
,2 10*069.010*)636.0705.0( =−=−=
Compressed reinforcement stresses:
MPaddd
cesc 275707.0*511.0
045.0707.0*511.0*15*94.20*
'***1
1 =−
=−
=α
ασασ
Compressed reinforcement area:
( ) ²79.3275*045.0707.0
069.0*)'(
' 2 cmddMA
sc
=−
=−
=σ
Complementary tensioned reinforcement area:
²47.321.300
275*79.3*' cmAAs
scs ===
σσ
Section area:
Tensioned reinforcement: 36.09+3.47=39.56 cm2
Compressed reinforcement: 3.79 cm2
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Reference solution for minimal reinforcement area
The minimum percentage for a rectangular beam in pure bending is:
⎪⎩
⎪⎨⎧
=db
dbff
MaxA
w
wyk
effct
s
**0013.0
***26.0 ,
min,
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2
Where:
MPaff ctmeffct 56.2, == from cracking conditions
Therefore:
⎪⎩
⎪⎨⎧
==
==−
−
²77.3²10*68.3707.0*40.0*0013.0
²10*77.3707.0*40.0*500
56.2*26.0max4
4
min, cmm
mAs
Finite elements modeling
■ Linear element: S beam, ■ 9 nodes, ■ 1 linear element.
ULS and SLS load combinations(kNm)
Simply supported beam subjected to bending
ULS (reference value: 1002kNm)
SLS (reference value: 705kNm)
SLS –Quasi-permanent (reference value: 637kNm)
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Theoretical reinforcement area(cm2)
For Class A reinforcement steel ductility (reference value: A=39.56cm2 and A’=3.79cm2)
Minimum reinforcement area(cm2)
(reference value: 3.77cm2)
5.24.2.4 Reference results
Result name Result description Reference value My,ULS My corresponding to the 101 combination (ULS) [kNm] 1001 kNm My,SLS,cq My corresponding to the 102 combination (SLS) [kNm] 705 kNm My,SLS,qp My corresponding to the 102 combination (SLS) [kNm] 637 cm2
Az (Class A) Theoretical reinforcement area [cm2] 39.56 cm2
Amin Minimum reinforcement area [cm2] 3.77 cm2
5.24.3 Calculated results
Result name
Result description Value Error
My My USL -985.226 kN*m 1.6662 % My My SLS cq -692.949 kN*m 1.6694 % My My SLS qp -626.623 kN*m 1.6760 % Az Az -39.5901 cm² 0.0008 % Amin Amin 3.77193 cm² 0.0001 %
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5.25 EC2 Test 10: Verifying a T concrete section, without compressed reinforcement - Inclined stress-strain diagram
Test ID: 4984
Test status: Passed
5.25.1 Description Simple Bending Design for Ultimate Limit State - The purpose of this test is to verify the My resulted stresses for the ULS load combination, the results of the theoretical reinforcement area, "Az" and the minimum reinforcement percentage, "Amin".
This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.
5.25.2 Background Verifies the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.
5.25.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 500 kN/m ■ Exploitation loadings (category A): Q = 300kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q ■ 30ψ2 ,=
■ Reinforcement steel ductility: Class B ■ The calculation is performed considering inclined stress-strain diagram The objective is to verify:
■ The stresses results ■ The theoretical reinforcement area ■ The reinforcement minimum percentage area
Simply supported beam
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Beam length: 6m ■ Concrete cover: c=4.00 cm ■ Effective height: d=h-(0.6*h+ebz)=0.900 m; d’=ebz=0.04m
Materials properties
Rectangular solid concrete C30/37 and S500B reinforcement steel is used. The following characteristics are used in relation to this material:
■ Exposure class XC2 ■ Reinforcement steel ductility: Class B ■ The calculation is made considering the inclined stress-strain diagram ■ Concrete C16/20:
MPa,
ff
c
ckcd 20
5130
γ===
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N
MPa.*.f*.f //ckctm 90230300300 3232 ===
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1
■ Steel S400B :
MPa.,
ff
s
ykyd 78434
151500
γ===
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x = 0) restrained in translation along X, Y and Z, ► Support at end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X.
■ Inner: None.
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Loading
The beam is subjected to the following load combinations:
Load combinations:
The ultimate limit state (ULS) combination is:
Cmax = 1.35 x G + 1.5 x Q=1.35*500+1.5*300=1125 kN/m
Characteristic combination of actions:
CCQ = 1.0 x G + 1.0 x Q=500+300=800 kN/m
Quasi-permanent combination of actions
CQP = 1.0 x G + 0.3 x Q=500+0.3*300=590 kN/m
Load calculations:
kNm.²*MEd 550628
61125==
kNm²*MEcq 36008
6800==
kNm²*MEqp 26558
6590==
5.25.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:
kNm*.*,,*,*.f*h
d*h*bM cdf
feffbtu310480420
220090200401
2=⎟
⎠
⎞⎜⎝
⎛ −=⎟⎟⎠
⎞⎜⎜⎝
⎛−=
Comparing Mbtu with MEd:
Therefore the concrete section is not entirely compressed;
There are required calculations considering the T section.
Reference reinforcement calculation:
For those calculations, the beam section will be divided in two theoretical section:
Section 1: For the calculation of the concrete only
Section 2: For the calculation of the compressed reinforcement
Theoretical section 2:
The moment corresponding to this section is:
kNm*.
..**.*)..(hd*f*h*)bb(M fcdfweffEd
3
2
1023
220090020200400401
2
=
=⎟⎠
⎞⎜⎝
⎛ −−=⎟⎠
⎞⎜⎝
⎛ −−=
Stress from the compressed steel reinforcement considering a steel grade S500B:
MPa*,, susu 466ε2772771432σ ≤+=
kNm*.MkNm*.M Edbtu33 1006255104804 =<=
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In order to determine the suε , the neutral axis position must be determined:
kNm*.*)..(MMM EdEdEd33
21 1086311020030635 =−=−=
28702090400
8631μ 1 .*².*.
.F*²d*b
M
cdw
Edcu ===
[ ] ( )[ ] 43502870211251μ211251α ..**.)*(*. cuu =−−=−−=
Depending of uα , the neutral axis position can be established:
5545034350
43501εαα1
ε 2 ..*.
.* cuu
usu =
−=
−= ‰
Then we calculate the stress in the tensioned steel reinforcement:
MPaMPa,.,,su 466024360045502772771432σ ≤=×+=
According to those above, the theoretical reinforcement can be calculated:
²cm..*..
.
F*h
d
MA
ydf
Ed 749102436
220090
2003
2
22 =
⎟⎠
⎞⎜⎝
⎛ −=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=
Theoretical section 1:
The theoretical section 1 corresponds to a calculation for a rectangular shape beam section
kNm*.*)..(MMM EdEdEd33
21 1086311020030635 =−=−=
3720μ28702090400
8631μ 1 ..*².*.
.F*²d*b
Mlu
cdw
Edcu =<===
For a S500B reinforcement and for a XC2 exposure class, there will be a: 3720μ2870μ .. lucu =<= , therefore there will be no compressed reinforcement.
There will be a calculation without compressed reinforcement:
[ ] ( )[ ] 43502870211251μ211251α ..**.)*(*. cuu =−−=−−=
m.).*.(*.)*.(*dz uc 74304350400190α4011 =−=−=
²cm..*.
.f*z
MAydc
Ed 4657024367430
8631
1
11 ===
Theoretical section 1:
In conclusion the entire reinforcement steel area is A=A1+A2=91.74+57.46=149.20cm2
Reference solution for minimal reinforcement area
The minimum percentage for a rectangular beam in pure bending is:
⎪⎩
⎪⎨
⎧=
d*b*.
d*b*f
f*.
MaxA
w
wyk
eff,ct
min,s00130
260
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2
Where:
MPa.ff ctmeff,ct 902== from cracking conditions
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Therefore:
⎪⎩
⎪⎨⎧
==
==
−
−²cm.
²m*..*.*.
²m*..*.*.*.maxA min,s 4251068490040000130
10425900400500
9022604
4
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
ULS and SLS load combinations (kNm)
Simply supported beam subjected to bending
ULS (reference value: 5062.5kNm)
Theoretical reinforcement area(cm2)
For Class B reinforcement steel ductility (reference value: A=149.20cm2)
Minimum reinforcement area(cm2)
(reference value: 5.42cm2)
5.25.2.3 Reference results
Result name Result description Reference value My,ULS My corresponding to the 101 combination (ULS) [kNm] 5062.5 kNm Az (Class B) Theoretical reinforcement area [cm2] 149.20 cm2
Amin Minimum reinforcement area [cm2] 5.42 cm2
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5.25.3 Calculated results
Result name
Result description Value Error
My My USL -5062.5 kN*m 0.0000 % Az Az -149.031 cm² 0.0001 % Amin Amin -5.42219 cm² 0.0000 %
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5.26 EC2 Test 12: Verifying a rectangular concrete beam subjected to uniformly distributed load, without compressed reinforcement- Bilinear stress-strain diagram (Class XD3)
Test ID: 4985
Test status: Passed
5.26.1 Description Simple Bending Design for Service State Limit
Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. The verification of the bending stresses at service limit state is performed.
During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
5.26.2 Background Simple Bending Design for Service State Limit
Verify the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During this test, the calculation of stresses, the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage are performed.
5.26.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 25 kN/m (including dead load), ■ Exploitation loadings (category A): Q = 15kN/m,
■ ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q
Simply supported beam
Units
Metric System
ADVANCE DESIGN VALIDATION GUIDE
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Geometry
Beam cross section characteristics:
■ Height: h = 0.65 m, ■ Width: b = 0.28 m, ■ Length: L = 6.40 m, ■ Section area: A = 0.182 m2 , ■ Concrete cover: c = 4.5 cm ■ Effective height: d = h-(0.6*h+ebz) = 0.57m; d’ = ebz = 0.045 m
Materials properties
Rectangular solid concrete C30/37 and S500A reinforcement steel is used. The following characteristics are used in relation to this material:
■ Exposure class XD3 ■ Concrete density: 25 kN/m3 ■ Stress-strain law for reinforcement: Bilinear stress-strain diagram ■ The concrete age t0 = 28 days ■ Humidity 50%
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x = 0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.80) restrained in translation along Y, Z and restrained in rotation along X.
■ Inner: None.
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 + 15 = 40 kN/ml
Quasi-permanent combination of actions:
CQP = 1.0 x G + 0.3 x Q = 25 + 0.3*15 = 29.5 kN/ml
■ Load calculations:
kNm².*)(MEcq 2058
4061525=
+=
kNm².*)*.(MEqp 1518
406153025=
+=
ADVANCE DESIGN VALIDATION GUIDE
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5.26.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 tft cmRH ββϕϕ =∞
Where:
MPaf
fcm
cm 725.2830
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ
t0 : concrete age t0=28days
210
3**
*1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
121 == αα if Mpafcm 35≤
If not 70
135α
.
cmf ⎟⎟⎠
⎞⎜⎜⎝
⎛= and
20
235α
.
cmf ⎟⎟⎠
⎞⎜⎜⎝
⎛=
In this case MpaMpaMpaff ckcm 35388 >=+= therefore
944.0383535 7.07.0
1 =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
cmfα
984.0383535 2.02.0
2 =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
cmfα
In this case:
Humidity RH = 50 %
( ) mmuAch 70.195
650280*2650*280*22
0 =+
==
37.2488.0*73.2*78.1)(*)(*),(78.1984.0*70.195*1.0
100501
1 003===∞⇒=
−+= tft cmRHRH ββϕϕϕ
Calculating the equivalence coefficient:
The coefficient of equivalence is determined by the following formula:
Ecar
Eqp
cm
se
MM
t
EE
*),(1 0∞+
=
ϕ
α
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386
Where:
MPafE ckcm 32837
10830*22
108*22
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
MPaEs 200000=
75.2205151*37.21*),(1 0 =+=∞+
Ecar
Eqp
MM
tϕ
76.16
205151*37.21
32837200000
*),(1 0
=
+
=
∞+
=
Ecar
Eqp
cm
se
MM
t
EE
ϕ
α
Material characteristics:
The maximum compression on the concrete is: Mpafckbc 1830*6,0*6,0 ===σ
For the maximum stress on the steel taut, we consider the constraint limit Mpaf yks 400*8,0 ==σ
Neutral axis position calculation:
The position of the neutral axis must be determined by calculating 1α (position corresponding to the state of maximum stress on the concrete and reinforcement):
430.040018*76.16
15*76.16*
*1 =
+=
+=
sce
ce
σσασαα
Moment resistance calculation:
Knowing the 1α value, it can be determined the moment resistance of the concrete section, using the following formulas:
243057004300α11 ..*.d*x === m
MNm...**.*.*.xd**x*b*M cwrb 297032430570018243028050
3σ
21 1
1 =⎟⎠
⎞⎜⎝
⎛ −=⎟⎟⎠
⎞⎜⎜⎝
⎛−=
Where:
Utile height : d = h – (0.06h + ebz) = 0.57 m
The moment resistance Mrb = 297 KNm
Because kNmMkNmM rbEcq 297205 =<= the supposition of having no compressed reinforcement is correct.
Calculation of reinforcement area with max constraint on steel and concrete
The reinforcement area is calculated using the SLS load combination
Neutral axis position: 4300α1 ,=
Lever arm: m..*.*dzc 4850343001570
3α1 1 =⎟
⎠
⎞⎜⎝
⎛ −=⎟⎟⎠
⎞⎜⎜⎝
⎛−=
Reinforcement section: ²cm.*.
.*z
MAsc
serser,s 5610
40048502050
σ1 ===
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Reference solution for minimal reinforcement area
The minimum percentage for a rectangular beam in pure bending is:
⎪⎩
⎪⎨
⎧=
d*b*.
d*b*f
f*.
MaxA
w
wyk
eff,ct
min,s00130
260
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2
Where:
MPa.ff ctmeff,ct 8962== from cracking conditions
Therefore:
⎪⎩
⎪⎨⎧
==
==
−
−²cm.
²m*..*.*.
²m*..*.*..maxA min,s 4021007257028000130
104025702805008962260
4
4
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
ULS and SLS load combinations (kNm)
Simply supported beam subjected to bending
SLS (reference value: 205kNm)
Theoretical reinforcement area (cm2)
(reference value: 10.50cm2)
ADVANCE DESIGN VALIDATION GUIDE
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Minimum reinforcement area (cm2)
(reference value: 2.40cm2)
5.26.2.3 Reference results
Result name Result description Reference value My,SLS My corresponding to the 102 combination (SLS) [kNm] 205 kNm Az Theoretical reinforcement area [cm2] 10.50 cm2
Amin Minimum reinforcement area [cm2] 2.39 cm2
5.26.3 Calculated results
Result name
Result description Value Error
My My SLS -200.533 kN*m 2.0833 % Az Az -10.2949 cm² 0.5220 % Amin Amin -2.38697 cm² 0.0001 %
ADVANCE DESIGN VALIDATION GUIDE
389
5.27 EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram
Test ID: 4998
Test status: Passed
5.27.1 Description Simple Bending Design for Service Limit State
The purpose of this test is to verify the My resulted stresses for the SLS load combination and the results of the theoretical reinforcement area Az and of the minimum reinforcement percentage.
5.27.2 Background This test performs the verification of the theoretical reinforcement area for a T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.
5.27.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 20 kN/m (including the dead load) ■ Exploitation loadings (category A): Q = 10kN/m, ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering bilinear stress-strain diagram The objective is to verify:
■ The stresses results ■ The theoretical reinforcement area ■ The minimum reinforcement percentage
Simply supported beam
ADVANCE DESIGN VALIDATION GUIDE
390
Units
Metric System
Geometry
Beam cross section characteristics:
■ Beam length: 8m ■ Beam height: h=0.76m ■ Concrete cover: c=4.50 cm ■ Effective height: d=h-(0.6*h+ebz)=0.669 m; d’=ebz=0.045m
Materials properties
Rectangular solid concrete C20/25 and S500A reinforcement steel is used. The following characteristics are used in relation to this material:
■ Exposure class XD1 ■ Concrete density: 16kN/m3 ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering bilinear stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50% ■ Concrete: fck = 20MPa
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X.
■ Inner: None.
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=20+10=30kN/ml
Quasi-permanent combination of actions:
CQP = 1.0 x G + 0.3 x Q=20+0.3*10=23kN/ml
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■ Load calculations:
( ) kNmM cqser 2408
²8*1020, =
+=
( ) kNmM qpser 1848
²8*10*3.020, =
+=
5.27.2.2 Reference results in calculating the concrete final value of the creep coefficient
)()(),( 00 tft cmRH ββϕϕ =∞
Where:
MPaf
fcm
cm 17.3820
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ
t0 : concrete age t0=28days
210
3**
*1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
121 == αα if Mpafcm 35≤
If not
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
In this case MpaMpaMpaff ckcm 35288 ≤=+= therefore
121 == αα In this case:
Humidity RH=50 %
mmuAch 24.162
3920318000*22
0 ===
97.2488.0*17.3*92.1)(*)(*),(92.124.162*1.0
100501
1 003===∞⇒=
−+= tft cmRHRH ββϕϕϕ
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Calculating the equivalence coefficient:
The coefficient of equivalence is determined by the following formula:
Ecar
Eqp
cm
se
MM
t
EE
*),(1 0∞+
=
ϕ
α
Where:
MPafE ckcm 29962
10820*22
108*22
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
MPaEs 200000=
28.3240184*97.21*),(1 0 =+=∞+
Ecar
Eqp
MM
tϕ
89.21
240184*97.21
2962200000
*),(1 0
=
+
=
∞+
=
Ecar
Eqp
cm
se
MM
t
EE
ϕ
α
Material characteristics:
The maximum compression on the concrete is: Mpafckbc 1220*6,0*6,0 ===σ
For the maximum stress on the steel taut, we consider the constraint limit Mpaf yks 400*8,0 ==σ
Neutral axis position calculation; Calculation of Mtser:
MNmhbhd
hd
M fefff
f
e
stser 122,010,0*20,1*
10,0669,0310,0669,0
*77.43
400**3**2
22 =−
−=
−
−=
ασ
kNmMkNmM tserser 122240 =>= the neutral axes is on the beam body
Concrete compressive stresses
MPahdhb
Mf
feff
serm 23,3
)210,0669,0(*10,0*20,1
240,0
)2
(**=
−=
−=σ
MPahd
de
s
f
e
sm
c 96,489.21
400
210,0669,0
89.2140023,3
*669,0
2
* =−⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
+=−
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
−
+=
ασα
σσσ
MPaMPa cc 1296.4 =<= σσ => there is no compressed reinforcement
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393
The calculation of the tension reinforcement theoretical section As1
cmdxsce
ce 30.14669,0*40096,4*89.21
96,4*89.21*
*1 =
+=
+=
σσασα
MNxbN ceffc 426,0
296,4*1430,0*20,1
2** 11 ===
σ
m.MN,,*,z*NM cc 265062204260111 === and m,,,zc 62203
1430066901 =−=
²65,10400*622,0
265,0*1
11 cmzMA
scs ===
σ
Calculation of the steel compressed reinforcement As2:
m,,,hxx f 043001001430012 =−=−=
MPa,,
,*,x
x*cc 491
1430004300964σσ
1
22 ===
MN,,*,*),,(x*)bb(N cweffc 0290
24910430030201
2σ 2
22 =−=−=
MNm,,*,z*NM cc 016055500290222 ===
with m,,,,zc 55503
0430010066902 =−−=
²cm,*,
,*z
MAsc
s 7204005550
0160σ2
22 ===
Notions of serviceability moment M0 :
²cm...AAA sss 9397206510210 =−=−=
m.MN,,,MMM 249001602650210 =−=−=
Theoretical steel reinforcement section :
²cm,,
,*,M
M*AA serss 589
24902400939
0
0 ===
Reference solution for minimal reinforcement area
The minimum percentage for a rectangular beam in pure bending is:
⎪⎩
⎪⎨
⎧=
d*b*.
d*b*f
f*.
MaxA
w
wyk
eff,ct
min,s00130
260
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2
Where:
MPa.ff ctmeff,ct 212== from cracking conditions
Therefore:
⎪⎩
⎪⎨⎧
==
==
−
−²cm.
²m*..*.*.
²m*..*.*.*.maxA min,s 61210612669030000130
103126690300500
2122604
4
ADVANCE DESIGN VALIDATION GUIDE
394
Finite elements modeling
■ Linear element: S beam, ■ 9 nodes, ■ 1 linear element.
SLS load combinations(kNm)
Simply supported beam subjected to bending
SLS (reference value: 240kNm)
Theoretical reinforcement area (cm2)
(reference value: As=9.58cm2)
Minimum reinforcement area (cm2)
(reference value: 2.57cm2)
5.27.2.3 Reference results
Result name Result description Reference value My,SLS My corresponding to the 102 combination (SLS) [kNm] 240 kNm Az Theoretical reinforcement area [cm2] 9.58 cm2
Amin Minimum reinforcement area [cm2] 2.57 cm2
5.27.3 Calculated results
Result name
Result description Value Error
My My SLS -240 kN*m 0.0000 % Az Az -9.64217 cm² -0.0000 % Amin Amin -2.574 cm² -0.0000 %
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5.28 EC2 Test 16: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram
Test ID: 4999
Test status: Passed
5.28.1 Description Simple Bending Design for Service State Limit
The purpose of this test is to verify the My resulted stresses for the SLS load combination and the results of the theoretical reinforcement area Az and of the minimum reinforcement percentage. This test performs verification for the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.
5.28.2 Background This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.
5.28.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 40 kN/m (including the dead load) ■ Exploitation loadings (category A): Q = 10kN/m, ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering bilinear stress-strain diagram The objective is to verify:
■ The stresses results ■ The theoretical reinforcement area ■ The minimum reinforcement percentage
Simply supported beam
ADVANCE DESIGN VALIDATION GUIDE
396
Units
Metric System
Geometry
Beam cross section characteristics:
■ Beam length: 8m ■ Beam height: h=0.67m ■ Concrete cover: c=4.50 cm ■ Effective height: d=h-(0.6*h+ebz)=0.585 m; d’=ebz=0.045m
Materials properties
Rectangular solid concrete C20/25 and S500A reinforcement steel is used. The following characteristics are used in relation to this material:
■ Exposure class XD1 ■ Concrete: fck = 20MPa ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering bilinear stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50%
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X.
■ Inner: None.
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=40+10=50kN/ml
Quasi-permanent combination of actions:
CQP = 1.0 x G + 0.3 x Q=40+0.3*10=43kN/ml
■ Load calculations: ( ) kNm²*M cq,ser 400
881040
=+
=
( ) kNm²**.M qp,ser 3448
8103040=
+=
ADVANCE DESIGN VALIDATION GUIDE
397
5.28.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 tft cmRH ββϕϕ =∞
Where:
MPaf
fcm
cm 17.3820
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ
t0 : concrete age t0=28days
210
3**
*1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
if
If not
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
In this case,
MpaMpaMpaff ckcm 35288 ≤=+= , therefore:
121 == αα In this case:
Humidity RH=50 %
mmuAch 123
3140192600*22
0 ===
11.3488.0*17.3*2)(*)(*),(2123*1.0
100501
1 003===∞⇒=
−+= tft cmRHRH ββϕϕϕ
Calculating the equivalence coefficient:
The coefficient of equivalence is determined by the following formula:
Ecar
Eqp
cm
se
MM
*)t,(
EE
01
α
∞+
=
Where:
MPa*f*E..
ckcm 29962
1082022
10822
3030=⎟
⎠
⎞⎜⎝
⎛ +=⎟
⎠
⎞⎜⎝
⎛ +=
MPaEs 200000=
68340034411311 0 .*.
MM
*)t,(Ecar
Eqp =+=∞+
ADVANCE DESIGN VALIDATION GUIDE
398
54.24
400344*11.31
29962200000
*),(1 0
=
+
=
∞+
=
Ecar
Eqp
cm
se
MM
t
EE
ϕ
α
Material characteristics:
The maximum compression on the concrete is:
For the maximum stress on the steel taut, we consider the constraint limit
Mpaf yks 400*8,0 ==σ
Neutral axis position calculation; Calculation of Mtser:
MNmhbhd
hd
M fefff
f
e
stser 083,010,0*90.10*
10,0585,0310,0585,0
*54.24*2
400**3**2
22 =−
−=
−
−=
ασ
kNmMkNmM tserser 83400 =>= the neutral axes is on the beam body
Concrete compressive stresses
MPahdhb
Mf
feff
serm 31.8
)210,0585,0(*10,0*90.0
400,0
)2
(**=
−=
−=σ
MPahd
de
s
f
e
sm
c 61.1054.24
400
210,0585,0
54.2440031.8
*585,0
2
* =−⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
+=−
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
−
+=
ασα
σσσ
MPaMPa cc 1261.10 =<= σσ => there is no compressed reinforcement
The calculation of the tension reinforcement theoretical section As1
and
ADVANCE DESIGN VALIDATION GUIDE
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Calculation of the steel compressed reinforcement As2:
mhxx f 1306,010,02306,012 =−=−=
MPaxxc
c 01.62306,0
1306,0*61.1*
1
22 ===
σσ
MNxbbN cweffc 282,0
201.6*1306,0*)18,090.0(
2*)( 2
22 =−=−=σ
MNmzNM cc 125,0441,0*282,0* 222 ===
with mzc 441,03
1306,010,0585,02 =−−=
²06.7400*441,0
125,0*2
22 cm
zMA
scs ===
σ
Notions of serviceability moment M0 :
Theoretical steel reinforcement section :
²83.18435,0
400,0*46.20*
0
0 cmMMAA sers
s ===
Reference solution for minimal reinforcement area
The minimum percentage for a rectangular beam in pure bending is:
⎪⎩
⎪⎨⎧
=db
dbff
MaxA
w
wyk
effct
s
**0013.0
***26.0 ,
min,
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2
Where:
MPaff ctmeffct 21.2, == from cracking conditions
Therefore:
⎪⎩
⎪⎨⎧
==
==−
−
²37.1²10*37.1585.0*18.0*0013.0
²10*21.1585.0*18.0*500
21.2*26.0max4
4
min, cmm
mAs
Finite elements modeling
■ Linear element: S beam, ■ 9 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
400
SLS load combinations(kNm)
Simply supported beam subjected to bending
SLS (reference value: 400kNm)
Theoretical reinforcement area(cm2)
(reference value: As=18.83cm2)
Minimum reinforcement area(cm2)
(reference value: 1.37cm2)
5.28.2.3 Reference results
Result name Result description Reference value My,SLS My corresponding to the 102 combination (SLS) [kNm] 400 kNm Az Theoretical reinforcement area [cm2] 18.83 cm2
Amin Minimum reinforcement area [cm2] 1.37 cm2
5.28.3 Calculated results
Result name
Result description Value Error
My My SLS -400 kN*m 0.0000 % Az Az -18.8948 cm² 0.0001 % Amin Amin -1.36843 cm² -0.0001 %
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5.29 EC2 Test 11: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement- Bilinear stress-strain diagram (Class XD1)
Test ID: 4983
Test status: Passed
5.29.1 Description Simple Bending Design for Service State Limit
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
5.30 EC2 Test 19: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1)
Test ID: 5033
Test status: Passed
5.30.1 Description Simple Bending Design for Serviceability State Limit
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section; the maximum spacing of cracks and the crack openings are verified.
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5.31 EC2 Test 20: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1)
Test ID: 5034
Test status: Passed
5.31.1 Description Simple Bending Design for Serviceability State Limit
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section; the maximum spacing of cracks and the crack openings are verified.
5.31.2 Background Simple Bending Design for Serviceability State Limit
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses will be made along with the calculation of compressing stresses in concrete section σc and compressing stresses in the steel reinforcement section σs; maximum spacing of cracks and the crack openings.
5.31.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 30 kN/m (including the dead load) ■ Exploitation loadings (category A): Q = 37.5 kN/m, ■ Structural class: S4 ■ Reinforcement steel ductility: Class B The objective is to verify:
■ The stresses results ■ The compressing stresses in concrete section σc ■ The compressing stresses in the steel reinforcement section σs. ■ The maximum spacing of cracks ■ The crack opening
Simply supported beam
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Units
Metric System
Geometry
Beam cross section characteristics:
■ Height: h = 0.80 m, ■ Width: b = 0.40 m, ■ Length: L = 8.00 m, ■ Section area: A = 0.32 m2 , ■ Concrete cover: c=4.5cm ■ Effective height: d=71cm;
Materials properties
Rectangular solid concrete C25/30 is used. The following characteristics are used in relation to this material:
■ Exposure class XD1 ■ Stress-strain law for reinforcement: Bilinear stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50%
■ Characteristic compressive cylinder strength of concrete at 28 days: Mpafck 25=
■ Characteristic yield strength of reinforcement: Mpafyk 500=
■ ²cm.Ast 1630= for 3 beds of 5HA16
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 8) restrained in translation along Y and Z
■ Inner: None.
Loading
The beam is subjected to the following load combinations:
■ Load calculations: ► M0Ed = 774 kNm ► Mcar = 540 kNm ► Mfq = 390 kNm ► Mqp = 330 kNm
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5.31.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 tft cmRH ββϕϕ =∞
Where:
MPaf
fcm
cm 92.2825
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ
t0 : concrete age t0=28days
210
3**
*1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
If Mpafcm 35≤ ,
121 == αα If not,
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
In this case,
MpaMpaMpaff ckcm 35338 ≤=+= , therefore 121 == αα
In this case:
Humidity RH=50 %
( ) mmuAch 67.266
800400*2800*400*22
0 =+
==
55.2488.0*92.2*78.1)(*)(*),(78.167.266*1.0
100501
1 003===∞⇒=
−+= tft cmRHRH ββϕϕϕ
The coefficient of equivalence is determined by the following formula:
Under quasi-permanent combinations:
),(1 0tEEcm
se
∞+
=
ϕ
α
Where:
MPafE ckcm 31476
10825*22
108*22
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
MPaEs 200000=
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55.2),( 0 =∞ tϕ
56.22
55.2131476200000
),(1 0
=
+
=
∞+
=
tEEcm
se
ϕ
α
Material characteristics:
The maximum compression on the concrete is: Mpafckbc 1525*6,06,0 ===σ
For the maximum stress on the steel taut, we consider the constraint limit Mpaf yks 400*8,0 ==σ
Neutral axis position calculation:
Neutral axis equation: 0αα21
111 =−+−− )'dx(**A)xd(**A²x*b* escestw
w
stscewstscestsce
b)A*dA'*d(**b*)²AA(*)AA(*
x+++++−
=α2αα 2
1
By simplifying the previous equation, by considering 0=scA , it will be obtained:
cm.**.**².*²..*.b
)A*d**b*²A*A*x stewsteste
3540
16307156224021630562216305622
α2αα 2
1
=++−
=
=++−
=
Calculating the second moment:
443
11
31
01450350071056221016303
350040
αα3
m.)²..(*.**,.*.
)²'dx(**A)²xd(**Ax*bI escestw
=⎥⎥⎦
⎤
⎢⎢⎣
⎡−+=
=⎥⎥⎦
⎤
⎢⎢⎣
⎡−+−+=
−
Stresses calculation:
MpaxIMser
c 96.7350,0*0145,0330,0* 1 ===σ
MpaMpaxxd
scest 4007.184350,0
350,071,0*96.7*56.22**1
1 =≤=−
=−
= σσασ
Maximum spacing of cracks:
Bottom reinforcement 3HA20+3HA16=15.46cm2
2, 06.015.0*4.0
4.028.0
15.03
)350.08.0(255.0)71.08.0(*5.2
min*40.0
2
3)(
)(*5.2
min* m
h
xhdh
bA effc ==
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
=
=−
=−
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
=−
−
=
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(2); Figure 7.1
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0500.006.0
10*16.30 4
,, ===
−
effc
seffp A
Aρ
mmnnnn
eq 16****
2211
222
211 =
++
=φφφφφ
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(3)
effpr
kkcks,
213max,
***425.0*ρ
φ+=
Where:
c=0.051m
113.25125*4.325*4.3
3/23/2
3 =⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=c
k
Therefore:
mmkkckseffp
r 162050.0
)10*16(*5.0*8.0*425.0051.0*113.2***425.0*3
,
213max, =+=+=
−
ρφ
Calculation of average strain:
41.631187200000
===cm
se E
Eα
44,
,
,
10*54.5*6,010*88.7200000
)050.0*41.61(*050.056.2*4.071.184).1(**
−− =≥=+−
=+−
=−s
s
s
effpeeffp
effctts
cmsm EE
fk
σρα
ρσ
εε
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(2)
Calculation of crack widths:
mmsw cmsmrk 128.0)1088.7(*162.0)(* 4max, =×=−= −εε
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(1)
For an exposure class XD1, using the French national annex, we retained an opening crack of 0.20mm max.
This criterion is satisfied.
Finite elements modeling
■ Linear element: S beam, ■ 11 nodes, ■ 1 linear element.
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SLS load combinations (kNm) and stresses (MPa)
Simply supported beam subjected to bending
ULS (reference value: MEd=774kNm)
SLS characteristic (reference value: Mser-cq=540kNm)
SLS quasi-permanent (reference value: Mser-qp=330kNm)
Compressing stresses in concrete section σc-qp
(reference value: σc-qp =7.96MPa )
Compressing stresses in the steel reinforcement section σs-qp
(reference value: σs-qp=185MPa)
Maximum cracking space Sr,max
(reference value: Sr,max=162mm)
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Maximum crack opening Wk
(reference value: Wk=0.128mm)
5.31.2.3 Reference results
Result name Result description Reference value MEd My corresponding to the 102 combination (ULS) [kNm] 774 kNm Mser-cq My corresponding to the 104 combination (SLS) [kNm] 540 kNm Mser-qp My corresponding to the 108 combination (SLS) [kNm] 330 kNm σc Compressing stresses in concrete section σc [MPa] 7.96 MPa
σs Compressing stresses in the steel reinforcement section σs [MPa] 185 Mpa
Sr,max Maximum cracking space Sr,max [mm] 162 mm
Wk Maximum crack opening Wk 0.128 mm
5.31.3 Calculated results
Result name
Result description Value Error
My My ULS -774 kN*m 0.0000 % My My SLS cq -540 kN*m 0.0000 % My My SLS qp -330 kN*m 0.0000 % Sc QP Sc QP 7.76924 MPa 0.0000 % Ss QP Ss QP -181.698 MPa 0.0001 % Sr,max Sr,max 16.1577 cm -0.0003 % wk Wk -0.0125137 cm 0.0000 %
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5.32 EC2 Test 14: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram (Class XD1)
Test ID: 4987
Test status: Passed
5.32.1 Description Simple Bending Design for Service State Limit - Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending.
During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
The verification of the bending stresses at service limit state is performed.
5.32.2 Background Simple Bending Design for Service State Limit
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
5.32.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Linear loads: Loadings from the structure: G = 50 kN/m + dead load,
Exploitation loadings (category A): Q = 60kN/m,
■ Punctual loads G=30kN
Q=25kN
3,02 =ψ
Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q
Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q
Simply supported beam
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.80 m, ■ Width: b = 0.40 m, ■ Length: L = 6.30 m, ■ Section area: A = 0.320 m2 , ■ Concrete cover: c=4.5cm ■ Effective height: d=h-(0.6*h+ebz)=0.707m; d’=ebz=0.045m
Materials properties
Rectangular solid concrete C25/30 and S500A reinforcement steel is used. The following characteristics are used in relation to this material:
■ Exposure class XD1 ■ Concrete density: 25kN/m3 ■ Stress-strain law for reinforcement: Bilinear stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50%
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.80) restrained in translation along Y, Z and restrained in rotation along X.
■ Inner: None.
Loading
The beam is subjected to the following load combinations:
■ Dead load: 0.40*0.80*25 = 8kN/ml
■ Load combinations: Characteristic combination of actions:
CCQ = 1.0 x G + 1.0 x Q=8+50+60=118kN/ml
Quasi-permanent combination of actions:
CQP = 1.0 x G + 0.3 x Q=8+50+0.3*60=76kN/ml
■ Load calculations:
( ) ( ) mkNM cqser .05.6728
²3.6*605084
3.6*2530, =
+++
+=
( ) ( ) mkNM qpser .11.4368
²3.660*3.05084
3.6*25*3.030, =
×+++
+=
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5.32.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 tft cmRH ββϕϕ =∞
Where:
MPaf
fcm
cm 92.2825
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ
t0 : concrete age t0=28days
210
3**
*1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
121 == αα if Mpafcm 35≤
If not
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
In this case MpaMpaMpaff ckcm 35338 ≤=+= therefore
121 == αα In this case:
Humidity RH=50 %
( ) mmuAch 267
800400*2800*400*22
0 =+
==
54.2488.0*92.2*78.1)(*)(*),(78.1267*1.0
100501
1 003===∞⇒=
−+= tft cmRHRH ββϕϕϕ
Calculating the equivalence coefficient:
The coefficient of equivalence is determined by the following formula:
Ecar
Eqp
cm
se
MM
t
EE
*),(1 0∞+
=
ϕ
α
Where:
MPafE ckcm 31476
10825*22
108*22
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
MPaEs 200000=
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65.2672435*54.21*),(1 0 =+=∞+
Ecar
Eqp
MM
tϕ
82.16
672435*54.21
31476200000
*),(1 0
=
+
=
∞+
=
Ecar
Eqp
cm
se
MM
t
EE
ϕ
α
Material characteristics:
The maximum compression on the concrete is: Mpafckbc 1515*6,0*6,0 ===σ
For the maximum stress on the steel taut, we consider the constraint limit Mpaf yks 400*8,0 ==σ
Neutral axis position calculation:
The position of the neutral axis must be determined by calculating (position corresponding to the state of maximum stress on the concrete and reinforcement):
387.040018*82.16
15*82.16*
*1 =
+=
+=
sce
ce
σσασαα
Moment resistance calculation:
Knowing the 1α value, it can be determined the moment resistance of the concrete section, using the following formulas:
273.0707.0*387.0*11 === dx α m
MNmxdxbM cwrb 505.03273.0707.0*15*273.0*4.0*5.0)
3(****
21 1
1 =⎟⎠⎞
⎜⎝⎛ −=−= σ
Where:
Utile height : d = h – (0.06h + ebz) = 0.707m
The moment resistance Mrb = 505KNm
Because kNmMkNmM rbEcq 505672 =>= , the supposition of having no compressed reinforcement is
incorrect.
The calculation of the tension reinforcement theoretical section A1
mxdzc 616.03273.0707.0
31 =−=−=
²51.20400*616.0
505.0*1 cm
zMA
sc
rb ===σ
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Stress calculation for steel reinforcement σsc:
064.0707.0045.0'' ===
ddδ
MPacesc 78.210387.0
067.0387.0*15*82.16'**1
1 =−
=−
=α
δασασ
Calculation of the steel compressed reinforcement A’:
²96.1178.210*)045.0707.0(
505.0672.0*)'(
' cmddMMA
sc
rbser =−
−=
−−
=σ
Calculation of the steel tensioned reinforcement A2:
²30.6400
78.210*96.11'*2 cmAAs
sc ===σσ
Calculation of the steel reinforcement :
²81.2630.651.2021 cmAAA =+=+=
Reference solution for minimal reinforcement area
The minimum percentage for a rectangular beam in pure bending is:
⎪⎩
⎪⎨⎧
=db
dbff
MaxA
w
wyk
effct
s
**0013.0
***26.0 ,
min,
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2
Where:
MPaff ctmeffct 56.2, == from cracking conditions
Therefore:
⎪⎩
⎪⎨⎧
==
==−
−
²76.3²10*68.3707.0*40.0*0013.0
²10*76.3707.0*40.0*500
56.2*26.0max4
4
min, cmm
mAs
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
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ULS and SLS load combinations (kNm)
Simply supported beam subjected to bending
SLS (reference value: 672kNm)
Theoretical reinforcement area (cm2)
(reference value: As=26.81cm2; A’=11.96cm2)
Minimum reinforcement area (cm2)
(reference value: 3.76cm2)
5.32.2.3 Reference results
Result name Result description Reference value My,SLS My corresponding to the 102 combination (SLS) [kNm] 672 kNm Az Theoretical reinforcement area [cm2] 26.81 cm2
Amin Minimum reinforcement area [cm2] 3.77 cm2
5.32.3 Calculated results
Result name Result description Value Error My My SLS -658.615 kN*m 1.9964 % Az Az -26.5647 cm² 0.9099 % Amin Amin -3.77193 cm² -0.0001 %
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5.33 EC2 Test 18: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement - Bilinear stress-strain diagram (Class XD1)
Test ID: 5011
Test status: Passed
5.33.1 Description Simple Bending Design for Serviceability State Limit - Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending.
During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section.
5.33.2 Background Simple Bending Design for Serviceability State Limit
Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of compressing stresses in concrete section σc and compressing stresses in the steel reinforcement section σs.
5.33.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 37.5 kN/m (including the dead load), ■ Exploitation loadings (category A): Q = 37.5 kN/m, ■ Structural class: S4 ■ Reinforcement steel ductility: Class A ■ For the stress calculation the French annexes was used The objective is to verify:
■ The stresses results ■ The compressing stresses in concrete section σc ■ The compressing stresses in the steel reinforcement section σs.
Simply supported beam
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.80 m, ■ Width: b = 0.35 m, ■ Length: L = 8.00 m, ■ Section area: A = 0.28 m2 , ■ Concrete cover: c=4 cm ■ Effective height: d=72cm;
Materials properties
Rectangular solid concrete C25/30 is used. The following characteristics are used in relation to this material:
■ Exposure class XD1 ■ Stress-strain law for reinforcement: Bilinear stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50%
■ Characteristic compressive cylinder strength of concrete at 28 days: Mpafck 25=
■ Characteristic yield strength of reinforcement: Mpafyk 500=
■ ²cm.Ast 7037= for 3 HA20
■ ²cm.Asc 286= for 2 HA10
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 8) restrained in translation along Y and Z
■ Inner: None.
Loading
The beam is subjected to the following load combinations:
■ Load combinations: ► Characteristic combination of actions:
CCQ = 1.0*G+1.0*Q =75 kN/ml
■ Load calculations: ► M0Ed = 855 kNm ► Mcar = 600 kNm ► Mqp = 390 kNm
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5.33.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 tft cmRH ββϕϕ =∞
Where:
MPaf
fcm
cm 92.2825
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ
t0 : concrete age t0=28days
210
3**
*1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
if
If not,
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
In this case,
MpaMpaMpaff ckcm 35338 ≤=+= , therefore 121 == αα
In this case,
Humidity RH=50 %
( ) mmuAch 48.243
800350*2800*350*22
0 =+
==
56.2488.0*92.2*80.1)(*)(*),(80.148.243*1.0
100501
1 003===∞⇒=
−+= tft cmRHRH ββϕϕϕ
The coefficient of equivalence is determined by the following formula:
Ecar
Eqp
cm
se
MM
t
EE
*),(1 0∞+
=
ϕ
α
Where:
MPafE ckcm 31476
10825*22
108*22
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
MPaEs 200000=
664.2600390*56.21*),(1 0 =+=∞+
Ecar
Eqp
MM
tϕ
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90.16
664.231476200000
*),(1 0
==
∞+
=
Ecar
Eqp
cm
se
MM
t
EE
ϕ
α
Material characteristics:
The maximum compression on the concrete is: Mpafckbc 1525*6,06,0 ===σ
For the maximum stress on the steel taut, we consider the constraint limit Mpaf yks 320*8,0 ==σ
Neutral axis position calculation:
Neutral axis equation: 0αα21
111 =−+−− )'dx(**A)xd(**A²x*b* escestw
cm,).*.*(*.**)²..(*².)..(*.
b)A'*dA*d(**b*)²AA(*²)AA(*
x scstewscstescste
413435
286470377290163522867037901628670379016
α2αα1
=+++++−
=
=+++++−
=
Calculating the second moment:
443
11
31
0147201472097445349016286453472901670373
413435
αα3
m.cm)².(*.*,)²,(.*,,*
)²'dx(**A)²xd(**Ax*bI escestw
==⎥⎥⎦
⎤
⎢⎢⎣
⎡−+−+=
=⎥⎥⎦
⎤
⎢⎢⎣
⎡−+−+=
Stresses calculation:
MpaMpaxIM
cser
c 12143441,0*01472,0
600,0* 1 =≥=== σσ
MpaMpaxxd
scest 4002593441,0
3441,072,0*14*90.16**1
1 =≤=−
=−
= σσασ
Finite elements modeling
■ Linear element: S beam, ■ 9 nodes, ■ 1 linear element.
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SLS load combinations (kNm) and stresses (MPa)
Simply supported beam subjected to bending
SLS (reference value: Mscq=600kNm)
Compressing stresses in concrete section σc
(reference value: σc =14MPa )
Compressing stresses in the steel reinforcement section σs
(reference value: σs=260.15MPa)
5.33.2.3 Reference results
Result name Result description Reference value My,SLS My corresponding to the 104 combination (SLS) [kNm] 600 kNm σc Compressing stresses in concrete section σc (MPa) 14 MPa
σs Compressing stresses in the steel reinforcement section σs (MPa) 260.15 MPa
5.33.3 Calculated results
Result name
Result description Value Error
My My SLS -600 kN*m 0.0000 % Sc CQ Sc CQ 14.122 MPa 0.0142 % Ss CQ Ss CQ -260.116 MPa 0.0014 %
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5.34 EC2 Test 24: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Test ID: 5058
Test status: Passed
5.34.1 Description Verifies a rectangular cross section beam made from concrete C20/25 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the cross-sectional area of the shear reinforcement (Asw) calculation.
5.35 EC2 Test 25: Verifying the shear resistance for a rectangular concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Test ID: 5065
Test status: Passed
5.35.1 Description Verifies the shear resistance for a rectangular concrete beam C20/25 with inclined transversal reinforcement - Bilinear stress-strain diagram (Class XC1).
For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) is calculated, along with the cross-sectional area of the shear reinforcement (Asw).
5.36 EC2 Test 23: Verifying the shear resistance for a rectangular concrete - Bilinear stress-strain diagram (Class XC1)
Test ID: 5053
Test status: Passed
5.36.1 Description Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the cross-sectional area of the shear reinforcement (Asw) calculation.
5.37 EC2 Test 13: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1)
Test ID: 4986
Test status: Passed
5.37.1 Description Simple Bending Design for Service State Limit - Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending.
During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.
The verification of the bending stresses at service limit state is performed.
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5.38 EC2 Test 17: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Inclined stress-strain diagram (Class XD1)
Test ID: 5000
Test status: Passed
5.38.1 Description Simple Bending Design for Serviceability State Limit
Verifies the adequacy of a rectangular cross section made from concrete C20/25 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section.
5.38.2 Background Simple Bending Design for Serviceability State Limit
Verifies the adequacy of a rectangular cross section made from concrete C20/25 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of compressing stresses in concrete section σc and compressing stresses in the steel reinforcement section σs.
5.38.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 9.375 kN/m (including the dead load), ■ Mfq = Mcar = Mqp = 75 kNm ■ Structural class: S4 ■ Characteristic combination of actions: CCQ = 1.0 x G ■ Reinforcement steel ductility: Class B The objective is to verify:
■ The stresses results ■ The compressing stresses in concrete section σc ■ The compressing stresses in the steel reinforcement section σs.
Simply supported beam
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.50 m, ■ Width: b = 0.20 m, ■ Length: L = 8.00 m, ■ Section area: A = 0.10 m2 , ■ Concrete cover: c=4.5cm ■ Effective height: d=44cm;
Materials properties
Rectangular solid concrete C20/25 is used. The following characteristics are used in relation to this material:
■ Exposure class XD1 ■ Stress-strain law for reinforcement: Inclined stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50%
■ Characteristic compressive cylinder strength of concrete at 28 days: Mpafck 20=
■ Characteristic yield strength of reinforcement: Mpaf yk 400=
■ ²42,9 cmAst = 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
5.38.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 tft cmRH ββϕϕ =∞
Where:
MPaf
fcm
cm 17.3820
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ
t0 : concrete age t0=28days
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210
3**
*1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
121 == αα if Mpafcm 35≤
If not,
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
In this case, MpaMpaMpaff ckcm 35288 ≤=+=
Therefore 121 == αα
In this case:
Humidity RH=50 %
( ) mmuAch 86.142
500200*2500*200*22
0 =+
==
03.3488.0*17.3*96.1)(*)(*),(96.186.142*1.0
100501
1 003===∞⇒=
−+= tft cmRHRH ββϕϕϕ
The coefficient of equivalence is determined by the following formula:
Ecar
Eqp
cm
se
MM
t
EE
*),(1 0∞+
=
ϕ
α
Where:
MPafE ckcm 29962
10820*22
108*22
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
MPaEs 200000=
03.41*03.31*),(1 0 =+=∞+Ecar
Eqp
MM
tϕ
90.26
1*03.3129962200000
*),(1 0
=
+
=
∞+
=
Ecar
Eqp
cm
se
MM
t
EE
ϕ
α
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Material characteristics:
The maximum compression on the concrete is: Mpa*,f, ckbc 15256060σ ===
For the maximum stress on the steel taut, we consider the constraint limit Mpaf*, yks 32080σ ==
Checking inertia cracked or not:
Before computing the constraints, check whether the section is cracked or not. For this, we determine the cracking moment which corresponds to a tensile stress on the concrete equal to
ctmf :
vIfM ctm
cr*
=
Where:
433
00208,012
50,0*20,012* mhbI ===
mhv 25,02
==
The average stress in concrete is:
Mpaff ckctm 21,220*30.0*30.0 32
32
===
The critical moment of cracking is therefore:
MNmvIfM ctm
cr 018,025,000208,0*21,2*
===
The servility limit state moment is 0.075MNm therefore the cracking inertia is present.
Neutral axis position calculation:
Neutral axis equation: 0)'(**)(**²**21
111 =−+−− dxAxdAxb escestw αα
w
stscewstscestsce
bAdAdbAAAA
x)*'*(***2)²(*)(* 2
1+++++−
=ααα
By simplifying the previous equation, by considering 0=scA , it will be obtained:
w
stewsteste
bAdbAA
x)****2²*)(* 2
1ααα ++−
=
cmx 04,2320
)42,9/44(*90,26*20*2²42,9*²90.2642,9*90.261 =
++−=
Calculating the second moment:
443
1
31 001929,0)²230,044,0(*90.26*10*42,9
3230,0*20,0)²(**
3* mxdAxbI est
w =⎥⎦
⎤⎢⎣
⎡−+=⎥
⎦
⎤⎢⎣
⎡−+= −α
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Stresses calculation:
MpaMpaxIM
cser
c 1294,8230,0*001929,0
075,0* 1 =≤=== σσ
MpaMpaxxd
scest 32057.219230,0
230,044,0*94,8*90.26**1
1 =≤=−
=−
= σσασ
Finite elements modeling
■ Linear element: S beam, ■ 9 nodes, ■ 1 linear element.
SLS load combinations (kNm) and stresses (MPa)
Simply supported beam subjected to bending
SLS (reference value: Mscq=75kNm)
Compressing stresses in concrete section σc
(reference value: σc =8.94MPa )
Compressing stresses in the steel reinforcement section σs
(reference value: σs=218.57MPa)
5.38.2.3 Reference results
Result name Result description Reference value My,SLS My corresponding to the 103 combination (SLS) [kNm] 75 kNm σc Compressing stresses in concrete section σc (MPa) 8.94 MPa
σs Compressing stresses in the steel reinforcement section σs (MPa) 219.67 MPa
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5.38.3 Calculated results
Result name
Result description Value Error
My My SLS -75 kN*m -0.0000 % Sc CQ Sc CQ 8.99322 MPa 0.0359 % Ss CQ Ss CQ -219.639 MPa 0.0006 %
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5.39 EC2 Test28: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class X0)
Test ID: 5083
Test status: Passed
5.39.1 Description Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max. will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, calculation.
The test will not use the reduced shear force value.
5.39.2 Background Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value.
5.39.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete: C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: The dead load will be represented by a linear load of 13.83kN/m
■ Exploitation loadings: The live load will be considered from one linear load of 26.60kN/m
■ Structural class: S1 ■ Reinforcement steel ductility: Class B ■ Exposure class: X0 ■ Concrete density: 25kN/m3
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■ 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.
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Loading:
For a beam under a uniformly distributed load Pu, the shear force is defined by the following equation:
2**)( lPxPxV u
u −=
For the beam end, (x=0) the shear force will be:
kNlPV uEd 9,292
210*57,58
2*
−=−=−=
In the following calculations, the negative sign of the shear will be neglected, as this has no effect in the calculations.
Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreduced shear, that will be used for further calculations.
5.39.2.2 Reference results in calculating the maximum design shear resistance
θθ cot***1
max, +=
tgbzfvV wucd
Rd
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 6.2.3.(3)
In this case:
°= 30θ and °= 90α
1v strength reduction factor for concrete cracked in shear
54.0250251*6.0
2501*6,01 =⎥⎦
⎤⎢⎣⎡ −=⎥⎦
⎤⎢⎣⎡ −= ckfv
mdzu 688.0764.0*9.0*9.0 ===
MNtg
VRd 590.030cot30
22.0*688.0*67.16*54.0max, =
+=
MNVMNV RdEd 590.0293.0 max, =<=
Calculation of transverse reinforcement:
The transverse reinforcement is calculated using the following formula:
mlcmtgfztgV
sA
ywdu
Edsw /²66.5
15,1500*688.0
30*293.0**. =
°==
θ
Finite elements modeling
■ Linear element: S beam, ■ 11 nodes, ■ 1 linear element.
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Advance Design gives the following results for Atz (cm2/ml)
5.39.2.3 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
5.39.3 Calculated results
Result name
Result description Value Error
Fz Fz -292.852 kN -0.0002 % Atz Atz 5.65562 cm² 0.0000 %
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5.40 EC2 Test32: Verifying a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (Class XC1) Test ID: 5102
Test status: Passed
5.40.1 Description Verifies a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (Class XC1). The column made of concrete C25/30. The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed. The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced.
5.40.2 Background Method based on nominal curvature. Verifies the adequacy of a square cross section made from concrete C25/30.
5.40.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: ► 15t axial force ► 1.5tm rotation moment applied to the column top ► The self-weight is neglected
■ Exploitation loadings: ► 6.5t axial force ► 0.7tm rotation moment applied to the column top
► 3,02 =ψ
► The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ► Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ► Concrete cover 5cm ► Transversal reinforcement spacing a=30cm ► Concrete C25/30 ► Steel reinforcement S500B ► The column is considered isolated and braced
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Units
Metric System
Geometry
Beam cross section characteristics:
■ Height: h = 0.40 m, ■ Width: b = 0.40 m, ■ Length: L = 6.00 m, ■ Concrete cover: c = 5 cm
Boundary conditions
The boundary conditions are described below:
The column is considered connected to the ground by a fixed connection and free at the top.
Loading
The beam is subjected to the following load combinations:
■ Load combinations: The ultimate limit state (ULS) combination is:
NEd =1.35*15+1.5*6.5=30t=0.300MN
MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm
■ m...
NM
eEd
Ed 10030000310
0 ===
5.40.2.2 Reference results in calculating the concrete column
Geometric characteristics of the column:
The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
mll 12*20 ==
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b)
Calculating the slenderness of the column:
10440.0
12*32*32 0 ===alλ
Effective creep coefficient calculation:
The creep coefficient is calculated using the next formula:
( )Ed
EQPef M
Mt ., 0∞= ϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
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Where:
( )0,t∞ϕ creep coefficient
EQPM serviceability firs order moment under quasi-permanent load combination
EdM ULS first order moment (including the geometric imperfections)
First order eccentricity evaluation:
ieee += 01
mei 03.0=
The first order moment provided by the quasi-permanent loads:
meNM
eee iEqp
Eqpi 13.030.0
65.0*30.01570.0*30.050.1
0
001 =+
++
=+=+=
tNEqp 95.1670.0*30.050.11 =+=
MNmtmeNM EqpEqp 022.020.213.0*95.16* 111 ====
The first order ULS moment is defined latter in this example:
The creep coefficient ( )0,t∞ϕ is defined as follows:
)(*)(*),( 00 tft cmRH ββϕϕ =∞
92.2825
8.168.16)( =+
==cm
cm ffβ
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ (for t0= 28 days concrete age).
30*1.0
1001
1h
RH
RH
−+=ϕ
( ) 85.1200*1.0
100501
1200400400*2400*400*2*2
30 =−
+=⇒=+
== RHmmuAch ϕ
64.2488.0*92.2*85.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ
The effective creep coefficient calculation:
( ) 49.1039.0022.0*64.2*, 0 ==∞=
Ed
EQPef M
Mtϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
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The necessity of buckling calculation (second order effect):
For an isolated column, the slenderness limit verification is done using the next formula:
nCBA ***20
lim =λ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)
Where:
112.067.16*²40.0
300.0*
===cdc
Ed
fANn
( ) 77.049.1*2.01
1*2,01
1=
+=
+=
ef
Aϕ
1.1*21 =+= ωB because the reinforcement ratio in not yet known
70.07,1 =−= mrC because the ratio of the first order moment is unknown
43.35112.0
7.0*1.1*77.0*20lim ==λ
43.35104 lim =>= λλ
Therefore, the second order effects most be considered.
The second order effects; The buckling calculation:
The stresses for the ULS load combination are:
NEd= 1.35*15 + 1.50*6.50= 30 t = 0,300MN
MEd= 1.35*1.50 + 1.50*0.7= 3.075 t = 0,031MNm
Therefore, it must be determined:
■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:
mNMeEd
Ed 10.0300.0031.0
0 ===
Additional eccentricity:
mlei 03.040012
4000 ===
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The first order eccentricity: stresses correction:
The forces correction, used for the combined flexural calculations:
MNNEd 300.0=
meee i 13.001 =+=
MNmNeM EdEd 039.0300.0*)03.010.0(*1 =+==
0*eNM Ed=
⎩⎨⎧
=⎪⎩
⎪⎨⎧
=⎪⎩
⎪⎨⎧
== mmmmmmmm
mmhmm
e 203.13
20max
3040020
max30
20max0
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4)
Reinforcement calculation in the first order situation:
To play the nominal rigidity method a starting section frame is needed. For this it will be used a concrete section considering only the first order effects.
Advance Design iterates as many time as necessary.
The needed frames will be determined assuming a compound bending with compressive stress. All the results above were obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid of the tensioned steel.
MNmhdNMM Gua 084.0240.035.0*300.0039.0)
2(*0 =⎟
⎠⎞
⎜⎝⎛ −+=−+=
Verification if the section is partially compressed:
496,0)35,040,0*4,01(*
35,040,0*8,0)*4,01(**8,0 =−=−=
dh
dh
BCμ
103.067.16*²35.0*40.0
084.0*²*
===cdw
uacu fdb
Mμ
BCcu μμ =<= 496.0103.0 therefore the section is partially compressed.
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Calculations of steel reinforcement in pure bending:
Calculations of steel reinforcement in combined bending:
For the combined bending:
24 06.178.434
300.010*84.5' cmfNAAyd
−=−=−= −
The minimum reinforcement percentage:
2min, 02.3*002.0²69.0
78.434300.0*10.0*10,0 cmAcm
fNA c
yd
Eds =≥===
The reinforcement will be 4HA10 representing a 3.14cm2 section
The second order effects calculation:
The second order effect will be determined by applying the method of nominal curvature:
Calculation of nominal curvature:
Considering a symmetrical reinforcement 4HA10 (3.14cm ²), the curvature can be determined by the following formula:
0
1**1r
KKr r ϕ=
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.8.3 (1)
With:
1
0
0138.035.0*45.0
20000078.434
*45,0*45,01 −==== m
dEf
drs
yd
ydε
rK : is a correction factor depending on axial load,=> 1≤−−
=balu
ur nn
nnK
112.067.16*²40.0
300.0*
===cdc
Ed
fANn
0512.067.16*²40.0
78.434*10.14,3** 4
===−
cdc
yds
fAfA
ω
0512.10512.011 =+=+= ωun
4,0=baln
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1144.140.00512.1112.00512.1
=⇒≤=−−
= rr KK
ϕK : is a factor for taking account of creep=> 1*1 ≥+= efK ϕβϕ
218.0150104
2002535.0
15020035,0 −=−+=−+=
λβ ckf
11675.049.1*218.01*1 =⇒≥=−=+= ϕϕ ϕβ KK ef
Therefore the curvature becomes:
1
0
0138.01**1 −== mr
KKr r ϕ
Calculation moment:
20 MMM EdEd +=
Where:
EdM 0 : first order moment including the geometrical imperfections.
2M : second order moment
The second order moment must be calculated from the curvature:
22 *eNM Ed=
mcl
re 199.0
10²12*0138.0*1 2
02 ===
MNmeNM Ed 0597.0199.0*300.0* 22 ===
MNmMMM EdEd 099.00597.0039.020 =+=+=
The frame must be sized corresponding the demands of the second order effects:
MNNEd 300.0=
MNmMEd 099.0=
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Reinforcement calculation corresponding to the second order:
The input parameters in the diagram are:
093.067.16*²40.0*40.0
099.0** 2 ===
cd
Ed
fhbMμ
112.067.16*40.0*40.0
300.0**
===cd
Ed
fhbNv
150ω .= ; => ∑ === ²cm..
.*².*.f
f*h*b*A
yd
cds 209
784346716400150ω ; which means 4.60cm2 per side.
Buckling checking
The verification will be made considering the reinforcement found previously (9.20cm2)
The curvature evaluation:
The frames have an influence on the rK parameter only:
1
0
0138.01 −= mr
1≤−−
=balu
ur nn
nnK
112.0=n
15.067.16*²40.0
78.434*10*20,9** 4
===−
cdc
yds
fAfA
ω
15.115.011 =+=+= ωun
4,0=baln
11384.140.015.1112.015.1
=⇒≤=−
−= rr KK
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1=ϕK : coefficient that takes account of the creep => 1*1 ≥+= efK ϕβϕ
The curvature becomes:
1
0
0138.01**1 −== mr
KKr r ϕ
It was obtained the same curvature and therefore the same second order moment, which validates the section reinforcement found.
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
Theoretical reinforcement area (cm2)
(reference value: 9.20cm2)
Theoretical value (cm2)
(reference value: 9.28 cm2)
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5.40.2.3 Reference results
Result name Result description Reference value Az Reinforcement area [cm2] 4.64 cm2
R Theoretical reinforcement area [cm2] 9.28 cm2
5.40.3 Calculated results
Result name
Result description Value Error
Az Az -4.64 cm² -0.0000 %
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5.41 EC2 Test29: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Inclined stress-strain diagram (Class XC1)
Test ID: 5092
Test status: Passed
5.41.1 Description Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max., will be determined, along with the cross-sectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value.
The beam model was provided by Bouygues.
5.41.2 Background Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with the cross-sectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value.
The beam model was provided by Bouygues.
5.41.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete: C35/40 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: The dead load will be represented by two linear loads of 22.63kN/m and 47.38kN/m
■ Exploitation loadings: The live load will be considered from one linear load of 80.00kN/m
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■ Structural class: S3 ■ Reinforcement steel ductility: Class B ■ Exposure class: XC1
■ Characteristic compressive cylinder strength of concrete at 28 days:
■ Partial factor for concrete: ■ Relative humidity: RH=50% ■ Concrete age: t0=28days ■ Design value of concrete compressive strength: fcd=23 MPa ■ Secant modulus of elasticity of concrete: Ecm=34000 MPa
■ Concrete density: ■ Mean value of axial tensile strength of concrete: fctm=3.2 MPa
■ Final value of creep coefficient:
■ Characteristic yield strength of reinforcement: ■ Steel ductility: Class B ■ K coefficient: k=1.08
■ Design yield strength of reinforcement: ■ Design value of modulus of elasticity of reinforcing steel: Es=200000MPa
■ Steel density:
■ Characteristic strain of reinforcement or prestressing steel at maximum load:
■ Strain of reinforcement or prestressing steel at maximum load: ■ Slenderness ratio: 80.0=λ The objective is to verify:
■ The of shear stresses results ■ The longitudinal reinforcement corresponding to a 5cm concrete cover ■ The transverse reinforcement corresponding to a 2.7cm concrete cover
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Length: L = 10.00 m, ■ Stirrup slope: α= 90° ■ Strut slope: θ=29.74˚
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 10.00) restrained in translation along Y, Z and rotation along X
■ Inner: None.
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Loading:
Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreduced shear, that will be used for further calculations.
5.41.2.2 Reference results in calculating the longitudinal reinforcement For a 5 cm concrete cover and dinit=1.125m, the reference value will be 108.3 cm2:
For a 2.7 cm concrete cover and dinit=1.148m, the reference value will be 105.7 cm2:
5.41.2.3 Reference results in calculating the transversal reinforcement
ywdu
Edsw
fztgV
sA
**. θ
=
Where:
mdzu 0125.1125.1*9.0*9.0 ===
°=⇒= 74.2975.1cot θθ
°= 90α
mlcmtgfztgV
sA
ywdu
Edsw /²50.19
15,1500*0125.1
74.29*502.1**. =
°==
θ
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The minimum reinforcement percentage calculation:
mlcmbff
bsA
wyk
ckww
sw /²21.590sin*55.0*500
35*08,0sin***08,0
sin**min, =°==≥ ααρ
Finite elements modeling
■ Linear element: S beam, ■ 15 nodes, ■ 1 linear element. Advance Design gives the following results for Atz (cm2/ml)
5.41.2.4 Reference results
Result name Result description Reference value My 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 [cm2/ml] 19.10 cm2/ml
At,z,2 At,z transversal reinforcement for the middle of the beam [cm2/ml] 5.21 cm2/ml
5.41.3 Calculated results
Result name
Result description Value Error
My My -5255.58 kN*m -0.0000 % Fz Fz -1501.59 kN -0.0003 % Az Az 5cm -105.694 cm² 2.4071 % Az Az 2.7cm -105.694 cm² -0.0000 % Atz Atz end 19.0973 cm² 0.0000 % Atz Atz middle 5.20615 cm² 0.0000 %
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5.42 EC2 Test33: Verifying a square concrete column subjected to compression by nominal rigidity method- Bilinear stress-strain diagram (Class XC1)
Test ID: 5109
Test status: Passed
5.42.1 Description Verifies a square concrete column subjected to compression by nominal rigidity method- Bilinear stress-strain diagram (Class XC1). The column is made of concrete C30/37. The verification of the axial force, applied on top, at ultimate limit state is performed. Nominal rigidity method - The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part, the translations along X and Y axis are also blocked. This test is based on the example from "Applications of Eurocode 2" (J. & JA Calgaro Cortade).
5.42.2 Background Nominal rigidity method.
Verify the adequacy of a square cross section made from concrete C30/37.
5.42.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: ► 1260kN axial force ► The self-weight is neglected
■ Concrete cover 5cm ■ Concrete C30/37 ■ Steel reinforcement S500B ■ Concrete age 28 days ■ Relative humidity 50% ■ Buckling length: L0=0.7*4.47=3.32mm
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.30 m, ■ Width: b = 0.30 m, ■ Length: L = 4.74 m, ■ Concrete cover: c=5cm
Boundary conditions
The boundary conditions are described below:
The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part, the translations along X and Y axis are also blocked.
Loading
The beam is subjected to the following load combinations:
■ Load combinations: The ultimate limit state (ULS) combination is:
NEd =1260kN
■ Proposed reinforcement: 2*6.28cm2=12.56cm2
5.42.2.2 Reference results in calculating the concrete column
Load calculation:
NEd= 1.35*NG = 1700 KN
Initial eccentricity: cmMMe
u
u 07.1
00 ===
Additional eccentricity due to geometric imperfections:
( ) cmcmcmcmLcmei 2805.0;2max400322;2max
400;2max 0 ==⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.2(7)
The first order eccentricity : meee i 02.001 =+=
The necessity of buckling calculation (second order effect):
For an isolated column, the slenderness limit check is done using the next formula:
nCBA ***20
lim =λ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)
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Where:
944.020*²30.0
7.1*
===cdc
Ed
fANn
7.0=A if efϕ is not known, if it is, ef
Aϕ*2,01
1+
=
1.1=B if ω (reinforcement ratio) is not known, if it is, cdc
yds
fAfA
B**
*21*21 +=+= ω
In this case 27.120*²3.0
78.434*10*56.12*214
=+=−
B
70.0=C if rm is not known, if it is, mrC −= 7,1
In this case:
81.12944.0
7.0*27.1*7.0*20***20lim ===
nCBAλ
34.383.0
12*32.312*0 ===h
Lλ
81.1234.38 lim =>= λλ
Therefore, the second order effects most be considered
Effective creep coefficient calculation:
The creep coefficient is calculated using the next formula:
( )Ed
EQPef M
Mt ., 0∞= ϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
Where:
( )0,t∞ϕ creep coefficient
EQPM serviceability firs order moment under quasi-permanent load combination
EdM ULS first order moment (including the geometric imperfections)
The ratio of the moment is in this case:
74.07.126.1
**
1
1
0
0 ====ed
eqp
ed
eqp
Ed
Eqp
NN
eNeN
MM
),( 0t∞ϕis the final value of the creep:
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)(*)(*),( 00 tft cmRH ββϕϕ =∞
With:
t0 is the concrete age
2130
***1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
121 == αα if MPafcm 35≤ if not
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
RH relative humidity: RH=50%
h0= mean radius of the element in mm
( ) mmuAch 150
300300*2300*300*2*2
0 =+
==
u=column section perimeter
⇒≥= MPaMPafcm 3538944.0
3835 7.0
1 =⎟⎠⎞
⎜⎝⎛=α
and 984.0
3835 2.0
2 =⎟⎠⎞
⎜⎝⎛=α
86.1984.0*944.0*150*1.0
100501
1***1.0100
11
32130
=⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+= ααϕ
h
RH
RH
48.2488.0*73.2*86.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ
835.174.0*48.2*),(0
00 ==∞=
Ed
Eqpef M
Mtϕϕ
835.2835.111 =+=+ efϕ
Calculation of nominal rigidity:
The nominal rigidity of a post or frame member, it is estimated from the following formula:
sssccdc IEKIEKEI **** +=
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.2 (1)
ec
kkKϕ+
=1
* 21
835.21 =+ efϕ
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22.12030
201 === ckfk
170*
2λnk =
944.020*²3.0
7.1*
===cdc
Ed
fANn
34.38=λ
20.0213.0170
34.38*944.0170*
22 =⇒=== knk λ
086.0835.2
20.022.11
21 =×
=+
=e
ckkKϕ
CE
cmcd
EEγ
=
MPafE ckcm 32837
10825*22
108*22
3.03.0
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=
MPaEECE
cmcd 27364
2.132837
===γ
433
000675.012
3.0*3.012* mhbIc ===
Ks=1
Es = 200 Gpa
45242
10*256.12
05.025.0*2
10*56.12*22
'*2
*2 mddAI theos
−−
=⎟⎠⎞
⎜⎝⎛ −
=⎟⎠⎞
⎜⎝⎛ −
=
²1.410*256.1*200000*1000675.0*27364*086.0 5 MNmEI =+= −
MNLEINf
B 67.3²32.310.4*²
²*²
===ππ
234.18²²
0
===ππβ
c
MNm
NNMM
Ed
BEdEd 070.0
17.167.3234.11*034.0
11*0 =
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+=
β
The calculation made with flexural:
MNNEd 7.1=and
MNmMEd 070.0=,
Therefore, a 2*6.64cm2 reinforcement area is obtained.
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The calculated reinforcement is very close to the initial assumption (2*6.28cm2)
It is not necessary to continue the calculations; it retains a section of 2*6.64cm2
It sets up 4HA20 (2*6.28=12.57cm2)
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
Theoretical reinforcement area(cm2)
(reference value: 6.54cm2)
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Theoretical value (cm2)
(reference value: 13.08 cm2)
5.42.2.3 Reference results
Result name Result description Reference value Az Reinforcement area [cm2] 6.54 cm2
R Theoretical reinforcement area [cm2] 13.08 cm2
5.42.3 Calculated results
Result name
Result description Value Error
Az Az -6.53808 cm² 0.0001 %
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5.43 EC2 Test 27: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Test ID: 5076
Test status: Passed
5.43.1 Description Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram is generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max., will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, and the theoretical reinforcement. For the calculation, the reduced shear force values will be used.
5.43.2 Background Description: Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram is generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max. will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, and the theoretical reinforcement.
5.43.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ 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:
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The objective is to verify:
■ The shear stresses results ■ The theoretical reinforcement value
Units
Metric System
Geometry
Beam cross section characteristics:
■ Height: h = 0.70 m, ■ Width: b = 0.35 m, ■ Length: L = 5.75 m, ■ Concrete cover: c=3.5cm ■ Effective height: d=h-(0.06*h+ebz)=0.623m; d’=ehz=0.035m ■ Stirrup slope: α= 90° ■ Strut slope: θ=45˚
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.30) restrained in translation along Y, Z and rotation along X
■ Inner: None.
Loading:
For a beam subjected to a point load, Pu, the shear stresses are defined by the formula below:
2* lPV u
Ed =
According to EC2 the Pu point load is defined by the next formula:
Pu = 1,35*Pg + 1,50*Pq=1.35*40kN+1.50*25kN=91.5kN
In this case :
KNVEd 2632
5.91*75.5==
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5.43.2.2 Reference results in calculating the lever arm zc: The lever arm is calculated using the design formula for pure bending:
mlkNPu /5.91=
kNmMEd 15.3788
²75.5*5.91==
167.067.16*²623.0*35.0
378.0*²*
===cdw
Edcu fdb
Mμ
( ) ( ) 230.0167.0*211*25.1*211*25.1 =−−=−−= cuu μα
( ) ( ) mdz uc 566.0230.0*4.01*623.0*4.01* =−=−= α
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 θcot.zx = .
In the case of a member subjected to a distributed load, the equation of the shear force is:
2**)( lPxPxV u
u −=
Therefore:
mzx 566.045cot*566.0cot* =°== θ
kNV redEd 211263566.0*5.91, −=−=
Warning: Advance Design does not apply the reduction of shear force corresponding to the distributed loads (cutting edge at x = d).
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Calculation of maximum design shear resistance:
( )θ1θανα 1 2wucdcwmax,Rd
cotcotcot*b*z*f**V
+
+=
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 6.2.3.(3)
Where:
1=cwα coefficient taking account of the state of the stress in the compression chord and ⎥⎦⎤
⎢⎣⎡ −=
2501*6,01
ckfv
When the transverse frames are vertical, the above formula simplifies to:
θθ cot***1
max, +=
tgbzfvV wucd
Rd
In this case:
°= 45θ and °= 90α
1v strength reduction factor for concrete cracked in shear
54.0250251*6.0
2501*6,01 =⎥⎦
⎤⎢⎣⎡ −=⎥⎦
⎤⎢⎣⎡ −= ckfv
mdzu 56.0623.0*9.0*9.0 ===
kNMNVRd 891891.02
35.0*566.0*67.16*54.0max, ===
kNVkNV RdEd 891236 max, =<=
Calculation of transversal reinforcement:
Given the vertical transversal reinforcement (α = 90°), the transverse reinforcement is calculated using the following formula:
mlcmtgfztgV
sA
ywdu
Edsw /²67,8
15,1500*566,0
45*211,0**. ===
θ
Calculation of theoretical reinforcement value:
The French national annex indicates the formula:
αρ sin**min, wwsw bsA
≥
With:
4min, 1015.7
50020*08,0*08,0 −×===
yk
ckw f
fρ
mlcmbsA
wwsw /²43.190sin*2.0*10*15.7sin** 4
min, =°=≥ −αρ
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Finite elements modeling
■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. Advance Design gives the following results for At,min,z (cm2/ml)
5.43.2.3 Reference results
Result name Result description Reference value Fz Fz corresponding to the 101 combination (ULS) [kNm] 263 kN
At,min,z Theoretical reinforcement area [cm2/ml] 1.43 cm2/ml
5.43.3 Calculated results
Result name
Result description Value Error
Fz Fz -168.937 kN -0.0003 % Atminz Atmin,z 1.43108 cm² 0.0002 %
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5.44 EC2 Test31: Verifying a square concrete column subjected to compression and rotation moment to the top - Bilinear stress-strain diagram (Class XC1)
Test ID: 5101 Test status: Passed
5.44.1 Description Nominal rigidity method. Verifies the adequacy of a rectangular cross section column made from concrete C25/30. The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. Verifies the column to resist simple bending. The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed. The column is considered connected to the ground by a fixed connection and free to the top part.
5.44.2 Background Nominal rigidity method.
Verifies the adequacy of a rectangular cross section made from concrete C25/30.
5.44.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: ► 15t axial force ► 1.5tm rotation moment applied to the column top ► The self-weight is neglected
■ Exploitation loadings: ► 6.5t axial force ► 0.7tm rotation moment applied to the column top
■ 3,02 =ψ
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Concrete cover 5cm ■ Transversal reinforcement spacing a=30cm ■ Concrete C25/30 ■ Steel reinforcement S500B ■ The column is considered isolated and braced
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.40 m, ■ Width: b = 0.40 m, ■ Length: L = 6.00 m, ■ Concrete cover: c=5cm
Boundary conditions
The boundary conditions are described below:
The column is considered connected to the ground by a fixed connection and free to the top part.
Loading
The beam is subjected to the following load combinations:
■ Load combinations: The ultimate limit state (ULS) combination is:
NEd =1.35*15+150*6.5=30t=0.300MN
MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm
■ m...
NM
eEd
Ed 10030000310
0 ===
5.44.2.2 Reference results in calculating the concrete column
Geometric characteristics of the column:
The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
mll 12*20 ==
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b)
Calculating the slenderness of the column:
10440.0
12*32*32 0 ===alλ
Effective creep coefficient calculation:
The creep coefficient is calculated using the next formula:
( )Ed
EQPef M
Mt ., 0∞= ϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
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Where:
( )0,t∞ϕ creep coefficient
serviceability first order moment under quasi-permanent load combination
ULS first order moment (including the geometric imperfections)
First order eccentricity evaluation:
ieee += 01
mei 03.0=
The first order moment provided by the quasi-permanent loads:
meNM
eee iEqp
Eqpi 13.030.0
65.0*30.01570.0*30.050.1
0
001 =+
++
=+=+=
tNEqp 95.1670.0*30.050.11 =+=
MNmtmeNM EqpEqp 022.020.213.0*95.16* 111 ====
The first order ULS moment is defined latter in this example:
)(*)(*),( 00 tft cmRH ββϕϕ =∞
92.2825
8.168.16)( =+
==cm
cm ffβ
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ (for t0= 28 days concrete age).
30*1.0
1001
1h
RH
RH
−+=ϕ
( ) 85.1200*1.0
100501
1200400400*2400*400*2*2
30 =−
+=⇒=+
== RHmmuAch ϕ
64.2488.0*92.2*85.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ
The effective creep coefficient calculation:
( ) 49.1039.0022.0*64.2*, 0 ==∞=
Ed
EQPef M
Mtϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
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The second order effects; The buckling calculation:
For an isolated column, the slenderness limit check is done using the next formula:
nCBA ***20
lim =λ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)
Where:
112.067.16*²40.0
300.0*
===cdc
Ed
fANn
( ) 77.049.1*2.01
1*2,01
1=
+=
+=
ef
Aϕ
1.1*21 =+= ωB because the reinforcement ratio in not yet known
70.07,1 =−= mrC because the ratio of the first order moment is not known
43.35112.0
7.0*1.1*77.0*20lim ==λ
43.35104 lim =>= λλ
Therefore, the second order effects most be considered.
The second order effects; The buckling calculation:
The stresses for the ULS load combination are: NEd= 1.35*15 + 1.50*6.50= 30 t = 0,300MN
MEd= 1.35*1.50 + 1.50*0.7= 3.075 t = 0,031MNm
Therefore it must be determined:
■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:
mNMeEd
Ed 10.0300.0031.0
0 ===
Additional eccentricity:
mlei 03.040012
4000 ===
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The first order eccentricity: stresses correction:
The forces correction, used for the combined flexural calculations:
MNNEd 300.0=
meee i 13.001 =+=
MNmNeM EdEd 039.0300.0*)03.010.0(*1 =+==
0*eNM Ed=
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4)
Reinforcement calculation in the first order situation:
To play the nominal rigidity method a starting section frame is needed. For this it will be used a concrete section considering only the first order effects.
Advance Design iterates as many time as necessary.
The needed frames will be determined assuming a compound bending with compressive stress. All the results above were obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid of the tensioned steel.
MNmhdNMM Gua 084.0240.035.0*300.0039.0)
2(*0 =⎟
⎠⎞
⎜⎝⎛ −+=−+=
Verification if the section is partially compressed:
496,0)35,040,0*4,01(*
35,040,0*8,0)*4,01(**8,0 =−=−=
dh
dh
BCμ
103.067.16*²35.0*40.0
084.0*²*
===cdw
uacu fdb
Mμ
BCcu μμ =<= 496.0103.0 therefore the section is partially compressed.
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Calculations of steel reinforcement in pure bending:
103.0=cuμ
[ ] 136,0)103,0*21(1*25,1 =−−=uα
mdz uc 331,0)136,0*4,01(*35,0)*4,01(* =−=−= α
Calculations of steel reinforcement in combined bending:
For the combined bending:
24 06.178.434
300.010*84.5' cmfNAAyd
−=−=−= −
The minimum reinforcement percentage:
2min, 02.3*002.0²69.0
78.434300.0*10.0*10,0 cmAcm
fNA c
yd
Eds =≥===
The reinforcement will be 4HA10 representing a 3.14cm2 section
The second order effects calculation:
The second order effect will be determined by applying the method of nominal rigidity:
Calculation of nominal rigidity:
The nominal rigidity of a post or frame member, it is estimated from the following formula:
sssccdc IEKIEKEI **** +=
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.2 (1)
With:
2.1cm
cdEE =
MpaMpaff ckcm 338 =+=
MpafE cmcm 31476
1033*22000
10*22000
3.03.0
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
MpaEE cmcd 26230
2.131476
2.1===
410.133,21240.0
12* 3
43
mhbIc−===
(cross section inertia)
MpaEs 200000=
sI : Inertia
002.040.040.0
10.14,3 4
=×
==−
c
s
AAρ
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01.0002.0 <=≤c
s
AAρ
Mpafk ck 12.12025
201 ===
112.067.16*²40.0
300.0*
===cdc
Ed
fANn
20.0069.0170104*112.0
170*2 ≤===
λnk
410*06,705.0240.0*
210*14,3*2
2*
2*2 6
242
mchAI ss
−−
=⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=
1=sK and
031.049.11069.0*12.1
1* 21 =
+=
+=
efc
kkKϕ
Therefore:
²15.310*06,7*200000*110*133,2*26230*031.0 63 MNmEI =+= −−
Corrected stresses:
The total moment, including the second order effects is defined as a value and is added to the first order moment value:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+=
11*0
Ed
BEdEd
NNMM β
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.3 (1)
MNmM Ed 039.00 = (time of first order (ULS) taking into account the geometric imperfections, relative to the center of gravity of concrete).
MNNEd 300.0= (normal force acting at ULS).
And:
0
²cπβ = with 80 =c because the moment is constant (no horizontal force at the top of post).
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.3 (2)
234.18²
==πβ
MNlEINB 216.0
²1215.3*²*² 2
0
=== ππ
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Therefore the second order moment is:
MNmMEd 133.01
300.0216.0
234.11*039.0 −=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+=
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.
sssccdc IEKIEKEI **** +=
With:
MpaEcd 26230= ; 410.133,2 3mIc
−=
MpaEs 200000=
sI : Inertia
²8²40,0*005,0*005,0005,0 cmAA cs ===⇒=ρ
It sets up : 4HA16 => ²04,8005,0 cmAs =⇒=ρ
01.0002.0 <=≤c
s
AAρ
45242
10*81,105.0240.0*
210*04,8*2
2*
2*2 mchAI s
s−
−
=⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=
1=sK and
031.049.11069.0*12.1
1* 21 =
+=
+=
efc
kkKϕ
Therefore:
²35.510.81,1*200000*110.133,2*26230*031.0 53 MNmEI =+= −−
The second order effects must be recalculated:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+=
11*0
Ed
BEdEd
NNMM β
234.1=β
MNlEINB 367.0
²1235.5*²*² 2
0
=== ππ
MNmMEd 254.01
300.0367.0
234.11*039.0 =⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+=
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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:
238.067.16*²40.0*40.0
254.0** 2 ===
cd
Ed
fhbMμ
112.067.16*40.0*40.0
300.0**
===cd
Ed
fhbNv
485.0=ω is obtained, which gives: ∑ === ²75.2978.434
67.16*²40.0*485.0*** cmf
fhbAyd
cds
ω
Therefore set up a section 14.87cm ² per side must be set, or 3HA32 per side (by excess)
Buckling checking
The column in place will be verified without buckling. The new reinforcement area must be considered for the previous calculations: 6HA32 provides As=48.25cm2
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The normal rigidity evaluation:
It is estimated nominal rigidity of a post or frame member from the following formula:
sssccdc IEKIEKEI .... +=
With:
sI : Inertia
03.040.0*40.0
10*25,48 4
===−
c
s
AAρ
44242
10.086,105.0240.0*
210*25,48*2
2*
2*2 mchAI s
s−
−
=⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=
12.11 =k
069.02 =k
1=sK and
031.049.11069.0*12.1
1* 21 =
+=
+=
efc
kkKϕ
The following conditions are not implemented in Advance Design:
If: 01.0≥=
c
s
AAρ
efcK ϕ*5,01
3,0+
=
Then:
²45.2310*086,1*20000010*133,2*26230*031.0 43 MNmEI =+= −−
Corrected stresses
The total moment, including second order effects, is defined as a value plus the moment of the first order:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+=
11*0
Ed
BEdEd
NNMM β
MNmM Ed 039.00 =
MNNEd 300.0= (normal force acting at ULS)
234.1=β
MNlEINB 607.1
²1245.23*²*² 2
0
=== ππ
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It was therefore a moment of second order which is:
MNmMEd 05.01
300.0607.1
234.11*039.0 =⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+=
Reinforcement calculation:
047.067.16*²40.0*40.0
05.0**
05.0 2 ====>=cd
EdEd fhb
MMNmM μ
112.067.16*40.0*40.0
300.0**
300.0 ====>=cd
EdEd fhb
NvMNN
The minimum reinforcement percentage conditions are not satisfied therefore there will be one more iteration.
Additional iteration:
The additional iteration will be made for a section corresponding to 1%; As=0.009*0.40=14.4cm2, which is 7.2cm2 per side. A 3HA16 reinforcement will be chosen by either side (6HA16 for the entire column), which will give As=12.03cm2.
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Nominal rigidity evaluation:
sssccdc IEKIEKEI .... +=.
sI : Inertia
00754.040.0*40.0
10*06,12 4
===−
c
s
AAρ
if 01.0002.0 <=≤
c
s
AAρ
45242
10.71,205.0240.0*
210*06,12*2
2*
2*2 mchAI s
s−
−
=⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=
1=sK and
031.049.11069.0*12.1
1* 21 =
+=
+=
efc
kkKϕ
Therefore:
²15.710*71,2*200000*110*133,2*26230*031.0 53 MNmEI =+= −−
Second order loads calculation:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+=
11*0
Ed
BEdEd
NNMM β
MNmM Ed 039.00 =
MNNEd 300.0= (normal force acting at ULS).
234.1=β
MNlEINB 49.0
²1215.7*²*² 2
0
=== ππ
MNmMEd 115.01
300.049.0234.11*039.0 =
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+=
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Reinforcement calculation:
Frames are calculated again from the interaction diagram:
108.067.16*²40.0*40.0
115.0** 2 ===
cd
Ed
fhbMμ
112.067.16*40.0*40.0
300.0**
===cd
Ed
fhbNv
18.0=ω is obtained, which gives: ∑ === ²04.1178.434
67.16*²40.0*18.0*** cmf
fhbAyd
cds
ω
Therefore set up a section 5.52cm ² per side must be set; this will be the final column reinforcement
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
Theoretical reinforcement area(cm2)
(reference value: 11.04cm2)
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Theoretical value (cm2)
(reference value: 11.16 cm2)
5.44.2.3 Reference results
Result name Result description Reference value Az Reinforcement area [cm2] 5.58 cm2
R Theoretical reinforcement area [cm2] 11.16 cm2
5.44.3 Calculated results
Result name Result description Value Error Az Az -5.58 cm² -0.0000 %
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5.45 EC2 Test35: Verifying a rectangular concrete column subjected to compression to top – Based on nominal rigidity method - Bilinear stress-strain diagram (Class XC1)
Test ID: 5123
Test status: Passed
5.45.1 Description Verifies the adequacy of a rectangular concrete column made of concrete C30/37 subjected to compression to top – Based on nominal rigidity method- Bilinear stress-strain diagram (Class XC1).
Based on nominal rigidity method
The purpose of this test is to determine the second order effects by applying the nominal rigidity method, and then calculate the frames by considering a section symmetrically reinforced.
The column is considered connected to the ground by a fixed connection and free to the top part.
5.45.2 Background Based on nominal rigidity method
Verify the adequacy of a rectangular cross section made from concrete C30/37.
The purpose of this test is to determine the second order effects by applying the nominal rigidity method, and then calculate the frames by considering a section symmetrically reinforced.
5.45.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: ► 0.45 MN axial force ► 0.10 MNm rotation moment applied to the column top ► The self-weight is neglected
■ Exploitation loadings: ► 0.50 MN axial force ► 0.06 MNm rotation moment applied to the column top
■ 3,02 =ψ
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 0.3 x Q ■ Concrete cover 5cm ■ Concrete C30/37 ■ Steel reinforcement S500B ■ The column is considered isolated and braced
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.45 m, ■ Width: b = 0.60 m, ■ Length: L = 4.50 m, ■ Concrete cover: c = 5cm along the long section edge and 3cm along the short section edge
Boundary conditions
The boundary conditions are described below:
The column is considered connected to the ground by a fixed connection and free to the top part.
Loading
The beam is subjected to the following load combinations:
■ Load combinations: The ultimate limit state (ULS) combination is:
NEd =1.35*0.45+1.5*0.50=1.3575MN
MED=1.35*0.10+0.30*0.06=0.225MNm
■ me 166.03575.1225.0
0 ==
5.45.2.2 Reference results in calculating the concrete column
Geometric characteristics of the column:
The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
mll 950.4*2*20 ===
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b)
Calculating the slenderness of the column:
28.6945.0
9*32*32 0 ===alλ
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Effective creep coefficient calculation:
The creep coefficient is calculated using the next formula:
( )Ed
EQPef M
Mt ., 0∞= ϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
Where:
( )0,t∞ϕ creep coefficient
EQPM serviceability firs order moment under quasi-permanent load combination
EdM ULS first order moment (including the geometric imperfections)
MNmM Eqp 118.006.0*3.010.00 =+=
MNmM Ed 225.00 =
The moment report becomes:
524.0225.0118.0
0
0 ==Ed
Eqp
MM
The creep coefficient ( )0,t∞ϕ
is defined as follows:
)(*)(*),( 00 tft cmRH ββϕϕ =∞
MPaf
fcm
cm 73.2830
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ (for t0= 28 days concrete age).
213
0
***1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
RH =relative humidity; RH=50%
Where 121 == αα if MPafcm 35≤ if not
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
( ) mmuAch 274
700450*2700*450*2*2
0 =+
==
MPaMPafcm 3538 >= ,
Therefore:
944.0383535 7.07.0
1 =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
cmfα and 984.0
383535 2.02.0
2 =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
cmfα
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70.1984.0*944.0*274*1.0
100501
1***1.0100
11
3213
0
=⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+= ααϕ
h
RH
RH
26.2488.0*73.2*70.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ
The effective creep coefficient calculation:
( ) 18.1524.0*26.2*, 0 ==∞=Ed
EQPef M
Mtϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
18.218.111 =+=+ efϕ
The necessity of buckling calculation (second order effect):
For an isolated column, the slenderness limit check is done using the next formula:
nCBA ***20
lim =λ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)
Where:
215.020*70.0*45.0
3575.1*
===cdc
Ed
fANn
( ) 81.018.1*2.01
1*2,01
1=
+=
+=
ef
Aϕ
because the reinforcement ratio in not yet known
because the ratio of the first order moment is not known
90.26215.0
7.0*1.1*85.0*20lim ==λ
90.2628.69 lim =>= λλ
Therefore, the second order effects most be considered
5.45.2.3 The eccentricity calculation and the corrected loads on ULS:
Initial eccentricity:
me 166.03575.1225.0
0 ==
Additional eccentricity:
cmlei 25.24009
4000 ===
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First order eccentricity- stresses correction:
MNNEd 3575.1=
cmmeee i 85.181885.001 ==+=
MNmMEd 256.0=
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.
MNmhdNMM Gua 494.0245.040.0*3575.1256.0)
2(*0 =⎟
⎠⎞
⎜⎝⎛ −+=−+=
Verification if the section is partially compressed:
495,0)40,045,0*4,01(*
40,045,0*8,0)*4,01(**8,0 =−=−=
dh
dh
BCμ
220.020*²40.0*70.0
495.0*²*
===cdw
uacu fdb
Mμ
BCcu μμ =<= 495.0220.0 therefore the section is partially compressed.
Calculations of steel reinforcement in pure bending:
220.0=cuμ
[ ] 315,0)220,0*21(1*25,1 =−−=uα
mdz uc 350,0)315,0*4,01(*40,0)*4,01(* =−=−= α
²46.3278,434*350,0
495,0*
cmfz
MAydc
ua ===
Calculations of steel reinforcement in combined bending:
For the combined bending:
24 24.178.434
3575.110*46.32' cmfNAAyd
−=−=−= −
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The minimum reinforcement percentage:
2min, 3.6*002.0²12.3
78.4343575.1*10.0*10,0 cmAcm
fNA c
yd
Eds =≥===
The reinforcement will be 8HA10 representing a 6.28cm2 section
The second order effects calculation:
The second order effect will be determined by applying the method of nominal rigidity:
Calculation of nominal rigidity:
It is estimated nominal rigidity of a post or frame member from the following formula:
sssccdc IEKIEKEI **** +=
With:
2.1cm
cdEE =
MPaMPaff ckcm 388 =+=
MPafE cmcm 32837
1038*22000
10*22000
3.03.0
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
MPaEE cmcd 27364
2.132837
2.1===
410*316.512
45.0*70.012* 3
33
mhbIc−=== (concrete only inertia)
MPaEs 200000=
sI : Inertia
002.040.0*70.0
10*28,6 4
===−
c
s
AAρ
01.0002.0 <=≤c
s
AAρ
Mpafk ck 22.12030
201 ===
215.020*70.0*45.0
3575.1.
===cdc
Ed
fANn
20.0088.0170
28.69*215.0170
*2 ≤===λnk
45242
10.92,105.0245.0*
210.28,6*2
2*
2*2 mchAI s
s−
−
=⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=
1=sK and 049.018.11088.0*22.1
1* 21 =
+=
+=
efc
kkKϕ
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Therefore:
²97.1010*92.1*200000*110*316.5*27364*049.0 53 MNmEI =+= −−
Stresses correction:
The total moment, including second order effects, is defined as a value plus the time of the first order:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+=
11*0
Ed
BEdEd
NNMM β
MNmM Ed 0256.00 = (moment of first order (ULS) taking into account geometric imperfections
MNNEd 3575.1= (normal force acting at ULS).
0
²cπβ = and 80 =c the moment is constant (no horizontal force at the top of post).
234.18²
==πβ
MNlEINB 32.1
²982.10*²*² 2
0
=== ππ
The second order moment is:
MNmMEd 18.111
3575.132.1234.11*256.02 −=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−+=
An additional iteration must be made by increasing the ratio of reinforcement.
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Additional iteration:
The iteration is made considering 8HA12 or As=9.05cm2
Therefore 0029.0
45.0*70.010*05,9 4
==−
ρ
Is obtained:
Therefore:
MNNEd 3575.12 =
MNmMEd 48.22 =
Given the mentioned reports, there must be another reiteration:
The reinforcement section must be increase to 8HA20 or As=25.13cm2 and 008.0=ρ :
Is obtained:
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Therefore:
MNNEd 3575.12 =
MNmMEd 563.02 =
Using the EC2 calculation tools, combined bending analytical calculation, a 28.80cm2 value is found.
There must be another iteration:
The reinforcement section must be increase to As=30cm2 and 0095.0=ρ :
Therefore:
MNNEd 3575.12 =
MNmMEd 499.02 =
Using the EC2 calculation tools, combined bending analytical calculation, a 22.22cm2 value is found.
The theoretical reinforcement section of 30cm2 will be adopted.
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
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Theoretical reinforcement area(cm2)
(reference value: 16cm2 )
Theoretical value (cm2)
(reference value: 31.99cm2)
5.45.2.4 Reference results
Result name Result description Reference value Az Reinforcement area [cm2] 16 cm2
R Theoretical reinforcement area [cm2] 31.99 cm2
5.45.3 Calculated results
Result name Result description Value Error Az Az -15.995 cm² 0.0000 %
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5.46 EC2 Test36: Verifying a rectangular concrete column using the method based on nominal curvature- Bilinear stress-strain diagram (Class XC1)
Test ID: 5125
Test status: Passed
5.46.1 Description Verifying a rectangular concrete column using the method based on nominal curvature - Bilinear stress-strain diagram (Class XC1)
Verifies the adequacy of a rectangular cross section column made from concrete C30/37.
Method based on nominal curvature
The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced.
The column is considered connected to the ground by an articulated connection (all the translations are blocked and all the rotations are permitted) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.
This example is provided by the “Calcul des Structures en beton” book, by Jean-Marie Paille, edition Eyrolles.
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5.47 EC2 Test 37: Verifying a square concrete column using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1)
Test ID: 5126
Test status: Passed
5.47.1 Description Verifies the adequacy of a square concrete column made of concrete C25/30, using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1).
Simplified Method
The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.
5.47.2 Background Simplified Method
Verifies the adequacy of a square cross section made from concrete C25/30.
5.47.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: ► 800kN axial force ► The self-weight is neglected
■ Exploitation loadings: ► 800kN axial force
■ Concrete cover 5cm ■ Concrete C25/30 ■ Steel reinforcement S500B ■ Relative humidity RH=50% ■ Buckling length L0=0.70*4=2.80m
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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:
kNNED 2280800*5.1800*35.1 =+=
5.47.2.2 Reference results in calculating the concrete column
Scope of the method:
25.244.0
12*8.212*
121212
002
0
2
4000 ========
aL
aL
aL
a
aL
AIL
iLλ
According to Eurocode 2 EN 1992-1-1 (2004) Chapter 5.8.3.2(1)
The method of professional rules can be applied as:
120<λ
MPafck 5020 <<
mh 15.0>
Reinforcement calculation:
6025.24 <=λ , therefore:
746.0
6225.241
86.0
621
86.022 =
⎟⎠⎞
⎜⎝⎛+
=
⎟⎠⎞
⎜⎝⎛+
=λ
α
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 93.0=hk
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1500500*6.06.1
500*6.06.1 =−=−= yk
s
fk
²24,1467.16*4.0*4.0746.0*1*93.0
280.2*78.434
1****
*1 cmfhbkkN
fA cd
sh
ed
yds =⎟
⎠⎞
⎜⎝⎛ −=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
α
According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283
Finite elements modeling
■ Linear element: S beam, ■ 5 nodes, ■ 1 linear element.
Theoretical reinforcement area(cm2)
(reference value: 14.24cm2=4*3.56cm2)
5.47.2.3 Reference results
Result name Result description Reference value Ay Reinforcement area [cm2] 3.57cm2
R Theoretical reinforcement area [cm2] 14.26 cm2
5.47.3 Calculated results
Result name
Result description Value Error
Ay Az -3.56562 cm² 0.0001 %
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5.48 EC2 Test 26: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Test ID: 5072
Test status: Passed
5.48.1 Description Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectional area of the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used.
5.48.2 Background Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectional area of the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used.
5.48.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: The dead load will be represented by a linear load of 40kN/m
■ Exploitation loadings: The live load will be considered from one linear load of 25kN
■ Structural class: S1 ■ Reinforcement steel ductility: Class B ■ The reinforcement will be displayed like in the picture below:
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The objective is to verify:
■ The shear stresses results ■ The cross-sectional area of the shear reinforcement, Asw
Units
Metric System
Geometry
Beam cross section characteristics:
■ Height: h = 0.70 m, ■ Width: b = 0.35 m, ■ Length: L = 5.75 m, ■ Concrete cover: c=3.5cm ■ Effective height: d=h-(0.06*h+ebz)=0.623m; d’=ehz=0.035m ■ Stirrup slope: α= 90° ■ Strut slope: θ=45˚
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 5.30) restrained in translation along Y, Z and rotation along X
■ Inner: None.
Loading:
For a beam subjected to a point load, Pu, the shear stresses are defined by the formula below:
2* lPV u
Ed =
According to EC2 the Pu point load is defined by the next formula:
Pu = 1,35*Pg + 1,50*Pq=1.35*40kN+1.50*25kN=91.5kN
In this case :
KNVEd 2632
5.91*75.5==
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5.48.2.2 Reference results in calculating the lever arm zc: The lever arm will be calculated from the design formula for pure bending:
mlkNPu /5.91=
kNmMEd 15.3788
²75.5*5.91==
167.067.16*²623.0*35.0
378.0*²*
===cdw
Edcu fdb
Mμ
( ) ( ) 230.0167.0*211*25.1*211*25.1 =−−=−−= cuu μα
( ) ( ) mdz uc 566.0230.0*4.01*623.0*4.01* =−=−= α
Calculation of reduced shear force
When the shear force curve has no discontinuities, the EC2 allows to consider, as a reduction, the shear force to a
horizontal axis
In case of a member subjected to a distributed load, the equation of the shear force is:
2**)( lPxPxV u
u −=
Therefore:
mzx 566.045cot*566.0cot* =°== θ
kNV redEd 211263566.0*5.91, −=−=
Warning: Advance Design does not apply the reduction of shear force corresponding to the distributed loads (cutting edge at x = d).
Calculation of maximum design shear resistance:
( )θ1θανα 1 2wucdcwmax,Rd
cotcotcot*b*z*f**V
+
+=
According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 6.2.3.(3)
Where:
1=cwα coefficient taking account of the state of the stress in the compression chord and
⎥⎦⎤
⎢⎣⎡ −=
2501*6,01
ckfv
When the transverse frames are vertical, the above formula simplifies to:
θθ1
cottgb*z*f*v
V wucdmax,Rd +
=
In this case:
°= 45θ and °= 90α
1v strength reduction factor for concrete cracked in shear
54.0250251*6.0
2501*6,01 =⎥⎦
⎤⎢⎣⎡ −=⎥⎦
⎤⎢⎣⎡ −= ckfv
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mdzu 56.0623.0*9.0*9.0 ===
kNMNVRd 891891.02
35.0*566.0*67.16*54.0max, ===
kNVkNV RdEd 891236 max, =<=
Calculation of transversal reinforcement:
Given the vertical transversal reinforcement (α = 90°), the following formula is used:
mlcmtgfztgV
sA
ywdu
Edsw /²67,8
15,1500*566,0
45*211,0**. ===
θ
Finite elements modeling
■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. Advance Design gives the following results for Atz (cm2/ml)
5.48.2.3 Reference results
Result name Result description Reference value Fz Fz corresponding to the 101 combination (ULS) [kNm] 263 kN
At,z Theoretical reinforcement area [cm2/ml] 8.67 cm2/ml
5.48.3 Calculated results
Result name
Result description Value Error
Fz Fz -263.062 kN -0.0002 % Atz Atz 8.68636 cm² 0.0000 %
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5.49 EC2 Test30: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Test ID: 5098
Test status: Passed
5.49.1 Description Verifies a T cross section beam made from concrete C30/37 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with the cross-sectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value.
5.50 EC2 Test34: Verifying a rectangular concrete column subjected to compression on the top – Method based on nominal curvature - Bilinear stress-strain diagram (Class XC1)
Test ID: 5114
Test status: Passed
5.50.1 Description Verifies the adequacy of a rectangular cross section column made of concrete C30/37. The verification of the axial stresses applied on top, at ultimate limit state is performed.
Method based on nominal curvature - The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced.
The column is considered connected to the ground by a fixed connection and free at the top part.
5.50.2 Background Method based on nominal curvature
Verifies the adequacy of a rectangular cross section made from concrete C30/37.
The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced.
5.50.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: ► 300kN axial force ► The self-weight is neglected
■ Exploitation loadings: ► 500kN axial force
■ 3,02 =ψ
■ 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
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■ The column is considered isolated and braced
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.40 m, ■ Width: b = 0.60 m, ■ Length: L = 4.00 m, ■ Concrete cover: c = 5 cm along the long section edge and 3cm along the short section edge
Boundary conditions
The boundary conditions are described below:
The column is considered connected to the ground by a fixed connection and free to the top part.
Loading
The beam is subjected to the following load combinations:
■ Load combinations: The ultimate limit state (ULS) combination is:
NEd =1.35*0.30+1.5*0.50=1.155MN
NQP=1.35*0.30+0.30*0.50=0.450MN
■
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5.50.2.2 Reference results in calculating the concrete column
Geometric characteristics of the column:
The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:
mll 84*2*20 ===
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b)
Calculating the slenderness of the column:
28.6940.0
8*32*32 0 ===alλ
Effective creep coefficient calculation:
The creep coefficient is calculated using the next formula:
( )Ed
EQPef M
Mt ., 0∞= ϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
Where:
( )0,t∞ϕ creep coefficient
EQPM serviceability firs order moment under quasi-permanent load combination
EdM ULS first order moment (including the geometric imperfections)
The moment report becomes:
39.0155.1450.0
**
1
1
0
0 ====ed
eqp
ed
eqp
Ed
Eqp
NN
eNeN
MM
The creep coefficient ( )0,t∞ϕ is defined as:
)(*)(*),( 00 tft cmRH ββϕϕ =∞
MPaf
fcm
cm 73.2830
8.168.16)( =+
==β
488.0281.0
11.0
1)( 20.020.00
0 =+
=+
=t
tβ (for t0= 28 days concrete age).
213
0
***1.0100
11 ααϕ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
h
RH
RH
RH = relative humidity; RH=50%
Where 121 == αα if MPafcm 35≤ if not
7.0
135
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα and
2.0
235
⎟⎟⎠
⎞⎜⎜⎝
⎛=
cmfα
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( ) mmuAch 240
600400*2600*400*2*2
0 =+
==
MPaMPafcm 3538 >= ,
Therefore:
944.0383535 7.07.0
1 =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
cmfα and 984.0
383535 2.02.0
2 =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
cmfα
73.1984.0*944.0*240*1.0
100501
1***1.0100
11
3213
0
=⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛ −+= ααϕ
h
RH
RH
30.2488.0*73.2*73.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ
The effective creep coefficient calculation:
( ) 90.039.0*30.2*, 0 ==∞=Ed
EQPef M
Mtϕϕ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)
The necessity of buckling calculation (second order effect):
For an isolated column, the slenderness limit check is done using the next formula:
nCBA ***20
lim =λ
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)
Where:
241.020*60.0*40.0
155.1*
===cdc
Ed
fANn
( ) 85.090.0*2.01
1*2,01
1=
+=
+=
ef
Aϕ
1.1*21 =+= ωB because the reinforcement ratio in not yet known
70.07,1 =−= mrC because the ratio of the first order moment is not known
66.26112.0
7.0*1.1*85.0*20lim ==λ
66.2628.69 lim =>= λλ
Therefore, the second order effects most be considered
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5.50.2.3 The eccentricity calculation and the corrected loads on ULS:
Initial eccentricity:
No initial eccentricity because the post is only applied in simple compression.
Additional eccentricity:
mlei 02.04008
4000 ===
First order eccentricity- stresses correction:
⎩⎨⎧
=⎪⎩
⎪⎨⎧
=⎪⎩
⎪⎨⎧
== mmmmmmmm
mmhmm
e 203.13
20max
3040020
max30
20max0
According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4)
Solicitations corrected to take into account when calculating flexural combined, are:
NEd= 1.155MN
MEd= 1.155*0.02=0.0231MNm
Reinforcement calculation in the first order situation:
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.
MNmhdNMM Gua 196.0240.035.0*155.10231.0)
2(*0 =⎟
⎠⎞
⎜⎝⎛ −+=−+=
Verification if the section is partially compressed:
496,0)35,040,0*4,01(*
35,040,0*8,0)*4,01(**8,0 =−=−=
dh
dh
BCμ
133.020*²35.0*60.0
196.0*²*
===cdw
uacu fdb
Mμ
BCcu μμ =<= 496.0133.0 therefore the section is partially compressed.
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Calculations of steel reinforcement in pure bending:
133.0=cuμ
[ ] 179,0)133,0*21(1*25,1 =−−=uα
mdz uc 325,0)179,0*4,01(*35,0)*4,01(* =−=−= α
²87.1378,434*325,0
196,0*
cmfz
MAydc
ua ===
Calculations of steel reinforcement in combined bending:
For the combined bending:
24 69.1278.434
155.110*87.13' cmfNAAyd
−=−=−= −
The minimum reinforcement percentage:
2min, 80.4*002.0²66.2
78.434155.1*10.0*10,0 cmAcm
fNA c
yd
Eds =≥===
The reinforcement will be 8HA10 representing a 6.28cm2 section
The second order effects calculation:
The second order effect will be determined by applying the method of nominal rigidity:
Calculation of nominal rigidity:
It is estimated nominal rigidity of a post or frame member from the following formula:
sssccdc IEKIEKEI **** +=
With:
2.1cm
cdEE =
MPaMPaff ckcm 388 =+=
MPafE cmcm 32837
1038*22000
10*22000
3.03.0
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
MPaEE cmcd 27364
2.132837
2.1===
410.2,312
40.0*60.012* 3
33
mhbIc−===
(concrete only inertia)
MPaEs 200000=
sI : Inertia
0026.040.0*60.0
10.28,6 4
===−
c
s
AAρ
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01.0002.0 <=≤c
s
AAρ
Mpafk ck 22.12030
201 ===
241.020*60.0*40.0
155.1.
===cdc
Ed
fANn
20.0098.0170
28.69*241.0170
*2 ≤===λnk
410.41,105.0240.0*
210.28,6*2
2*
2*2 5
242
mchAI ss
−−
=⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛ −=
1=sK and 063.090.01098.0*22.1
1* 21 =
+=
+=
efc
kkKϕ
Therefore:
²34.810*41,1*200000*110*2,3*27364*063.0 53 MNmEI =+= −−
Stresses correction:
The total moment, including second order effects, is defined as a value plus the time of the first order:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−+=
11*0
Ed
BEdEd
NNMM β
MNmM Ed 0231.00 = (moment of first order (ULS) taking into account geometric imperfections
(normal force acting at ULS).
0
²cπβ = and 80 =c the moment is constant (no horizontal force at the top of post).
234.18²
==πβ
MNlEINB 27.1
²825.8*²*² 2
0
=== ππ
The second order efforts are:
MNNEd 155.12 =
mMNMEd .31.02 =
The reinforcement calculations in the second order effect:
The reinforcement calculations for the combined flexural, under the second order effect, are done using the Graitec EC2 tools. The obtained section is null, therefore the minimum reinforcement defined above is sufficient to absorb all the forces.
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Finite elements modeling
■ Linear element: S beam, ■ 5 nodes, ■ 1 linear element.
Theoretical reinforcement area (cm2) and minimum reinforcement area (cm2)
(reference value: 2.4cm2 and 4.8cm2)
Theoretical value (cm2)
(reference value: 4.80 cm2)
5.50.2.4 Reference results
Result name Result description Reference value Az Reinforcement area [cm2] 2.40 cm2
Amin Minimum reinforcement area 4.80 cm2
R Theoretical reinforcement area [cm2] 4.80 cm2
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5.50.3 Calculated results
Result name
Result description Value Error
Az Az -2.4 cm² 0.0000 % Amin Amin 4.8 cm² 0.0000 %
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5.51 EC2 Test 38: Verifying a rectangular concrete column using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1)
Test ID: 5127
Test status: Passed
5.51.1 Description Verifies a rectangular cross section concrete C25/30 column using the simplified method –Professional rules - Bilinear stress-strain diagram (Class XC1).
The column is considered connected to the ground by an articulated connection (all the translations are blocked). At the top part, the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.
5.51.2 Background Simplified Method
Verifies the adequacy of a rectangular cross section made from concrete C25/30.
5.51.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: ► 800kN axial force ► The self-weight is neglected
■ Concrete cover 5cm ■ Concrete C25/30 ■ Steel reinforcement S500B ■ Relative humidity RH=50% ■ Buckling length L0=6.50m
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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:
kNNED 1080800*35.1 ==
5.51.2.2 Reference results in calculating the concrete column
Scope of the method:
06.753.0
12*5.612*
121212
002
0
2
4000 ========
aL
aL
aL
a
aL
AIL
iLλ
According to Eurocode 2 EN 1992-1-1 (2004) Chapter 5.8.3.2(1)
The method of professional rules can be applied as:
120<λ
MPafck 5020 <<
mh 15.0>
Reinforcement calculation:
12006.7560 ≤=< λ
33.006.75
3232 3.13.1
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
λα
According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283
93.0)**61(*)*5.075.0( =−+= δρhkh
1500500*6.06.1
500*6.06.1 =−=−= yk
s
fk
²43.2367.16*5.0*3.033.0*1*93.0
08.1*78.434
1****
*1 cmfhbkkN
fA cd
sh
ed
yds =⎟
⎠⎞
⎜⎝⎛ −=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
α
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Finite elements modeling
■ Linear element: S beam, ■ 8 nodes, ■ 1 linear element.
Theoretical reinforcement area(cm2)
(reference value: 23.43cm2=4*5.86cm2)
5.51.2.3 Reference results
Result name Result description Reference value Ay Reinforcement area [cm2] 5.85cm2
R Theoretical reinforcement area [cm2] 23.40 cm2
5.51.3 Calculated results
Result name
Result description Value Error
Ay Ay -5.85106 cm² 0.0001 %