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


15

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


16

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


17

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


18

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


19

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


22

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


23

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


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


24

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%


25

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,


26

■ 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.


■ 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


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


27

Eigen mode shapes


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


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%


28

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 )


29


■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: ρ = 7800 kg/m3.

Boundary conditions

■ Outer: AB side fixed, ■ Inner: None.

Loading


1.3.2.2 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.


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

Eigen mode shapes


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


30


Result name


Eigen mode 1 frequency [Hz] 8.95 Hz -0.54% Eigen mode 2 frequency [Hz] 21.69 Hz -2.17%


31

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,


32

■ Point coordinates: ► O (0 ;0),

► A (-0.05 3 ; -0.05),

► B (0.05 3 ; -0.05).


■ 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



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


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


33

Eigen mode shapes


34

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


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%


35


Test ID: 2442

Test status: Passed


1.5.2 Background



Units

I. S.

Geometry

■ Thickness: t = 0.01 m, ■ Side: a = 1 m, ■ α = 45°


36

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

► C ( 22

a ; 22

a ; 0 )

► D (2

22 +a ;

22

a ; 0 )



Boundary conditions


Loading


1.5.2.2 Eigen mode frequencies relative to the α angle

Reference solution


fj = ⋅⋅π 2a21

λi2

)1(12Et

2

2


2 = g(α).

α = 45° λ1

2 4.4502 λ2





37

Eigen mode shapes




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%


38

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.


39

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



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



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,


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


40

Eigen mode shapes


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


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%


41

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.


42

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 ).



Boundary conditions

■ Outer: Fixed in A, ■ Inner: None.

Loading



Reference solutions

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




43

Eigen mode shapes


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


44


Result name


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%


45

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,


46

■ AD, DF, FB Beams: ► Section: A = 0.01516 m2, ► Shear area: Ar = A / 2.5, ► Inertia moment: I = 2.174 x 10-4 m4,

■ CE Bar: ► Section: A1 = 4.5 x 10-3 m2,

■ AC, EB bar: ► Section: A2 = 4.5 x 10-3 m2,

■ CD, EF bars: ► Section: A3 = 3.48 x 10-3 m2.


■ Isotropic linear elastic material, ■ 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


Linear element: without meshing,

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


47

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


Linear element: without meshing,

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


48

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


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

■ 6 nodes.


49

Deformed shape

Tied (sub-tensioned) beam Scale =1/31 Deformed


Solver Result name Result description Reference value CM2 FX Tension force on CE bar [N] 584584


Result name Result description Value Error Fx Tension force on CE bar [N] 584580 N 0.00%


50

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.


■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3.


51

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


■ Planar element: plate, imposed mesh, ■ 70 nodes, ■ 58 planar elements.

Circular plate under uniform load Scale =1.5 Meshing 01-0003SSLSB_FEM


52

Deformed shape

Circular plate under uniform load Scale =1.5 Deformed 01-0003SSLSB_FEM


Solver Result name Result description Reference value CM2 DZ Vertical displacement on the plate center [mm] -6.50


Result name


DZ Vertical displacement on the plate center [mm] -6.47032 mm 0.46%


53

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


■ 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.


54

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


■ 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




55

Modal deformations

Double fixed beam Mode 1





56


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


Result name


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%


57

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.


58



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.


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

Eigen modes shape

Short beam on simple supports (eccentric) Mode 1


59





60



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


Result name


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%


61


Test ID: 2441

Test status: Passed


1.12.2 Background



Units

I. S.

Geometry


► A ( 0 ; 0 ; 0 ) ► B ( a ; 0 ; 0 )


62

► C ( 0.5a ; 3 2 a ; 0 )

► D ( 1.5a ; 3 2 a ; 0 )



Boundary conditions


Loading



Reference solution


fj = ⋅⋅π 2a21

λi2

)1(12Et

2

2


2 = g(α).

α = 30° λ1

2 3.961 λ2




Eigen mode shapes


63




Result name


Eigen mode 1 frequency [Hz] 9.82 Hz -0.80% Eigen mode 2 frequency [Hz] 23.44 Hz -7.95%


64


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


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.


65




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



Reference solution



fj = 2

2i

R2 ⋅π

λ

AEIz

ρ where i = 1,2,




66

Eigen mode shapes


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


Result name


Eigen mode frequency in plane 1 [Hz] 24.96 Hz -1.34% Eigen mode frequency in plane 2 [Hz] 26.71 Hz 100%


67

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.


68



Boundary conditions

■ Outer: ► Hinged at A (null horizontal and vertical displacements), ► Simple support in B.

■ Inner: None.

Loading



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( −


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


69

Eigen mode shapes


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


70


Result name


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%


71


Test ID: 2439

Test status: Passed


1.15.2 Background



Units

I. S.

Geometry


► A ( 0 ; 0 ; 0 ) ► B ( a ; 0 ; 0 ) ► C ( 0 ; a ; 0 ) ► D ( a ; a ; 0 )


72



Boundary conditions


Loading



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.



Eigen mode shapes




73


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%


74


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


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,


75

■ 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 )



Boundary conditions

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

Loading



Reference solution



fj = 2

2i

R2 ⋅π

λ

AEIz

ρ where i = 1,2,



Eigen mode shapes


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


76


Result name


Eigen mode frequency in plane 1 [Hz] 120.09 Hz 0.92% Eigen mode frequency in plane 2 [Hz] 227.1 Hz 100%


77

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



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.


78


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


79

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) →


■ 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


Solver Result name Result description Reference value CM2 Dz Vertical displacement on the middle of the beam [mm] -0.11905


Result name


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


80

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


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)


81



Boundary conditions

■ Outer: ► Simple support on all sides, ► For the modeling: hinged at A, B and D.

■ Inner: None.

Loading



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)


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


82

Eigen mode shapes


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


83


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%


84

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


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.


85


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

Boundary conditions


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




86

Results shape

Slender beam on two fixed supports Scale = 1/5 Shear force

1.19.2.3 Bending moment in G

Reference solution


■ Bending moment at G: MG

MG = pL2

24 - 0.045LF – 0.3C




87

Results shape

Slender beam on two fixed supports Scale = 1/5 Bending moment

1.19.2.4 Vertical displacement at G

Reference solution


■ Vertical displacement at G: vG

vG = pl4

384EI + 0.003375FL3

EI + 0.015CL2

EI




88

Results shape

Slender beam on two fixed supports Scale = 1/4 Deformed

1.19.2.5 Horizontal reaction at A

Reference solution


■ Horizontal reaction at A: HA HA = -0.7F1 –0.3F2




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


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%


89

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



90

Boundary conditions

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

Loading



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


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


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


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%


91

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


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.


92


■ 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



Results shape

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


93

1.21.2.3 Vertical reaction at A and C

Reference solution

Analytical solution.



1.21.2.4 Bending moment at A and C

Reference solution



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


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


Result name


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%


94

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


■ 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.


95

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

=


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

Deformed shape


Solver Result name Result description Reference value CM2 Fx Critical load on node 5. [N] -98696


Result name


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


96

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.


97


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(

++−



Results shape

Slender beam on three supports Scale = 1/49 Bending moment


98

1.23.2.3 Reaction in B

Reference solution

■ Compression force in the spring: VB

VB = -11F8 + k



1.23.2.4 Vertical displacement at B

Reference solution

■ Deflection at the spring location: vB

vB = 11F

Ky(8 + k)



Results shape

Slender beam on three supports Deformed


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


99


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%


100

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.


101

Geometry

■ Medium radius: R = 1 m , ■ Circular hollow section:

► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.


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



1.24.2.3 Moments at θ = 15°

Reference solution

■ Torsion moment: Mx’ = Mt = FR(1 - sinθ) ■ Bending moment: Mz’ = Mf = -FRcosθ




102


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


Result name


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%


103

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


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


104

■ 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


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




105

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




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


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%


106

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


► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.


107


■ 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



Displacements shape

Fixed thin arc in planar bending Scale = 1/11 Deformed


108


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


Result name


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%


109

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


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)


110



Boundary conditions

■ Outer: Edge AD fixed. ■ Inner: None.

Loading



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




111

Eigen mode shapes

Thin square plate fixed on one side Mode 1




112


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


Result name


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%


113

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,


114

■ 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



Boundary conditions


Loading



Reference solution

Dynamic radius method (slender beams theory).



Deformed shape

Bending effects of a symmetrical portal frame Scale = 1/7 Mode 13


115


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


Result name


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%


116

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


Fixed thin arc in planar bending Scale = 1/24 01-0027SSLLB_FEM

Units

I. S.

Geometry


► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.


117


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]




118

Results shape

Fixed thin arc in planar bending Scale = 1/19 Deformed


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


Result name


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%


119

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


Beam on elastic soil, hinged ends Scale = 1/27 01-0034SSLLB_FEM

Units

I. S.

Geometry

■ L = (π 10 )/2, ■ I = 10-4 m4.


120



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




121

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




122

Bending moment diagram

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


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


Result name


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%


123

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.




124

Boundary conditions

■ Outer: Fixed on 3 sides, ■ Inner: None.

Loading

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


Reference solution

The reference displacements are calculated on nodes 7 and 9.

v9 = -6ph(3 + 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


■ 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


125

Deformed shape


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


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%


126

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


Beam on elastic soil, free ends Scale = 1/21 01-0032SSLLB_FEM

Units

I. S.

Geometry

■ L = (π 10 )/2, ■ I = 10-4 m4.


127



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



Bending moment diagram

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


128

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




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


Result name


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%


129

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



Boundary conditions

■ Outer: ► Hinged support, ► For the modeling, a fixed restraint and 4 beams were added at the pylon supports level.

■ Inner: None.


130

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.


Reference solution


131

Software ANSYS 5.3 NE/NASTRAN 7.0 Max deflection (m) 0.714 0.714 λ dominating mode 2.77 2.77



Deformed shape


132

Buckling modal deformation (dominating mode)


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


Result name


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%


133

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.




134

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



1.34.2.3 Cylinder deformation in all points ■ Radial deformation:

δR = pR2Eh

■ Longitudinal deformation:

δL = -pRνL

Eh


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


Result name


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%


135

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.




136

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).



Deformed shape

Caisson beam in torsion Scale = 1/4 Deformed


137


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


Result name


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%


138

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


■ Longitudinal elastic modulus: E = 2285938 daN/cm2, ■ Transverse elastic modulus G = 879207 daN/cm2 ■ Poisson's ratio: ν = 0.3.


139

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


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

Deformed shape


140


Solver Result name Result description Reference value CM2 DZ Vertical displacement at node 6 [cm] -1.017


Result name


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


141

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.




142

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


■ 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




143

Deformation shape

Thin cylinder under a hydrostatic pressure Deformed


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


Result name


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%


144

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





145

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


σxx = qh

σyy = 0



1.38.2.3 Cylinder deformation at the free end

Reference solution


δL = qLEh

■ δR radial deformation of the cylinder:

δR = -qνREh




146

Deformation shape

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


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


Result name


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%


147

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


148



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.



Displacements shape

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


149


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


Result name


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%


150

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


Simply supported square plate Scale = 1/9 01-0036SSLSB_FEM

Units

I. S.

Geometry

■ Side = 1 m, ■ Thickness h = 0.01m.




151

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



Deformed shape

Simply supported square plate Scale = 1/6 Deformed


152


Solver Result name Result description Reference value CM2 Dz Vertical displacement in point O [μm] -0.158


Result name


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


153

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.


■ Longitudinal elastic modulus: E = 30 x 106 N/cm2, ■ Poisson's ratio: ν = 0.3.

Boundary conditions

■ Outer: Fixed on 3 sides, ■ Inner: None.


154

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


■ Planar element: membrane, imposed mesh, ■ 6 nodes, ■ 2 quadrangle planar elements and 1 bar.

Deformed shape


155


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


Result name


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%


156

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


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.


157



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




Reference solution

■ δR radial deformation of the torus:

δR = pb

2Eh (r - ν(r + a))




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


158


Result name


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%


159

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


Spherical dome under a uniform external pressure

01-0050SSLSB_FEM

Units

I. S.


160

Geometry

■ Radius: a = 2.54 m, ■ Thickness: h = 0.0127 m, ■ Angle: θ = 75°.



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.


■ Planar element: shell, imposed mesh, ■ 401 nodes, ■ 400 planar elements.

Deformed shape


161


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


Result name


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%


162

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.


163

Geometry

■ Length: L = 10.35 m (total length), ■ Radius: R = 4.953 m, ■ Thickness: h = 0.094 m.



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




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


Result name


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


164

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


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,




165

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.



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


Result name


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%


166

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


Spherical shell under internal pressure 01-0046SSLSB_FEM

Units

I. S.


167

Geometry

■ Thickness: h = 0.02 m, ■ Radius: R2 = 1 m, ■ θ = 90° (hemisphere).



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




Reference solution

■ δR radial deformation of the calotte:

δR = pR22 (1 - ν) sin θ

2Eh




168

Deformed shape

Spherical shell under internal pressure Scale = 1/11 Deformed


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


Result name


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%


169

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


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,




170

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.


Reference solution



■ 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


Result name


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%


171

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


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

Units

I. S.

Geometry

■ Width: DA = CB = 20 m, ■ Length: AB = DC = 5 m, ■ Thickness: h = 1 m,


■ Longitudinal elastic modulus: E =1000 Pa, ■ Poisson's ratio: ν = 0.3.

Boundary conditions

■ Outer: Punctual support at A, B and D (null displacement along z-axis), ■ Inner: None.


172

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




Deformed shape

Simply supported rectangular plate loaded with punctual force and moments Deformed


Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -12.480


Result name Result description Value Error Dz Vertical displacement in point C [m] -12.6677 m -1.50%


173

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.


■ 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.


174

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.


■ 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.


■ 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


175


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


Result name


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%


176

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,



Boundary conditions

■ Outer: Fixed AD edge, ■ Inner: None.


177

Loading

■ External: ► At B: Fz = -1.0 N, ► At C: FZ = 1.0 N,

■ Internal: None.


Reference solution




Deformed shape

Shear plate Scale = 1/35 Deformed


Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] 35.37 x 10-3


Result name


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


178

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



■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa, ■ Poisson's ratio: ν = 0.3, ■ Density: γ = 7.85 x 104 N/m3.


179

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


σxx = γz

σyy = 0




Reference solution


δL = γz2

2E

■ δR radial deformation of the cylinder:

δR = -γνRz

E


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.


Result name


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%


180

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°.


181



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




Deformed shape

Spherical shell with holes Scale = 1/79 Deformed


182

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


Result name


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


183

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


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,



Boundary conditions

■ Outer: Simple support on the plate perimeter (null displacement along z-axis), ■ Inner: None.


184

Loading

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


Reference solution





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


Result name


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%


185

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.


■ 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.


186

Loading

■ External: 1 MPa uniform pressure, ■ Internal: None.


Reference solution

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




Solver Result name Result description Reference value CM2 Dz Vertical displacement in point C [m] -2.8053 x 10-2


Result name


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


187


Test ID: 2490

Test status: Passed


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





Boundary conditions

■ Outer: Fixed edges: AB and BD,


■ Inner: None.


188

Loading



Reference solution







Result name


Dz Vertical displacement in point C [cm] -0.0549874 cm

0.88%


189

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


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.




190

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





Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.037454


Result name


DZ Vertical displacement in point C [m] -0.0369818 m 1.26%


191


Test ID: 2492

Test status: Passed


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".


Units

I. S.

Geometry




192



Boundary conditions


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





193




Result name




194


Test ID: 2489

Test status: Passed


1.58.2 Background


Units

I. S.

Geometry





Boundary conditions



■ Inner: None.


195

Loading



Reference solution







Result name




196


Test ID: 2493

Test status: Passed


1.59.2 Background



Units

I. S.

Geometry




197



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





198




Result name




199


Test ID: 2496

Test status: Passed


1.60.2 Background



Units

I. S.

Geometry





200

Boundary conditions


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





Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.42995 x 10-3


Result name


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

4.15%


201

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,


202


► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 ) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )



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



Reference solution


■ transverse bending:

fj = μi

2

2π R2 GIpρA where i = 1,2.



Eigen mode shapes


203


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


Result name


Eigen mode Transverse 1 frequency [Hz] 33.19 Hz -0.63% Eigen mode Transverse 2 frequency [Hz] 94.62 Hz -5.38%


204


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.


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,


205


► O ( 0 ; 0 ; 0 ) ► A ( 0 ; R ; 0 ) ► B ( R ; 0 ; 0 )



Boundary conditions

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

Loading



Reference solution



fj = μi

2

2π R2 GIpρA where i = 1,2.



Eigen mode shapes


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


206


Result name




207


Test ID: 2487

Test status: Passed


1.63.2 Background


Units

I. S.

Geometry





Boundary conditions

■ Outer: Fixed sides: AB and BD,


■ Inner: None.


208

Loading



Reference solution







Result name




209


Test ID: 2491

Test status: Passed


1.64.2 Background


Units

I. S.

Geometry





Boundary conditions



■ Inner: None.


210

Loading



Reference solution







Result name


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

0.61%


211


Test ID: 2495

Test status: Passed


1.65.2 Background



Units

I. S.

Geometry





212

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





Solver Result name Result description Reference value CM2 DZ Vertical displacement in point C [m] -0.2595 x 10-2


Result name




213

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.


214

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 +−=


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


Result name


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%


215

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

Test ID: 2504

Test status: Passed

1.67.1 Description Verifies the eigen modes (flexion) for a slender beam with variable rectangular section (fixed-fixed).

1.67.2 BackgroundOverview


Units

I. S.

Geometry

■ Length: L = 0.6 m, ■ Constant thickness: h = 0.01 m ■ Initial section:

► b0 = 0.03 m ► A0 = 3 x 10-4 m²

■ Section variation: ► with (α = 1) ► b = b0e-2αx ► A = A0e-2αx


■ 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



216

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


217


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


Result name


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%


218

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


■ E = 2 x 1011 Pa ■ ν = 0.3

Boundary conditions

■ Hinge at end x = 0, ■ Hinge at end x = 1.44 m.


219

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.


Reference values

Point Magnitudes and units Value C V, deflection (m) -1.25926 x 10-3


Solver Result name Result description Reference value CM2 Dz Deflection magnitude in point C [m] -1.25926


Result name


Dz Deflection magnitude in node C [m] -0.00125926 m

0.00%


220

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


■ 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



221

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=


Reference solution


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



Deformed shape of mode 1


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


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%


222

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.


223

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 −=


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


Result name


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%


224

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


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


■ Isotropic linear elastic material. ■ E = 2.1 x 1011 Pa


225

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


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


Result name


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%


226

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


■ Isotropic linear elastic materials ■ Longitudinal elastic modulus: E = 2.1 E8 N/m2,


227

Boundary conditions

■ Outer: At node 5: K = 50000 kN/m ; ■ Inner: None.

Loading

■ External: Vertical load at node: P = -100 kN, ■ Internal: None.


System solution

2

22 BHL += . Also, U1 = V1 = U5 = U6 = V6 = 0

■ Stiffness matrix of bar 1


228

( ) ( )

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


229

■ 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


230

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


■ Linear element: beam, automatic mesh, ■ 5 nodes, ■ 4 linear elements.


231

Deformed shape

Normal forces diagram


232

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


Result name


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%


233

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


234


■ 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%


■ Linear element AB: Beam ■ Imposed mesh: 50 elements. ■ Linear element BC: Beam ■ Without meshing

Modal deformations


235


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.


236


Result name


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%


237

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


238


■ 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



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%


■ Linear element AB: Beam ■ Beam meshing: 20 elements.


239

Modal deformations


240

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:


241


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).


Result name


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%


242


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.


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.


243




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



Reference solution



fj = μi

2

2π R2 GIpρA where i = 1,2 with i = 1,2:



Eigen mode shapes


244


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


Result name




245

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


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


■ 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



246

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


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


247

MODE 2 Scale = 1/4

MODE 3 Scale = 1/4


248

MODE 4 Scale = 1/4

MODE 5 Scale = 1/4


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


249


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%


250

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


■ 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


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 )


251


Reference solution


98696.010000098696 98696

4P 2

2

critical ==⇒== λπ NLEI

Observation: in this case, the thermal load has no effect over the critical coefficient



Deformed shape


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


Result name


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%


252

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.


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


253


■ 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


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.


■ Straight elements: linear element ■ Imposed mesh: 5 meshes


254

Modal deformations


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.


Result name


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%


255

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.


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.


256

Units

I. S.

Geometry / meshing

A quarter of the structure is modeled by incorporating the terms of symmetries.

Thickness: 1 m


■ 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;


257

1.79.3 σyy stress at point A:

Reference solution:

Reference value: σyy = 50 MPa in A


■ Planar elements: membranes, ■ 28 planar elements, ■ 39 nodes.


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.


Result name


syy_mid Sigma yy in A [MPa] 50.8666 MPa 1.73%


258

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



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.


259

1.80.2.2 References solutions

Displacements

( )( )

( )( )

( )( )

( ) rad 000034.0Lk2EI3EI32

LkEI6PL

m 00058.0Lk2EI316

PL3v

rad 000106.0Lk2EI3EI16

LkEI3PL


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

−=−

+==

−=+

=

=


■ Linear element: S beam, automatic mesh, ■ 3 nodes, ■ 2 linear elements + 1 T/C.

Deformed shape


260

Moment diagram


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


Result name


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%


261

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


Units

I. S.

Geometry




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



262

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


LkEI3PL


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

−=−

+==

=+

=

=



Deformed shape


263

Moment diagrams


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


Result name


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%


264

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


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.


265

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

==

==−=−=

==


■ Linear element: bar, without meshing, ■ 4 nodes, ■ 6 linear elements.

Deformed shape


266

Normal forces


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


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%


267

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


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.


268

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

−=−===

==−=−=

===


■ Linear element: bar, without meshing, ■ 4 nodes, ■ 6 linear elements.

Deformed shape


269

Normal forces


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


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%


270

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,


■ 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 = -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


271

► Case 2 to 8 (LL):uniform linear load q = -9 kN/m (on spans 1, 3 to 7) uniform linear load q = -15 kN/m (on span 2)

Fx = 30 kN at x = 6m (case 2 span 1)

Fx = -50 kN at x = 6m (case 3 span 2)










Fx = -8 kN at x = 36m (case 8 span 7


► Case 9 (ACC):uniform linear load a = -25 kN/m (on 8th span)



Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107)

Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102)

Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103)

Comb BAELS: 1xDL + 1*LL (comb 108 to 114)

Comb BAELA: 1xDL + 1xACC with duration of less than 1h (comb 115)

■ Internal: None.

Reinforced concrete calculation hypothesis:

All concrete covers are set to 5 cm

BAEL 91 calculation (according to 99 revised version)

Span Concrete Reinforcement Application Concrete Cracking 1 B20 HA fe500 D>24h No 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


272

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


273

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


■ Linear elements: beams with imposed mesh ■ 29 nodes, ■ 28 linear elements.


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.


274


Result name


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%


275

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


276

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


277

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


■ 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


■ 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.


278

Deflection verification

Ratio

Max values on the element

■ Columns: L / 168 ■ Rafter: L / 96 ■ Column: L / 924


279

CM Stress diagrams

Work ratio

Stresses

Max values on the element

■ Columns: 375.16 MPa ■ Rafter: 339.79 MPa ■ Column: 180.98 MPa


280

Buckling lengths

Lfy

Lfz


281

Lateral-torsional buckling lengths

Ldi

Lds


282

Optimization


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


283

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


Result name


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%


284

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,


■ 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.


285

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


286

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


■ Linear elements: beams with imposed mesh ■ 29 nodes, ■ 28 linear elements.


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.


287


Result name


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%


288

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.


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


■ Longitudinal elastic modulus: E = 2 x 1011 Pa, ■ Density: ρ = 8000 kg/m3


289

Boundary conditions

■ Outer: A, B, C, D, E, F, G, H points restraint along x and y; ■ Inner: None.

Loading


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


■ Linear elements type: Beam ■ Imposed mesh: 4 Elements / Arms

Modal deformations


290


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


Result name


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%


291

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


Units

I. S.

Geometry




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



292

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


LkEI3PL


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

−=−

+==

−=+

=

=



Deformed shape


293

Moment diagram


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


Result name


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%


294

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


■ Longitudinal elastic modulus: E = 1.962 x 1010 N/m2, ■ Poisson's ratio: ν = 0.1, ■ Density: ρ = 25 kN/m3


295

Boundary conditions

■ Outer: Fixed in X = 0, Y = 0 m,

Loading

■ External: Seismic excitation on X direction


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


296

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


297

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


298

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


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


299


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%


300

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


301

Slab geometry

Support positions


302

Positions of punctual loads

Global loading overview


303

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


304

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.


305

Axi reinforcements

Ayi reinforcements


306

Axs reinforcements

Ays reinforcements


307


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.


Result name


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%


308

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


309

1.91.2.2 Effel Structure Results

Ridge displacements (combination 104)

Diagram of normal force envelope


310

Envelope of bending moments diagram

1.91.2.3 Effel Expert CM results

Main hypotheses

For columns


■ Buckling: XY plane: Automatic calculation of the structure on fixed nodes XZ plane: Automatic calculation of the structure on fixed nodes

Ka-Kb Method

■ Lateral-torsional buckling: Ldi automatic calculation: no restraints Lds imposed value: 2 m

For the rafters


■ Buckling: XY plane: Automatic calculation of the structure on fixed nodes 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)


311

Deflection verification

Ratio

CM Stress diagrams

Work ratio

Stresses


312

Buckling lengths

Lfy

Lfz

Lateral-torsional buckling lengths

Ldi


313

Lds

Optimization


314


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


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%


315

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.




316

Boundary conditions


Loadings

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


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



Deformed shape

Deformed cantilever rectangular plate Scale =1/4 01-0001SSLSB_FEM


317


Solver Result name Result description Reference value CM2 DZ Vertical displacement on the free extremity [cm] -9.71


Result name


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


318

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.


319

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.


320

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.


321

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.


322

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


324

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.


325

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.


326

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.


327

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.


328

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.


329

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


332

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.


333

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.


334

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.


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.


335

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.


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).


336

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.


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.



337

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.


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.


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.



338

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.


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.


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.



339

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.


340

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.


341

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).


342

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).


343

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).


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).


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).


344

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


346

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.


347

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.


348

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 !


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




3 - E Group (Base)

3 - EX

4 - EY

5 - EZ

Un-group 3 - E Group to independent loads : 3 - EX 4 - EY 5 - EZ



349


Test ID: 4391

Test status: Passed



2 - Q (Base)


4 - Q (Base)



Test ID: 4397

Test status: Passed




3 - E (Base) - only one seismic load



350


Test ID: 4407

Test status: Passed



2 - Q Group(Base or Acco)

2 - Q

3 - Q

4 - Q


6 - Snow (Base or Acco)


Un-group 2-Q Group to independent 2-Q, 3-Q, 4-Q loads.



Test ID: 4409

Test status: Passed




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.


351


Test ID: 4405

Test status: Passed

4.16.1 Description Creates loads:




4 - Q (Acco)

5 - Acc (Base)


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




3 - Q (Acco)



352

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.


Test ID: 4384

Test status: Passed





3 - Q (Base)



353


Test ID: 4386

Test status: Passed







5 Concrete Design


356

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.


357

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.


358

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.


359

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.


360

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.


361

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.


362

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


363

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


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.


364

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 =<=


365


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 ===


In conclusion the entire reinforcement steel area is A=A1+A2=4.25+22.08=29.33cm2


■ 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)


366


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


Result name


My My USL -540.63 kN*m 2.0408 % Az Az -28.6439 cm² 2.3507 %


367

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.


■ 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


368


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


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===



MPafE ckcm 31476

1082522000

10822000

3.03.0

=⎟⎠⎞

⎜⎝⎛ +

=⎟⎠⎞

⎜⎝⎛ +

=

■ Steel S500 :



369

Boundary conditions


■ 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


■ 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


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


CCQ = 1.0 x G + 1.0 x Q=35+25=60 kN


CQP = 1.0 x G + 0.6 x Q=35+0.8*25=55 kN

■ 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=+=


370

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


at t0 = 28 days


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


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


371

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

‰


372

■ 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.


373

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


374

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


375

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



ULS and SLS load combinations(kNm)


ULS (reference value: 1002kNm)

SLS (reference value: 705kNm)

SLS –Quasi-permanent (reference value: 637kNm)


376


For Class A reinforcement steel ductility (reference value: A=39.56cm2 and A’=3.79cm2)

Minimum reinforcement area(cm2)

(reference value: 3.77cm2)


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


Result name


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 %


377

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.


■ 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



378

Units

Metric System

Geometry


■ Beam length: 6m ■ Concrete cover: c=4.00 cm ■ Effective height: d=h-(0.6*h+ebz)=0.900 m; d’=ebz=0.04m


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===


MPa.*.f*.f //ckctm 90230300300 3232 ===


■ Steel S400B :

MPa.,

ff

s

ykyd 78434

151500

γ===


Boundary conditions


■ 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.


379

Loading


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


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


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 =<=


380

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 =

⎟⎠

⎞⎜⎝

⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=


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 ===


In conclusion the entire reinforcement steel area is A=A1+A2=91.74+57.46=149.20cm2



⎪⎩

⎪⎨

⎧=

d*b*.

d*b*f

f*.

MaxA

w

wyk

eff,ct

min,s00130

260


Where:

MPa.ff ctmeff,ct 902== from cracking conditions


381

Therefore:

⎪⎩

⎪⎨⎧

==

==

−

−²cm.

²m*..*.*.

²m*..*.*.*.maxA min,s 4251068490040000130

10425900400500

9022604

4



ULS and SLS load combinations (kNm)


ULS (reference value: 5062.5kNm)


For Class B reinforcement steel ductility (reference value: A=149.20cm2)




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



382


Result name


My My USL -5062.5 kN*m 0.0000 % Az Az -149.031 cm² 0.0001 % Amin Amin -5.42219 cm² 0.0000 %


383

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.


■ 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


Units

Metric System


384

Geometry


■ 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



■ 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


■ 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


■ Load combinations: Characteristic combination of actions:

CCQ = 1.0 x G + 1.0 x Q = 25 + 15 = 40 kN/ml


CQP = 1.0 x G + 0.3 x Q = 25 + 0.3*15 = 29.5 kN/ml


kNm².*)(MEcq 2058

4061525=

+=

kNm².*)*.(MEqp 1518

406153025=

+=


385

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:


Ecar

Eqp

cm

se

MM

t

EE

*),(1 0∞+

=

ϕ

α


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 ===


387



⎪⎩

⎪⎨

⎧=

d*b*.

d*b*f

f*.

MaxA

w

wyk

eff,ct

min,s00130

260


Where:


Therefore:

⎪⎩

⎪⎨⎧

==

==

−

−²cm.

²m*..*.*.

²m*..*.*..maxA min,s 4021007257028000130

104025702805008962260

4

4






Theoretical reinforcement area (cm2)



388

Minimum reinforcement area (cm2)



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



Result name


My My SLS -200.533 kN*m 2.0833 % Az Az -10.2949 cm² 0.5220 % Amin Amin -2.38697 cm² 0.0001 %


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.


■ 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



390

Units

Metric System

Geometry


■ 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



■ 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



■ Inner: None.

Loading



CCQ = 1.0 x G + 1.0 x Q=20+10=30kN/ml


CQP = 1.0 x G + 0.3 x Q=20+0.3*10=23kN/ml


391


( ) 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


Where:

MPaf

fcm

cm 17.3820

8.168.16)( =+

==β

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t

tβ


210

3**

*1.0100

11 ααϕ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

h

RH

RH


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===∞⇒=



392



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




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


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 ===



⎪⎩

⎪⎨

⎧=

d*b*.

d*b*f

f*.

MaxA

w

wyk

eff,ct

min,s00130

260


Where:


Therefore:

⎪⎩

⎪⎨⎧

==

==

−

−²cm.

²m*..*.*.

²m*..*.*.*.maxA min,s 61210612669030000130

103126690300500

2122604

4


394



SLS load combinations(kNm)




(reference value: As=9.58cm2)







Result name


My My SLS -240 kN*m 0.0000 % Az Az -9.64217 cm² -0.0000 % Amin Amin -2.574 cm² -0.0000 %


395

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

Test ID: 4999

Test status: Passed


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.


■ 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



396

Units

Metric System

Geometry


■ 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



■ 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



■ Inner: None.

Loading



CCQ = 1.0 x G + 1.0 x Q=40+10=50kN/ml


CQP = 1.0 x G + 0.3 x Q=40+0.3*10=43kN/ml

■ Load calculations: ( ) kNm²*M cq,ser 400

881040

=+

=

( ) kNm²**.M qp,ser 3448

8103040=

+=


397



Where:

MPaf

fcm

cm 17.3820

8.168.16)( =+

==β

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t

tβ


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:


Humidity RH=50 %

mmuAch 123

3140192600*22

0 ===

11.3488.0*17.3*2)(*)(*),(2123*1.0

100501

1 003===∞⇒=




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 =+=∞+


398

54.24

400344*11.31

29962200000

*),(1 0

=

+

=

∞+

=

Ecar

Eqp

cm

se

MM

t

EE

ϕ

α


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


399

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 ===



⎪⎩

⎪⎨⎧

=db

dbff

MaxA

w

wyk

effct

s

**0013.0

***26.0 ,

min,


Where:


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




400

SLS load combinations(kNm)




(reference value: As=18.83cm2)







Result name


My My SLS -400 kN*m 0.0000 % Az Az -18.8948 cm² 0.0001 % Amin Amin -1.36843 cm² -0.0001 %


401

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


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.


402

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


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.


■ 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



403

Units

Metric System

Geometry


■ 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;


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


■ 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


■ Load calculations: ► M0Ed = 774 kNm ► Mcar = 540 kNm ► Mfq = 390 kNm ► Mqp = 330 kNm


404



Where:

MPaf

fcm

cm 92.2825

8.168.16)( =+

==β

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t

tβ


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===∞⇒=



Under quasi-permanent combinations:

),(1 0tEEcm

se

∞+

=

ϕ

α

Where:

MPafE ckcm 31476

10825*22

108*22

3.03.0

=⎟⎠⎞

⎜⎝⎛ +

=⎟⎠⎞

⎜⎝⎛ +

=

MPaEs 200000=


405

55.2),( 0 =∞ tϕ

56.22

55.2131476200000

),(1 0

=

+

=

∞+

=

tEEcm

se

ϕ

α


The maximum compression on the concrete is: Mpafckbc 1525*6,06,0 ===σ



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


406

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


Calculation of crack widths:

mmsw cmsmrk 128.0)1088.7(*162.0)(* 4max, =×=−= −εε


For an exposure class XD1, using the French national annex, we retained an opening crack of 0.20mm max.

This criterion is satisfied.




407

SLS load combinations (kNm) and stresses (MPa)


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)


408

Maximum crack opening Wk

(reference value: Wk=0.128mm)


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


Result name


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 %


409

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.


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.


■ 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



410

Units

Metric System

Geometry


■ 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



■ 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


■ 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


■ Dead load: 0.40*0.80*25 = 8kN/ml


CCQ = 1.0 x G + 1.0 x Q=8+50+60=118kN/ml


CQP = 1.0 x G + 0.3 x Q=8+50+0.3*60=76kN/ml


( ) ( ) mkNM cqser .05.6728

²3.6*605084

3.6*2530, =

+++

+=

( ) ( ) mkNM qpser .11.4368

²3.660*3.05084

3.6*25*3.030, =

×+++

+=


411



Where:

MPaf

fcm

cm 92.2825

8.168.16)( =+

==β

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t

tβ


210

3**

*1.0100

11 ααϕ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

h

RH

RH


If not

7.0

135

⎟⎟⎠

⎞⎜⎜⎝

⎛=

cmfα and

2.0

235

⎟⎟⎠

⎞⎜⎜⎝

⎛=

cmfα

In this case MpaMpaMpaff ckcm 35338 ≤=+= therefore


Humidity RH=50 %

( ) mmuAch 267

800400*2800*400*22

0 =+

==

54.2488.0*92.2*78.1)(*)(*),(78.1267*1.0

100501

1 003===∞⇒=




Ecar

Eqp

cm

se

MM

t

EE

*),(1 0∞+

=

ϕ

α

Where:

MPafE ckcm 31476

10825*22

108*22

3.03.0

=⎟⎠⎞

⎜⎝⎛ +

=⎟⎠⎞

⎜⎝⎛ +

=

MPaEs 200000=


412

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





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


413

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 =+=+=



⎪⎩

⎪⎨⎧

=db

dbff

MaxA

w

wyk

effct

s

**0013.0

***26.0 ,

min,


Where:


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




414





(reference value: As=26.81cm2; A’=11.96cm2)







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 %


415

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.


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.


■ 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.



416

Units

Metric System

Geometry


■ 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;



■ Exposure class XD1 ■ Stress-strain law for reinforcement: Bilinear stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50%


■ Characteristic yield strength of reinforcement: Mpafyk 500=

■ ²cm.Ast 7037= for 3 HA20

■ ²cm.Asc 286= for 2 HA10

Boundary conditions



■ Inner: None.

Loading


■ 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


417



Where:

MPaf

fcm

cm 92.2825

8.168.16)( =+

==β

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t

tβ


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===∞⇒=



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


418

90.16

664.231476200000

*),(1 0

==

∞+

=

Ecar

Eqp

cm

se

MM

t

EE

ϕ

α


The maximum compression on the concrete is: Mpafckbc 1525*6,06,0 ===σ



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

=+++++−

=

=+++++−

=


443

11

31

0147201472097445349016286453472901670373

413435

αα3

m.cm)².(*.*,)²,(.*,,*

)²'dx(**A)²xd(**Ax*bI escestw

==⎥⎥⎦

⎤

⎢⎢⎣

⎡−+−+=

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−+−+=


MpaMpaxIM

cser

c 12143441,0*01472,0

600,0* 1 =≥=== σσ

MpaMpaxxd

scest 4002593441,0

3441,072,0*14*90.16**1

1 =≤=−

=−

= σσασ




419



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)


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


Result name


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 %


420

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.


The verification of the bending stresses at service limit state is performed.


421

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


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.


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.


■ 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.



422

Units

Metric System

Geometry


■ 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;



■ Exposure class XD1 ■ Stress-strain law for reinforcement: Inclined stress-strain diagram ■ The concrete age t0=28 days ■ Humidity RH=50%


■ Characteristic yield strength of reinforcement: Mpaf yk 400=

■ ²42,9 cmAst = for 3 HA20

Boundary conditions



■ Inner: None.

Loading


■ Characteristic combination of actions: CCQ = 1.0 x G =9.375kN/ml

■ Load calculations: Mfq = Mcar = Mqp = 75 kNm



Where:

MPaf

fcm

cm 17.3820

8.168.16)( =+

==β

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t

tβ



423

210

3**

*1.0100

11 ααϕ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

h

RH

RH


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===∞⇒=



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


424


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 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 =

++−=


443

1

31 001929,0)²230,044,0(*90.26*10*42,9

3230,0*20,0)²(**

3* mxdAxbI est

w =⎥⎦

⎤⎢⎣

⎡−+=⎥

⎦

⎤⎢⎣

⎡−+= −α


425


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





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)


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


426


Result name


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 %


427

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


428

■ 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


■ 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


■ 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.


429

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


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




430

Advance Design gives the following results for Atz (cm2/ml)


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


Result name


Fz Fz -292.852 kN -0.0002 % Atz Atz 5.65562 cm² 0.0000 %


431

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.


■ 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


432

Units

Metric System

Geometry


■ Height: h = 0.40 m, ■ Width: b = 0.40 m, ■ Length: L = 6.00 m, ■ Concrete cover: c = 5 cm

Boundary conditions


The column is considered connected to the ground by a fixed connection and free at the top.

Loading



NEd =1.35*15+1.5*6.5=30t=0.300MN

MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm

■ m...

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)


433

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



434

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 ===


435

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.


436

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


437

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=


438

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


439

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.





Theoretical value (cm2)

(reference value: 9.28 cm2)


440


Result name Result description Reference value Az Reinforcement area [cm2] 4.64 cm2

R Theoretical reinforcement area [cm2] 9.28 cm2


Result name


Az Az -4.64 cm² -0.0000 %


441

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


442

■ 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


■ Length: L = 10.00 m, ■ Stirrup slope: α= 90° ■ Strut slope: θ=29.74˚

Boundary conditions



■ Inner: None.


443

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


444

The minimum reinforcement percentage calculation:

mlcmbff

bsA

wyk

ckww

sw /²21.590sin*55.0*500

35*08,0sin***08,0

sin**min, =°==≥ ααρ


■ Linear element: S beam, ■ 15 nodes, ■ 1 linear element. Advance Design gives the following results for Atz (cm2/ml)


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


Result name


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 %


445

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.


■ 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


446

Units

Metric System

Geometry


■ Height: h = 0.30 m, ■ Width: b = 0.30 m, ■ Length: L = 4.74 m, ■ Concrete cover: c=5cm

Boundary conditions


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



NEd =1260kN

■ Proposed reinforcement: 2*6.28cm2=12.56cm2


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 =+=


For an isolated column, the slenderness limit check is done using the next formula:

nCBA ***20

lim =λ



447

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



( )Ed

EQPef M

Mt ., 0∞= ϕϕ


Where:




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:


448

)(*)(*),( 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 **** +=


ec

kkKϕ+

=1

* 21

835.21 =+ efϕ


449

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.


450

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)






451







Result name


Az Az -6.53808 cm² 0.0001 %


452


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:


453


■ The shear stresses results ■ The theoretical reinforcement value

Units

Metric System

Geometry


■ 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



■ 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==


454

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).


455

Calculation of maximum design shear resistance:

( )θ1θανα 1 2wucdcwmax,Rd

cotcotcot*b*z*f**V

+

+=


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


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


456


■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. Advance Design gives the following results for At,min,z (cm2/ml)


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


Result name


Fz Fz -168.937 kN -0.0003 % Atminz Atmin,z 1.43108 cm² 0.0002 %


457

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.


■ 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


458

Units

Metric System

Geometry



Boundary conditions


The column is considered connected to the ground by a fixed connection and free to the top part.

Loading



NEd =1.35*15+150*6.5=30t=0.300MN

MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm

■ m...

NM

eEd

Ed 10030000310

0 ===




mll 12*20 ==



10440.0

12*32*32 0 ===alλ



( )Ed

EQPef M

Mt ., 0∞= ϕϕ



459

Where:


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


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


( ) 49.1039.0022.0*64.2*, 0 ==∞=

Ed

EQPef M

Mtϕϕ



460



nCBA ***20

lim =λ


Where:

112.067.16*²40.0

300.0*

===cdc

Ed

fANn

( ) 77.049.1*2.01

1*2,01

1=

+=

+=

ef

Aϕ


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 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 ===


mlei 03.040012

4000 ===


461

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=






MNmhdNMM Gua 084.0240.035.0*300.0039.0)

2(*0 =⎟

⎠⎞

⎜⎝⎛ −+=−+=


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



462


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



24 06.178.434

300.010*84.5' cmfNAAyd

−=−=−= −


2min, 02.3*002.0²69.0

78.434300.0*10.0*10,0 cmAcm

fNA c

yd

Eds =≥===



The second order effect will be determined by applying the method of nominal rigidity:


The nominal rigidity of a post or frame member, it is estimated from the following formula:



With:

2.1cm

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


463

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


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).


234.18²

==πβ

MNlEINB 216.0

²1215.3*²*² 2

0

=== ππ


464

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.


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 =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+=


465

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


466

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


467

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.


468

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 =


234.1=β

MNlEINB 49.0

²1215.7*²*² 2

0

=== ππ

MNmMEd 115.01

300.049.0234.11*039.0 =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+=


469


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






470







Result name Result description Value Error Az Az -5.58 cm² -0.0000 %


471

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.


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.


■ 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


472

Units

Metric System

Geometry


■ Height: h = 0.45 m, ■ Width: b = 0.60 m, ■ Length: L = 4.50 m, ■ Concrete cover: c = 5cm along the long section edge and 3cm along the short section edge

Boundary conditions



Loading



NEd =1.35*0.45+1.5*0.50=1.3575MN

MED=1.35*0.10+0.30*0.06=0.225MNm

■ me 166.03575.1225.0

0 ==




mll 950.4*2*20 ===



28.6945.0

9*32*32 0 ===alλ


473



( )Ed

EQPef M

Mt ., 0∞= ϕϕ


Where:




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


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


474

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


( ) 18.1524.0*26.2*, 0 ==∞=Ed

EQPef M

Mtϕϕ


18.218.111 =+=+ efϕ



nCBA ***20

lim =λ


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


5.45.2.3 The eccentricity calculation and the corrected loads on ULS:

Initial eccentricity:

me 166.03575.1225.0

0 ==


cmlei 25.24009

4000 ===


475

First order eccentricity- stresses correction:

MNNEd 3575.1=

cmmeee i 85.181885.001 ==+=

MNmMEd 256.0=





MNmhdNMM Gua 494.0245.040.0*3575.1256.0)

2(*0 =⎟

⎠⎞

⎜⎝⎛ −+=−+=


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



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 ===



24 24.178.434

3575.110*46.32' cmfNAAyd

−=−=−= −


476


2min, 3.6*002.0²12.3

78.4343575.1*10.0*10,0 cmAcm

fNA c

yd

Eds =≥===







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


477

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


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.


478

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:


479

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.




480


(reference value: 16cm2 )




Result name Result description Reference value Az Reinforcement area [cm2] 16 cm2



Result name Result description Value Error Az Az -15.995 cm² 0.0000 %


481

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.


482

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.



■ 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


483

Units

Metric System

Geometry



Boundary conditions


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


■ Load combinations:

kNNED 2280800*5.1800*35.1 =+=


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>


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


484

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





(reference value: 14.24cm2=4*3.56cm2)


Result name Result description Reference value Ay Reinforcement area [cm2] 3.57cm2



Result name


Ay Az -3.56562 cm² 0.0001 %


485


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:


486


■ The shear stresses results ■ The cross-sectional area of the shear reinforcement, Asw

Units

Metric System

Geometry


■ 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



■ 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==


487

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

+

+=


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


54.0250251*6.0

2501*6,01 =⎥⎦

⎤⎢⎣⎡ −=⎥⎦

⎤⎢⎣⎡ −= ckfv


488

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


■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. Advance Design gives the following results for Atz (cm2/ml)


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


Result name


Fz Fz -263.062 kN -0.0002 % Atz Atz 8.68636 cm² 0.0000 %


489

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


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.



■ 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


490

■ The column is considered isolated and braced

Units

Metric System

Geometry


■ Height: h = 0.40 m, ■ Width: b = 0.60 m, ■ Length: L = 4.00 m, ■ Concrete cover: c = 5 cm along the long section edge and 3cm along the short section edge

Boundary conditions



Loading



NEd =1.35*0.30+1.5*0.50=1.155MN

NQP=1.35*0.30+0.30*0.50=0.450MN

■


491




mll 84*2*20 ===



28.6940.0

8*32*32 0 ===alλ



( )Ed

EQPef M

Mt ., 0∞= ϕϕ


Where:




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


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


492

( ) 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


( ) 90.039.0*30.2*, 0 ==∞=Ed

EQPef M

Mtϕϕ




nCBA ***20

lim =λ


Where:

241.020*60.0*40.0

155.1*

===cdc

Ed

fANn

( ) 85.090.0*2.01

1*2,01

1=

+=

+=

ef

Aϕ


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



493

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.


mlei 02.04008

4000 ===

First order eccentricity- stresses correction:

⎩⎨⎧

=⎪⎩

⎪⎨⎧

=⎪⎩

⎪⎨⎧

== mmmmmmmm

mmhmm

e 203.13

20max

3040020

max30

20max0


Solicitations corrected to take into account when calculating flexural combined, are:

NEd= 1.155MN

MEd= 1.155*0.02=0.0231MNm





MNmhdNMM Gua 196.0240.035.0*155.10231.0)

2(*0 =⎟

⎠⎞

⎜⎝⎛ −+=−+=


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



494


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 ===



24 69.1278.434

155.110*87.13' cmfNAAyd

−=−=−= −


2min, 80.4*002.0²66.2

78.434155.1*10.0*10,0 cmAcm

fNA c

yd

Eds =≥===







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


495

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.


496



Theoretical reinforcement area (cm2) and minimum reinforcement area (cm2)

(reference value: 2.4cm2 and 4.8cm2)





Amin Minimum reinforcement area 4.80 cm2



497


Result name


Az Az -2.4 cm² 0.0000 % Amin Amin 4.8 cm² 0.0000 %


498

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




■ Concrete cover 5cm ■ Concrete C25/30 ■ Steel reinforcement S500B ■ Relative humidity RH=50% ■ Buckling length L0=6.50m


499

Units

Metric System

Geometry



Boundary conditions


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


■ Load combinations:

kNNED 1080800*35.1 ==


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>


12006.7560 ≤=< λ

33.006.75

3232 3.13.1

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

λα


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


500




(reference value: 23.43cm2=4*5.86cm2)


Result name Result description Reference value Ay Reinforcement area [cm2] 5.85cm2



Result name


Ay Ay -5.85106 cm² 0.0001 %

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Ad validation guide 2013 en parte1