VERiFiCATiON MANUALSOFiSTiK 2014 VERiFiCATiON MANUAL VERiFiCATiON MANUAL, Version 2014.4 Software Version: SOFiSTiK 2014 Copyright c 2013 by SOFiSTiK AG, Oberschleissheim, Germany. SOFiSTiK AG HQ Oberschleissheim Office Nuremberg Bruckmannring 38 Burgschmietstr. 40 85764 Oberschleissheim 90419 Nuremberg Germany Germany T +49 (0)89 315878-0 T +49 (0)911 39901-0 F +49 (0)89 315878-23 F +49(0)911 397904 info@sofistik.de www.sofistik.de This manual is protected by copyright laws. No part of it may be translated, copied or reproduced, in any form or by any means, without written permission from SOFiSTiK AG. SOFiSTiK reserves the right to modify or to release new editions of this manual. The manual and the program have been thoroughly checked for errors. However, SOFiSTiK does not claim that either one is completely error free. Errors and omissions are corrected as soon as they are detected. The user of the program is solely responsible for the applications. We strongly encourage the user to test the correctness of all calculations at least by random sampling. Front Cover Project: Yas Hotel, Abu Dhabi | Client: ALDAR Properties PJSC, Abu Dhabi | Structural Design and Engineering Gridshell: schlaich bergermann und partner | Architect: Asymptote Architecture | Photo: Bj ¨ orn Moermann Contents Contents Introduction 3 1 About this Manual 3 1.1 Layout and Organization of a Benchmark . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Finding the Benchmark of interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2 Index by Categories 7 2.1 Mechanical Benchmarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Design Code Benchmarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 I SOFiSTiK Software Quality Assurance (SQA) 11 3 SOFiSTiK SQA Policy 13 3.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.1.1 About SOFiSTiK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.1.2 Innovation and Reliability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.2 Organisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.2.1 Software Release Schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.2.2 SQA Modules - Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2.3 Responsibilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2.4 Software Release Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.3 Instruments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.3.1 CRM System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.3.2 Tracking System (internal) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.3.3 Continuous Integration – Continuous Testing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.4 Additional Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3.4.1 Training. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3.4.2 Academia Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.5 Disclaimer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 II Benchmark Examples 19 4 BE1: Joint Deflection of Plane Truss 21 4.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 4.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 5 BE2: Kinematic Coupling Conditions 23 5.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 SOFiSTiK 2014 | VERiFiCATiON MANUAL i Contents 5.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 5.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6 BE3: Beam Stresses and Deflections 29 6.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 6.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 6.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 6.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 6.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 7 BE4: Tie Rod with Lateral Loading 33 7.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 7.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 7.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 7.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 8 BE5: Bending of a T-beam 37 8.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 8.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 8.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 8.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 8.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 9 BE6: Warping Torsion Bar 39 9.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 9.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 9.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 9.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 9.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 10 BE7: Large Deflection of Cantilever Beams I 43 10.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 10.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 10.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 10.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 10.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 11 BE8: Large Deflection of Cantilever Beams II 47 11.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 11.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 11.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 11.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 11.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 12 BE9: Verification of Beam and Section Types I 51 12.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 12.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 12.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 12.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 12.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 13 BE10: Verification of Beam and Section Types II 59 13.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 13.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 ii VERiFiCATiON MANUAL | SOFiSTiK 2014 Contents 13.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 13.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 13.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 14 BE11: Plastification of a Rectangular Beam 63 14.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 14.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 14.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 14.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 14.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 15 BE12: Cantilever in Torsion 67 15.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 15.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 15.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 15.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 15.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 16 BE13: Buckling of a Bar with Hinged Ends I 71 16.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 16.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 16.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 16.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 16.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 17 BE14: Buckling of a Bar with Hinged Ends II 75 17.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 17.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 17.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 17.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 17.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 18 BE15: Flexural and Torsional Buckling 77 18.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 18.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 18.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 18.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 18.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 19 BE16: Torsion due to Biaxial Bending 81 19.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 19.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 19.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 19.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 19.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 20 BE17: Lateral Torsional Buckling 83 20.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 20.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 20.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 20.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 20.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 21 BE18: Three-storey Column under Large Compressive Force and Torsional Mo- ment 85 SOFiSTiK 2014 | VERiFiCATiON MANUAL iii Contents 21.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 21.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 21.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 21.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 21.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 22 BE19: Two-span Beam with Warping Torsion and Compressive Force 89 22.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 22.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 22.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 22.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 22.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 23 BE20: Passive Earth Pressure I 93 23.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 23.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 23.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 23.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 23.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 24 BE21: Passive Earth Pressure II 97 24.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 24.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 24.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 24.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 24.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 25 BE22: Tunneling - Ground Reaction Line 103 25.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 25.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 25.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 25.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 25.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 26 BE23: Undamped Free Vibration of a SDOF System 107 26.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 26.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 26.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 26.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 26.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 27 BE24: Free Vibration of a Under-critically Damped SDOF System 111 27.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 27.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 27.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 27.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 27.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 28 BE25: Eigenvalue Analysis of a Beam Under Various End Constraints 115 28.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 28.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 28.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 28.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 28.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 iv VERiFiCATiON MANUAL | SOFiSTiK 2014 Contents 29 BE26: Response of a SDOF System to Harmonic Excitation 117 29.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 29.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 29.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 29.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 29.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 30 BE27: Response of a SDOF System to Impulsive Loading 121 30.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 30.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 30.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 30.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 30.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 31 BE28: Cylindrical Hole in an Infinite Elastic Medium 125 31.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 31.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 31.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 31.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 31.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 32 BE29: Cylindrical Hole in an Infinite Mohr-Coulomb Medium 129 32.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 32.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 32.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 32.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 32.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 33 BE30: Strip Loading on an Elastic Semi-Infinite Mass 133 33.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 33.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 33.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 33.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 33.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 34 BE31: Snap-Through Behaviour of a Truss 137 34.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 34.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 34.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 34.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 34.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 35 BE32: Thermal Extension of Structural Steel in case of Fire 141 35.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 35.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 35.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 35.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 35.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 36 BE33: Work Laws in case of Fire for Concrete and Structural Steel 145 36.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 36.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 36.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 36.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 36.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 SOFiSTiK 2014 | VERiFiCATiON MANUAL v Contents 37 BE34: Ultimate Bearing Capacity of Concrete and Steel under Fire 151 37.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 37.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 37.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 37.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 37.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 38 BE35: Calculation of Restraining Forces in Steel Members in case of Fire 155 38.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 38.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 38.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 38.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 38.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 39 BE36: Pushover Analysis: Performance Point Calculation by ATC-40 Procedure 159 39.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 39.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 39.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 39.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 39.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 40 BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory 163 40.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 40.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 40.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 40.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 40.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 41 BE38: Calculation of Slope Stability by Phi-C Reduction 167 41.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 41.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 41.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 41.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 41.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 42 BE39: Natural Frequencies of a Rectangular Plate 171 42.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 42.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 42.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 42.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 42.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 43 BE40: Portal Frame 175 43.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 43.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 43.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 43.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 43.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 44 BE41: Linear Pinched Cylinder 179 44.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 44.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 44.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 44.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 vi VERiFiCATiON MANUAL | SOFiSTiK 2014 Contents 44.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 45 BE42: Thick Circular Plate 183 45.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 45.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 45.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 45.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 45.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 46 BE43: Panel with Circular Hole 187 46.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 46.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 46.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 46.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 46.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 47 BE44: Undrained Elastic Soil Layer Subjected to Strip Loading 191 47.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 47.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 47.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 47.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 47.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 48 BE45: One-Dimensional Soil Consolidation 195 48.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 48.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 48.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 48.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 48.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 49 BE46: Material Nonlinear Analysis of Reinforced Concrete Beam 199 49.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 49.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 49.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 49.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 49.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 50 BE47: Pushover Analysis: SAC LA9 Building 203 50.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 50.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 50.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 50.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 50.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 51 BE48: Triaxial Consolidated Undrained (CU) Test 209 51.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 51.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 51.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 51.3.1 Hostun-RF Sand, σ c = 200 kN/ m 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 51.3.2 Hostun-RF Sand, σ c = 300 kN/ m 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 51.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 51.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 52 BE49: Triaxial Drained Test 217 52.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 SOFiSTiK 2014 | VERiFiCATiON MANUAL vii Contents 52.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 52.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 52.3.1 Hostun-RF Sand, σ c = 100 kN/ m 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 52.3.2 Hostun-RF Sand, σ c = 300 kN/ m 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 52.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 52.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 53 BE50: A Circular Cavity Embedded in a Full-Plane Under Impulse Pressure 223 53.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 53.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 53.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 53.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 53.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 54 BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure 227 54.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 54.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 54.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 54.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 54.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 55 BE52: Verification of Wave Kinematics 233 55.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 55.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 55.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 55.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 55.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 56 BE53: Verification of Wave Loading 237 56.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 56.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 56.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 56.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 56.5 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 III Design Code Benchmark Examples 241 57 DCE-EN1: Design of Slab for Bending 243 57.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 57.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 57.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 57.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 57.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 57.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 58 DCE-EN2: Design of a Rectangular CS for Bending 247 58.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 58.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 58.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 58.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 58.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 58.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 59 DCE-EN3: Design of a T-section for Bending 253 viii VERiFiCATiON MANUAL | SOFiSTiK 2014 Contents 59.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 59.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 59.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 59.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 59.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 59.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 60 DCE-EN4: Design of a Rectangular CS for Bending and Axial Force 259 60.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 60.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 60.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 60.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 60.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 60.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 61 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force 263 61.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 61.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 61.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 61.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 61.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 61.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 62 DCE-EN6: Design of a Rectangular CS for Shear Force 269 62.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 62.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 62.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 62.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 62.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 62.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 63 DCE-EN7: Design of a T-section for Shear 273 63.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 63.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 63.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 63.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 63.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 63.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 64 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force 279 64.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 64.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 64.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280 64.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 64.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 64.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 65 DCE-EN9: Design of a Rectangular CS for Shear and Torsion 285 65.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 65.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 65.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 65.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 65.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 65.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 SOFiSTiK 2014 | VERiFiCATiON MANUAL ix Contents 66 DCE-EN10: Shear between web and flanges of T-sections 291 66.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 66.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 66.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 66.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 66.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 66.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 67 DCE-EN11: Shear at the interface between concrete cast 299 67.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 67.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 67.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 67.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 67.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 67.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 68 DCE-EN12: Calculation of crack widths 307 68.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 68.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 68.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 68.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 68.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 68.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 69 DCE-EN13: Design of a Steel I-section for Bending and Shear 315 69.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 69.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 69.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 69.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 69.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 69.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 70 DCE-EN14: Classification of Steel Cross-sections 321 70.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 70.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 70.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 70.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 70.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 70.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 71 DCE-EN15: Buckling Resistance of Steel Members 327 71.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 71.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 71.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328 71.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 71.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 71.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 72 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force 331 72.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 72.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 72.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 72.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 72.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 x VERiFiCATiON MANUAL | SOFiSTiK 2014 Contents 72.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 73 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS 337 73.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 73.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 73.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 73.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 73.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 73.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 74 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Con- crete CS 345 74.1 Problem Description. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 74.2 Reference Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 74.3 Model and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 74.4 Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 74.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 74.6 Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 SOFiSTiK 2014 | VERiFiCATiON MANUAL xi Contents xii VERiFiCATiON MANUAL | SOFiSTiK 2014 Contents Introduction SOFiSTiK 2014 | VERiFiCATiON MANUAL 1 About this Manual 1 About this Manual The primary objective of this manual is to verify the capabilities of SOFiSTiK by means of nontrivial problems which are bound to reference solutions. To this end, this manual contains a compilation of a number of selected computational benchmarks, each benchmark focusing on a specific (mechanical) topic. The obtained results from the SOFiSTiK analysis are contrasted with corresponding reference solutions 1 . The tasks covered by SOFiSTiK, address a broad scope of engineering applications and it is therefore not possible to validate all specific features with known reference solutions in terms of this Verification Manual. An attempt has been made though, to include most significant features of the software with re- spect to common problems of general static and dynamic analysis of structures (cf. Part II, Benchmarks Examples). Design examples, treating Design Code related tasks, are provided in Part III of this Verification Manual. 1.1 Layout and Organization of a Benchmark For the description of each Benchmark, a standard format is employed, where the following topics are always treated: • Problem Description • Reference Solution • Model and Results • Conclusion • Literature First, the problem description is given, where the target of the benchmark is stated, followed by the reference solution, where usually a closed-form analytical solution is presented, when available. The next section is the description of the model, where its properties, the loading configuration, the analysis method and assumptions, further information on the finite element model, are presented in detail. Finally, the results are discussed and evaluated with respect to the reference solution and a final conclusion for the response of the software to the specific problem is drawn. Last but not least, the textbooks and references used for the verification examples are listed, which are usually well known and come from widely acclaimed engineering literature sources. 1.2 Finding the Benchmark of interest There are several ways of locating a Benchmark that is of interest for the user. For each example a description table is provided in the beginning of the document, where all corresponding categories, that are treated by the specific benchmark, are tabulated, as well as the name of the corresponding input file. Such a description table with some example entries, follows next. 1 Where available, analytical solutions serve as reference. Where this is not feasible, numerical or empirical solutions are referred to. In any case, the origin of the reference solution is explicitly stated. SOFiSTiK 2014 | VERiFiCATiON MANUAL 3 About this Manual Overview Element Type(s): C2D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): SOIL Module(s): TALPA Input file(s): passive earth pressure.dat As it can be seen, the available categories are the element type, the analysis type, the procedure, the topics and the modules. For each category that is provided in the description table, a hyperlink is created, linking each example to the global categories tables. In this manner, the user has a direct overview of the attributes involved in each problem, and at the same time is able to browse by category through the Verification Manual focusing only on the one of his interest. Table 1.1 provides an overview of all the categories options that are available. Table 1.1: Categories Overview Categories Options Element Type Continuum 3D Continuum 2D (plane strain) Continuum axisymmetric Shell FE beam 3D Nonlinear FE beam 3D (AQB) Fiber beam 2D Fiber beam 3D Truss element Cable element Spring element Damping element Couplings Analysis Type Geometrically nonlinear Physically nonlinear Dynamic Static Potential problem Procedure Buckling analysis Eigenvalue/ Modal analysis Time stepping 4 VERiFiCATiON MANUAL | SOFiSTiK 2014 About this Manual Table 1.1: (continued) Categories Options Load stepping Phi-C reduction Topic Soil related Seismic Fire design Module AQB AQUA ASE BDK BEMESS DYNA SOFiLOAD SOFiMSHC STAR2 TALPA TENDON 1.3 Symbols For the purpose of this manual the following symbols and abbreviations apply. SOF. SOFiSTiK Ref. reference Tol. tolerance cs cross-section sect. section temp. temperature homog. homogeneous Be. benchmark con. construction SDOF single degree of freedom e r relative error of the approximate number |e r | absolute relative error of the approximate number e error of the approximate number SOFiSTiK 2014 | VERiFiCATiON MANUAL 5 About this Manual |e| absolute error of the approximate number ep() same as e () 6 VERiFiCATiON MANUAL | SOFiSTiK 2014 Index by Categories 2 Index by Categories Subsequent tables show all Benchmarks included in this Verification Manual, indexed by category. 2.1 Mechanical Benchmarks ELEMENT TYPE Keyword Benchmark Examples Continuum 3D C3D BE41, BE42, BE43 Continuum 2D C2D BE20, BE21, BE22, BE28, BE29, BE30, BE38, BE44, BE45, BE50 Continuum axisymmetric CAXI BE48, BE49 Shell SH3D BE7, BE8, BE11, BE14, BE32, BE33, BE34, BE35, BE39, BE46 FE beam 3D B3D BE3, BE4, BE5, BE6, BE7, BE8, BE9, BE10, BE11, BE12, BE13, BE15, BE16, BE17, BE18, BE19, BE25, BE37, BE40, BE46, BE47 Fiber beam 2D BF2D BE11, BE32, BE33, BE34, BE35 Truss element TRUS BE1, BE31 Spring element SPRI BE23, BE24, BE26, BE27 Damper element DAMP BE24, BE26 ANALYSIS TYPE Keyword Benchmark Examples Geometrically nonlinear GNL BE4, BE7, BE8, BE12, BE13, BE14, BE15, BE16, BE17, BE18, BE19, BE31, BE37, BE40 Physically nonlinear MNL BE11, BE20, BE21, BE22, BE29, BE32, BE33, BE34, BE35, BE38, BE46, BE47, BE48, BE49 Dynamic DYN BE23, BE24, BE25, BE26, BE27, BE39, BE50 Static STAT BE1, BE2, BE3, BE4, BE5, BE6, BE7, BE8, BE9, BE10, BE11, BE12, BE13, BE14, BE15, BE16, BE17, BE18, BE19, BE20, BE21, BE22, BE28, BE29, BE30, BE31, BE32, BE33, BE34, BE35, BE37, BE38, BE40, BE41, BE42, BE43, BE44, BE45, BE46 PROCEDURE Keyword Benchmark Examples Buckling analysis STAB BE13, BE14, BE15, BE37 Table continued on next page. SOFiSTiK 2014 | VERiFiCATiON MANUAL 7 Index by Categories PROCEDURE Keyword Benchmark Examples Eigenvalue / Modal analysis EIGE BE25, BE39, BE47 Time stepping TSTP BE23, BE24, BE26, BE27 Load stepping LSTP BE7, BE8, BE11, BE20, BE21, BE22, BE29, BE31, BE32, BE33, BE34, BE35, BE38, BE48, BE49 Phi-C reduction PHIC BE38 TOPIC Keyword Benchmark Examples Soil related SOIL BE20, BE21, BE22, BE28, BE29, BE30, BE38, BE44, BE45, BE48, BE49, BE50 Seismic EQKE BE36, BE47, BE51 Fire design FIRE BE32, BE33, BE34, BE35 Wave WAVE BE52, BE53 MODULE Keyword Benchmark Examples Design of Cross Sections and of Prestressed Concrete and Composite Cross Sections AQB BE5 Materials and Cross Sections AQUA BE9 General Static Analysis of Finite Element Structures ASE BE1, BE2, BE3, BE4, BE5, BE6, BE7, BE8, BE10, BE11, BE12, BE13, BE14, BE15, BE16, BE17, BE18, BE19, BE31, BE32, BE33, BE34, BE35, BE37, BE40, BE41, BE42, BE43, BE46, BE47 Dynamic Analysis DYNA BE18, BE19, BE23, BE24, BE25, BE26, BE27, BE37, BE39, BE50 Loadgenerator for Finite Elements and Frameworks SOFiLOAD BE36, BE47, BE51, BE52, BE53 Geometric Modelling SOFiMSHC BE2 Statics of Beam Structures 2nd Order Theory STAR2 BE11, BE37, BE46 2D Finite Elements in Geotechnical Engineering TALPA BE11, BE20, BE21, BE22, BE28, BE29, BE30, BE32, BE33, BE34, BE35, BE38, BE44, BE45, BE48, BE49 8 VERiFiCATiON MANUAL | SOFiSTiK 2014 Index by Categories 2.2 Design Code Benchmarks DESIGN CODE FAMILY Keyword Benchmark Examples Eurocodes EN DCE-EN6, DCE-EN7, DCE-EN13, DCE-EN14, DCE-EN15, DCE-EN16 German Standards DIN DCE-EN1, DCE-EN2, DCE-EN3, DCE-EN4, DCE- EN5, DCE-EN6, DCE-EN8, DCE-EN9, DCE-EN10, DCE-EN11, DCE-EN12, DCE-EN17, DCE-EN18 DESIGN CODE Keyword Benchmark Examples Design of concrete structures EN1992 DCE-EN1, DCE-EN2, DCE-EN3, DCE-EN4, DCE- EN5, DCE-EN6, DCE-EN7, DCE-EN8, DCE-EN9, DCE-EN10, DCE-EN11, DCE-EN12, DCE-EN17, DCE-EN18 Design of steel structures EN1993 DCE-EN13, DCE-EN14, DCE-EN15, DCE-EN16 MODULE Keyword Benchmark Examples Design of Cross Sections and of Prestressed Concrete and Composite Cross Sections AQB DCE-EN1, DCE-EN2, DCE-EN3, DCE-EN4, DCE- EN5, DCE-EN6, DCE-EN7, DCE-EN8, DCE-EN9, DCE-EN10, DCE-EN11, DCE-EN12, DCE-EN13, DCE-EN14, DCE-EN16, DCE-EN17, DCE-EN18 Lateral Torsional Buckling Check for Steel Cross Sections BDK DCE-EN15 Geometry of Prestressing Tendons TENDON DCE-EN17 SOFiSTiK 2014 | VERiFiCATiON MANUAL 9 Index by Categories 10 VERiFiCATiON MANUAL | SOFiSTiK 2014 Part I SOFiSTiK Software Quality Assurance (SQA) SOFiSTiK 2014 | VERiFiCATiON MANUAL 11 SOFiSTiK SQA Policy 3 SOFiSTiK SQA Policy 3.1 Objectives 3.1.1 About SOFiSTiK SOFiSTiK finite element software has been continuously developed since 1981. It is currently used by more than 10000 customers worldwide. SOFiSTiK is a multipurpose tool with extensive capabilities which fall into a wide spectrum of engineering analyses such as static and dynamic structural analysis, modal and buckling eigenvalue problems, nonlinearities and higher order effects, geotechnics and tunnel analysis, heat transfer and fire analysis, as well as numerous types of other applications. 3.1.2 Innovation and Reliability As a provider of cutting-edge engineering software, confidence in robustness and reliability of the prod- uct is an issue of outstanding relevance for SOFiSTiK. To some degree, however, innovation and reliabil- ity are conflicting targets, since every change introduces new possible sources of uncertainty and error. To meet both demands on a sustainable basis, SOFiSTiK has installed a comprehensive quality assur- ance system. The involved organizational procedures and instruments are documented in the following Sections. 3.2 Organisation 3.2.1 Software Release Schedule The SOFiSTiK software release schedule is characterized by a two-year major release cycle. The first customer shipment (FCS) of a SOFiSTiK major release is preceded by an extensive BETA testing period. In this phase - after having passed all internal test procedures (Section 3.2.4: Software Release Procedure) - the new product is adopted for authentic engineering projects both by SOFiSTiK and by selected customers. For a two-year transition period, subsequent major releases are fully supported in parallel, as shown in Fig. 3.1. 2009 2010 2011 2012 2013 2014 2015 2016 2017 BETA FCS SOFiSTiK 2010 Start of Transition Phase Discontinuation of Maintenance BETA SOFiSTiK 2012 BETA SOFiSTiK 2014 Figure 3.1: SOFiSTiK Release Schedule The major release cycle is supplemented by a two-month service pack cycle. Service packs are quality assured, which means they have passed both the continuous testing procedures and the functional tests (Section 3.2.2: SQA Modules - Classification). They are available for download via the SOFiSTiK update tool SONAR. Software updates for the current version (service packs) include bug-fixes and minor new features only; major new developments with increased potential regarding side- effects are reserved for major releases SOFiSTiK 2014 | VERiFiCATiON MANUAL 13 SOFiSTiK SQA Policy with an obligatory extensive testing period. 3.2.2 SQA Modules - Classification Figure 3.2 depicts the ”three pillars” of the SOFiSTiK SQA procedure. Preventive and analytic provisions can be differentiated. Preventive provisions essentially concern the organization of the development process. They aim at minimizing human errors by a high degree of automatism and by avoiding error-prone stress situations. These provisions comprise: • A thoroughly planned feature map and release schedule. • Strict phase differentiation: Prior to any software release (also for service packs), the development phase is followed up by a consolidation phase . This phase is characterized by extensive functional testing. No new features are implemented, only test feedback is incorporated. For major releases, an additional BETA test phase is scheduled. • Fully automated build and publishing mechanisms. Analytic provisions provide for the actual testing of the software products. Continuous Testing directly accompanies the development process: Automated and modular regression tests assure feedback at a very early stage of the development (Section 3.3.3: Continuous Testing). Functional Testing is carried out in particular during the consolidation phases. These tests essentially involve manual testing; they focus on comprehensive workflow tests and product oriented semantic tests. SQA (Software Quality Assurance) Development Process Phase differentiation, build- and publishing mechanism Organizational Provision (preventive) Continuous Testing Automated, modular and continuous regression testing Instruments (analytic) Functional Testing Focus: workflow tests, product oriented semantic tests Figure 3.2: SQA Modules 3.2.3 Responsibilities The consistent implementation of quality assurance procedures is responsibly coordinated by the man- aging board executive for products. The development divisions are in authority for: • The establishment, maintenance and checking of continuous testing procedures. • The implementation of corresponding feedback. 14 VERiFiCATiON MANUAL | SOFiSTiK 2014 SOFiSTiK SQA Policy The product management is responsible for: • The coordination and execution of functional testing. • The integration of customer feedback into the QA process. As a corporate activity is carried out: • Continuous review of processes. • The identification of supplemental objectives. • Identification and implementation of possible optimizations. Product Management Functional testing Integrating customer feedback Development Continuous Integration Continuous Testing Implementing feedback Corporate Activity Adaption of processes Definition of objectives Coordinated by managing board Figure 3.3: SQA Responsibilities 3.2.4 Software Release Procedure The defined minimumrequirements for software releases of type Hotfix, Service Pack and Major Release are illustrated by Figure 3.4. Approval of individual products is accomplished by the respective person in charge; the overall approval is in authority of the managing board executive for products. SOFiSTiK 2014 | VERiFiCATiON MANUAL 15 SOFiSTiK SQA Policy Release Requirements Hotfix Continuous Testing Passed Service Pack Continuous Testing Passed Functional Testing Passed Major Release Continuous Testing Passed Functional Testing Passed BETA Test Phase Passed Figure 3.4: Software Release Requirements 3.3 Instruments 3.3.1 CRM System Each request from our customers is traced by means of a Customer Relation Management (CRM) System assuring that no case will be lost. Detailed feedback to the customer is provided via this system. Possible bug fixes or enhancements of the software are documented with version number and date in corresponding log files. These log files are published via RSS-feed to our customers. In this way, announcement of available software updates (service-pack or hotfix) is featured proactively. Moreover, information is provided independent of and prior to the actual software update procedure. Further sources of information are the electronic newsletter/ newsfeeds and the internet forum (www.sofistik.de / www.sofistik.com). 3.3.2 Tracking System (internal) For SOFiSTiK-internal management and coordination of the software development process - both re- garding implementation of features and the fixing of detected bugs - a web-based tracking system is adopted. 3.3.3 Continuous Integration – Continuous Testing As mentioned above, the production chain is characterized by a high degree of automation. An important concern is the realization of prompt feedback cycles featuring an immediate response regarding quality of the current development state. 16 VERiFiCATiON MANUAL | SOFiSTiK 2014 SOFiSTiK SQA Policy Automated Continuous Integration procedure Automated Continuous Testing procedure Development/ PM Assessing feedback Committing modifications Figure 3.5: Feedback cycle: Continuous Integration – Continuous Testing Continuous integration denotes the automated process, assuring that all executed and committed mod- ifications of the program’s code basis are directly integrated via rebuild into the internal testing environ- ment. Upon completion of the integration, the continuous testing procedure is triggered automatically. This procedure executes a standardized testing scenario using the newly updated software. Test results are prepared in form of compact test protocols allowing for quick assessment. The executed tests are so-called regression tests. Regression tests examine by means of associated reference solutions wether the conducted modifications of the code basis cause undesired performance in other already tested parts of the program. Together, continuous integration and continuous testing form the basis for a quality control that directly accompanies the development process. This way, possibly required corrections can be initiated prompt- ly. SOFiSTiK has successfully implemented this procedure. Currently, the continuous test database comprises more than 3000 tests. 3.4 Additional Provisions 3.4.1 Training As a special service to our customers, SOFiSTiK provides for comprehensive and individually tailored training to support a qualified and responsible use of the software. This is complemented by offering a variety of thematic workshops which are dedicated to specific engineering topics. It is the credo of SOFiSTiK that a high-quality product can only be created and maintained by highly qual- ified personnel. Continuing education of the staff members is required by SOFiSTiK and it is supported by an education program which involves both in- house trainings and provisions of external trainings on a regular basis. SOFiSTiK 2014 | VERiFiCATiON MANUAL 17 SOFiSTiK SQA Policy 3.4.2 Academia Network Arising questions are treated by an intense discussion with customers, authorities and scientists to find the best interpretation. 3.5 Disclaimer Despite all efforts to achieve the highest possible degree of reliability, SOFiSTiK cannot assure that the provided software is bug-free or that it will solve a particular problem in a way which is implied with the opinion of the user in all details. Engineering skill is required when assessing the software results. 18 VERiFiCATiON MANUAL | SOFiSTiK 2014 Part II Benchmark Examples SOFiSTiK 2014 | VERiFiCATiON MANUAL 19 BE1: Joint Deflection of Plane Truss 4 BE1: Joint Deflection of Plane Truss Overview Element Type(s): TRUSS Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): ASE Input file(s): truss.dat 4.1 Problem Description The problem consists of a plane truss structure, as shown in Fig. 4.1. Determine the vertical deflection at the free node 8. P P P Figure 4.1: Problem Description 4.2 Reference Solution The problem of determining the displacements of trusses can be treated in various ways. Popular among engineers, is to apply energy methods, e.g. the method of virtual work or Castigliano’s theorem, to solve problems involving slope and deflection, that are based on the conservation of energy principle, and are more suitable for structures with complicated geometry such as trusses. Further information on this topic can be found in numerous engineering books, dealing with structural analysis [1]. 4.3 Model and Results The general properties of the model [2] are defined in Table 4.1. The total width of the truss is 60 ƒ t, consisting of four spaces of 15 ƒ t each, and the total height is 15 ƒ t. The load is applied equally at the three free nodes at the bottom of the truss. The results are presented in Table 4.2 and compared to the reference example [2]. Fig. 4.2 shows the deflections and the deformed shape of the structure. SOFiSTiK 2014 | VERiFiCATiON MANUAL 21 BE1: Joint Deflection of Plane Truss Table 4.1: Model Properties Material Properties Geometric Properties Loading E = 30 10 3 ks tot = 60 ƒ t = 18.288 m P = 20kp = 89.0kN = 206842.773 MP h tot = 15 ƒ t = 4.572 m ν = 0.3 2 = 11 = 7.5 ƒ t = 2.286 m 1 = 4 = 6 = 10 = 12 = 15 ƒ t = 4.572 m A 1 = A 4 = 2 n 2 = 12.90 cm 2 A 2 = A 11 = A 10 = A 12 = 1 n 2 = 6.45 cm 2 A 5 = A 9 = 1.5 n 2 = 9.68 cm 2 A 3 = A 6 = 3 n 2 = 19.35 cm 2 A 7 = A 8 = 4 n 2 = 25.81 cm 2 Table 4.2: Results SOF. Ref. [2] |e r | [%] δ 8 [mm] 66.809 66.802 0.011 Figure 4.2: Problem Description 4.4 Conclusion This example verifies the deflection of trusses. It has been shown that the behaviour of the truss is accurately captured. 4.5 Literature [1] R. C. Hibbeler. Structural Analysis. 8th. Prentice Hall, 2012. [2] STAAD Verification Problems. Bentley Systems. 2010. 22 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE2: Kinematic Coupling Conditions 5 BE2: Kinematic Coupling Conditions Overview Element Type(s): COUP Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): SOFiMSHC, ASE Input file(s): coupling.dat 5.1 Problem Description This problem verifies the kinematic coupling conditions for a structural point. Each coupling condition is tested on a pair of beams coupled with each other through structural points, as shown in Fig. 5.1. Four different cases are considered and the deflections of the beams are determined and compared to the analytical solution. A A Figure 5.1: Problem Description 5.2 Reference Solution In this example the problem of coupling structural points is treated. Through the definition of kinematic coupling conditions between structural points the constraint of one or multiple degrees of freedom is allowed. The displacement values of the given structural point A are defined according to the respective displacement values of the referenced (or master-) node A. Various cases are possible in SOFiSTiK for the coupling conditions. With the exception of the three conditions KPX, KPY and KPZ, which only couple the corresponding displacement. e.g. = o , all other coupling conditions satisfy the mechanical equilibrium conditions by taking the real distances between the two connected points into account, e.g. the conditions KPPX, KPPY, KPPZ correspond to the following expressions respectively [3] [4]: = o + ϕ yo (z − z o ) − ϕ zo (y − y o ) (5.1) SOFiSTiK 2014 | VERiFiCATiON MANUAL 23 BE2: Kinematic Coupling Conditions y = yo + ϕ zo ( − o ) − ϕ o (z − z o ) (5.2) z = zo + ϕ o (y − y o ) − ϕ yo ( − o ) (5.3) Mechanically they act like infinitely stiff structural members. A number of additional literals are provided in SOFiSTiK which allow to define a combination of coupling relations. For example, a rigid connection with hinged conditions at the reference node is described by KP = KPPX + KPPY + KPPZ (5.4) whereas KF = KP + KMX + KMY + KMZ = KPPX + KPPY + KPPZ + KMX + KMY + KMZ (5.5) describes mechanically a rigid connection with clamped support at the reference node. Further informa- tion on the topic are provided in SOFiSTiK manual of module SOFiMSHC [3]. 5.3 Model and Results The general properties of the model are defined in Table 5.1. All beams are of 4 m length and consist of a standard rectangular cross-section and a standard concrete material. The structural points A and A have a distance of 2 m in the axial direction. Four coupling conditions are considered : • KPPX, where only the displacement in the global x direction is connected • LPX, where only the displacement in the structural point’s local x direction is connected • KP, where the displacements in x, y and z direction are connected • KF, where the displacements and the rotations in x, y and z direction are connected All cases are tested for four loadcases, i.e. a horizontal P y , a longitudinal P , a vertical P z and a rotational M . Table 5.1: Model Properties Material Properties Geometric Properties Loading C 30/ 45 bem = 4 m P y = 50.0 kN h A = 0.4 m, b A = 0.2 m P = −50.0 kN h A = 0.3 m, b A = 0.15 m P z = 50.0 kN ( A − A ) = 2 m M = 10.0 kN (y A − y A ) = 0 m (z A − z A ) = 0 m 24 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE2: Kinematic Coupling Conditions In the cases, where only a displacement is transferred in the vertical z or horizontal direction y , a rotation results in the other direction. If for example, we consider a coupling of only the displacement in the y direction, then a rotation of ϕ z = 3 y / (2 bem ) will also result as the effect of a prescribed displacement of value y at the beam tip A . Table 5.2: Results for KPPX Coupling Condition Load [mm] Case SOF. Ref. P −0.107 −0.107 Table 5.3: Results for LPX Coupling Condition Load y [mm] ϕ z [mrd] Case SOF. Ref. SOF. Ref. P y 165.008 165.008 −61.878 −61.878 Table 5.4: Results for KP and KF Coupling Condition Coupling KP KF DOF / LC P y P P z P y P P z M [mm] SOF. 0.0 −0.107 165.01 0.0 −0.107 0.0 0.0 Ref. 0.0 −0.107 165.01 0.0 −0.107 0.0 0.0 y [mm] SOF. 165.008 0.0 0.0 49.059 0.0 0.0 0.0 Ref. 165.008 0.0 0.0 49.059 0.0 0.0 0.0 z [mm] SOF. 0.0 0.0 41.252 0.0 0.0 12.265 0.0 Ref. 0.0 0.0 41.252 0.0 0.0 12.265 0.0 ϕ [mrd] SOF. 0.0 0.0 0.0 0.0 0.0 0.0 6.661 Ref. 0.0 0.0 0.0 0.0 0.0 0.0 6.661 ϕ y [mrd] SOF. 0.0 0.0 15.470 0.0 0.0 0.175 0.0 Ref. 0.0 0.0 15.470 0.0 0.0 0.175 0.0 ϕ z [mrd] SOF. −61.878 0.0 0.0 −0.700 0.0 0.0 0.0 Ref. −61.878 0.0 0.0 −0.700 0.0 0.0 0.0 The results are presented in Tables 5.2 - 5.4, where they are compared to the reference results calcu- lated with the formulas provided in Section 5.2. Due to the extent of the results only non zero values will be presented in the result tables. Figures 5.2, 5.3 present the results for the KF coupling condition for the load cases 1 to 4 for both displacements and rotations, respectively. SOFiSTiK 2014 | VERiFiCATiON MANUAL 25 BE2: Kinematic Coupling Conditions LC 1 LC 2 LC 3 LC 4 Figure 5.2: Displacement Results for KF coupling for LC 1-4 LC 1 LC 2 LC 3 LC 4 Figure 5.3: Rotation Results for KF coupling for LC 1-4 26 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE2: Kinematic Coupling Conditions 5.4 Conclusion This example verifies the coupling of structural points. It has been shown that the behaviour is accurately captured. 5.5 Literature [3] SOFiMSHC Manual: Geometric Modelling. Version 12.01. SOFiSTiK AG. Oberschleißheim, Ger- many, 2012. [4] SOFiMSHA Manual: Import and Export of Finite Elements and Beam Structures. Version 16.01. SOFiSTiK AG. Oberschleißheim, Germany, 2012. SOFiSTiK 2014 | VERiFiCATiON MANUAL 27 BE2: Kinematic Coupling Conditions 28 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE3: Beam Stresses and Deflections 6 BE3: Beam Stresses and Deflections Overview Element Type(s): B3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): ASE Input file(s): rect beam.dat, I beam.dat 6.1 Problem Description A rectangular beam is supported as shown in Fig. 6.1 and loaded on the overhangs by a uniformly distributed load q. Determine the maximum bending stress σ in the middle portion of the beam and the deflection δ at the middle of the beam. q q | ←− −→| | ←− −→| ←− −→ Figure 6.1: Beam structure 6.2 Reference Solution The magnitude of the stresses at a cross-section is defined by the magnitude of the shearing force and bending moment at that cross-section. Under pure bending, the maximum tensile and compressive stresses occur in the outermost fibers. For any cross-section, which has its centroid at the middle of the depth h, and for a linear elastic material behaviour, the maximum stresses occur for z = ±h/ 2 [5]: σ m = Mh 2 and σ mn = − Mh 2 , (6.1) in which , is the moment of inertia of the cross-section with respect to the neutral axis and Mthe bending moment. For a beam overhanging equally at both supports with a uniformly distributed load applied at the overhangs (Fig. 6.1), assuming Bernoulli beam theory, the deflection at the middle of the beam is: δ = q 2 2 16E = M 2 8E , (6.2) where q is the value of the uniformly distributed load, the length of the overhangs, the length of the middle span and M the bending moment at the middle of the beam. SOFiSTiK 2014 | VERiFiCATiON MANUAL 29 BE3: Beam Stresses and Deflections 6.3 Model and Results The model is analysed for two different cross-sections, a rectangular and a general I-beamcross-section. The properties are defined in Table 6.1. The results are presented in Table 6.2. As to be expected, the analysis yields the same results for the maximum bending stress and deflection at the middle of the beam for the two models. Figure 6.2 shows the distribution of the stresses along the cross-sections for the two analysed examples. Figure 6.3 shows the deformed structure with the nodal displacements. Table 6.1: Model Properties Material Properties Geometric Properties Geometric Properties Loading Rectangular I-beam E = 30000 MP = 200 mm = 200 mm q = 10 kN/ m = 100 mm b = 16 mm h = 30 mm t eb = 2.174 mm b = 7 mm t ƒ nge = 2 mm y = 1.575 cm 4 y = 1.575 cm 4 Table 6.2: Results Rectangular I-beam Ref. σ m [MPa] 47.619 47.620 47.619 δ [mm] 0.529 0.529 0.529 1 7 2 . 1 4 1 7 2 . 1 4 1 7 2 . 1 4 1 7 2 . 1 4 0 2 6 . 7 4 0 2 6 . 7 4 0 2 6 . 7 4 0 2 6 . 7 4 1 7 2 . 1 4 1 7 2 . 1 4 1 7 2 . 1 4 1 7 2 . 1 4 0 2 6 . 7 4 0 2 6 . 7 4 0 2 6 . 7 4 - 0 2 6 . 7 4 - 0 2 6 . 7 4 - 0 2 6 . 7 4 - 1 7 2 . 1 4 - 1 7 2 . 1 4 - 1 7 2 . 1 4 - 1 7 2 . 1 4 - 1 7 2 . 1 4 - 1 7 2 . 1 4 - 1 7 2 . 1 4 - 1 7 2 . 1 4 - 0 2 6 . 7 4 - 0 2 6 . 7 4 - 0 2 6 . 7 4 - 9 1 6 . 7 4 9 1 6 . 7 4 9 1 6 . 7 4 9 1 6 . 7 4 9 1 6 . 7 4 9 1 6 . 7 4 - 9 1 6 . 7 4 - 9 1 6 . 7 4 - 9 1 6 . 7 4 - 9 1 6 . 7 4 - 9 1 6 . 7 4 - Figure 6.2: Distribution of stresses 1 . 3 2 0 . 5 2 9 1 . 3 2 0 . 9 7 0 0 . 9 7 0 0 . 6 2 3 0 . 6 2 3 0 . 4 9 6 0 . 4 9 6 0 . 3 9 7 0 . 3 9 70 . 2 9 2 0 . 2 9 2 0 . 2 3 1 0 . 2 3 1 Figure 6.3: Deformed Structure 30 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE3: Beam Stresses and Deflections 6.4 Conclusion This example adresses the computation of beam stresses and deflections. It has been shown that the behaviour of the beam is captured with an excellent accuracy. 6.5 Literature [5] S. Timoshenko. Strength of Materials, Part I, Elementary Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. SOFiSTiK 2014 | VERiFiCATiON MANUAL 31 BE3: Beam Stresses and Deflections 32 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE4: Tie Rod with Lateral Loading 7 BE4: Tie Rod with Lateral Loading Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): Topic(s): Module(s): ASE Input file(s): tie rod.dat 7.1 Problem Description A tie rod is subjected to the action of a tensile force N and a lateral load P applied at the middle as shown in Fig. (7.1). Determine the maximum deflection δ m , the slope θ at the left-hand end and the maximum bending moment M m . In addition, compare these three quantities for the case of the unstiffned tie rod (N = 0). | ←− −→| N N P Figure 7.1: Tie Rod 7.2 Reference Solution The combination of direct axial force and lateral load applied at a beam influences the reaction of the structure. Assuming that the lateral force acts in one of the principal planes of the beam and that the axial force is centrally applied by two equal and opposite forces, the expressions for the deflections can be derived from the differential equations of the deflection curve of the beam [6]. Under tension, the maximum deflections of a laterally loaded beam decrease whereas under compression they increase. The moments of the structure are influenced accordingly. For the simple problem of a beam with hinged ends, loaded by a single force P at the middle, the maximum deflection at the middle is: δ m = P 3 48E , (7.1) where is the lenght of the beam and E its flexural rigidity. The slope θ at both ends is: θ = ± P 2 16E . (7.2) SOFiSTiK 2014 | VERiFiCATiON MANUAL 33 BE4: Tie Rod with Lateral Loading The maximum value of the bending moment at the middle is: M m = P 4 . (7.3) When now the structure (Fig. 7.1) is submitted to the action of tensile forces N in addition to the initial lateral load P, the deflection at the middle becomes [6]: δ m = P 3 48E · − tanh 1 3 3 , (7.4) where 2 = N 2 / 4E. The first factor in Eq. (7.4) represents the deflection produced by the lateral load P acting alone. The second factor indicates in what proportion the deflection produced by P is magnified by the axial tensile force N, respectively. When N is small, it approaches unity, which indicates that under this condition the effect on the deflection of the axial force is negligible. The expressions for the moment and the slopes can be derived accordingly [6]. 7.3 Model and Results The properties of the model are defined in Table 7.1 and the results are presented in Table 7.2. Fig. 7.2 shows the deformed structure under tension and lateral loading. Table 7.1: Model Properties Material Properties Geometric Properties Loading E = 30000 MP = 2 m P = 0.1 kN h = 30 mm N = 0.1 kN b = 30 mm = 6.75 × 10 −8 m 4 Figure 7.2: Deformed Structure [mm]: N = 0 34 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE4: Tie Rod with Lateral Loading Table 7.2: Results N = 0 Ref. N = 0 Ref. δ m [m] 0.00823 0.00823 0.00807 0.00807 M m [kNm] 0.05000 0.05000 0.04919 0.04919 θ [rd] 0.01235 0.01235 0.01210 0.01210 7.4 Conclusion This example presents the influence of axial forces applied at a laterally loaded beam. The case of a tie rod is examined and the maximum deflections and moment are derived. It has been shown that the behaviour of a beam under the combination of direct axial force and lateral load can be adequately captured. 7.5 Literature [6] S. Timoshenko. Strength of Materials, Part II, Advanced Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. SOFiSTiK 2014 | VERiFiCATiON MANUAL 35 BE4: Tie Rod with Lateral Loading 36 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE5: Bending of a T-beam 8 BE5: Bending of a T-beam Overview Element Type(s): B3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): AQB, ASE Input file(s): t beam.dat 8.1 Problem Description An asymmetric T-beam is supported as shown in Fig. 8.1 and subjected to uniform bending M z . Deter- mine the maximum tensile and compressive bending stresses. M z M z z y h 2 h 1 | ←− −→| Figure 8.1: Model Properties 8.2 Reference Solution According to the discussion in Benchmark Example no. 3, it follows that the maximum tensile and compressive stresses in a beam in pure bending are proportional to the distances of the most remote fibers from the neutral axis of the cross-section. When the centroid of the cross-section is not at the middle of the depth, as, for instance, in the case of a T-beam, let h 1 and h 2 denote the distances from the neutral axis to the outermost fibers in the downward and upward directions (Fig. 8.1) respectively. Then for a bending moment M z , we obtain the maximum tensile and compressive stresses [5]: σ m = M z h 1 z and σ mn = − M z h 2 z . (8.1) 8.3 Model and Results The properties of the model are defined in Table 8.1. Distances from the centroid to the top and bottom of the beam are calculated as 14 cm and 6 cm respectively. The results are presented in Table 8.2. Figure 8.2 shows the distribution of the stresses along the cross-section. SOFiSTiK 2014 | VERiFiCATiON MANUAL 37 BE5: Bending of a T-beam Table 8.1: Model Properties Material Properties Geometric Properties Loading E = 30000 MP = 1 m M z = 100 kNm h = 20 cm h 1 = 6 cm, h 2 = 14 cm b = 9 cm t eb = 1.5 cm t ƒ nge = 4 cm z = 2000 cm 4 Table 8.2: Results SOF. Ref. σ m [MPa] 300 300 σ mn [MPa] −700 −700 0 0 . 0 0 0 . 0 0 1 0 0 . 0 0 1 0 0 . 0 0 1 0 0 . 0 0 1 0 0 . 0 0 3 0 0 . 0 0 3 0 0 . 0 0 3 0 0 . 0 0 3 0 0 . 0 0 1 0 0 . 0 0 1 0 0 . 0 0 1 0 0 . 0 0 1 0 0 . 0 0 3 0 0 . 0 0 3 0 0 . 0 0 7 - 0 0 . 0 0 7 - 0 0 . 0 0 7 - 0 0 . 0 0 7 - 0 0 . 0 0 7 - 0 0 . 0 0 7 - 0 0 . 0 0 7 - Figure 8.2: Distribution of Stresses 8.4 Conclusion This example shows the derivation of stresses for beams with asymmetric cross-section in which the centroid of the cross-section is not at the middle of the depth. It has been shown that the behaviour of the beam is captured with an excellent accuracy. 8.5 Literature [5] S. Timoshenko. Strength of Materials, Part I, Elementary Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. 38 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE6: Warping Torsion Bar 9 BE6: Warping Torsion Bar Overview Element Type(s): B3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): ASE Input file(s): warping.dat 9.1 Problem Description A cantilever I-bar is fixed at both ends, as shown in Fig. 9.1, and subjected to a uniformly distributed torque m T [7]. Determine the angle of twist ϕ at the midspan. | ←− −→| m T Figure 9.1: An I-bar with Uniformly Distributed Torque 9.2 Reference Solution In mechanics, torsion is the twisting of a structure due to an applied torque. There are two types of torsion: St. Venant torsion and warping torsion. St. Venant torsion exists always when an element is twisted, whereas the warping torsion occurs additionally under specific conditions. The warping of a section depends on the section geometry which means that there exist warping-free, such as circular, and warping-restrained sections. St. Venant torsion is based on the assumption that either the cross- section is warping-free or that the warping is not constrained. If at least one of these conditions is not met then the warping torsion appears [6]. y z ϕ T Figure 9.2: Circular Shaft A member undergoing torsion will rotate about its shear center through an angle of ϕ. Consider a circular shaft that is attached to a fixed support at one end. If a torque T is applied to the other end, the shaft SOFiSTiK 2014 | VERiFiCATiON MANUAL 39 BE6: Warping Torsion Bar will twist, with its free end rotating through an angle ϕ called the angle of twist [8]: ϕ = TL G T , (9.1) where G is the shear modulus and T the torsional moment of inertia. For a circular shaft subjected to torsion, each cross-section rotates along the shaft as a solid rigid slab (warping-free cross-section). The torsional moment resisted by the cross-section is: T = G T dϕ d , (9.2) For most cross-sections, e.g. non-circular, this rotation of the cross-section is accompanied by warping. Then the total torsional moment resisted by the cross-section becomes the sum of the pure torsion and warping torsion [9]. The stresses induced on the member is then classified into three categories: torsional shear stress, warping shear stress and warping normal stress. For example, when a bar of an I-cross-section is subjected to torsion, then the flanges of the cross-section experience bending in the flange planes. This means that torsion induces bending about the strong axis of the flanges. When the tendency for the cross-section to warp freely is prevented or restrained, it causes stresses to develop. The torque that the cross-section carries by bending is: T = EC M d 3 ϕ d 3 (9.3) where EC M , is the warping torsion stiffness. Furthermore, in warping torsion theory the bimoment is defined as an auxiliary quantity. The objective is to introduce a degree of freedom for beam elements that represents the torque due to restrained warping. The bimoment M ω is defined as: M ω = EC M d 2 ϕ d 2 (9.4) It should be noted, that the bimoment itself is not measurable, however it serves as a convenient param- eter to quantify this prevention of warping. m T y (a) m T () m T M T (b) M T + M T ´d d m T = m 0 + m 1 (c) d m 1 m 0 Figure 9.3: The Warping Torsion Problem Fig. 9.3 (a) shows the warping torsion problem of a bar subjected to a distributed external torque. The differential equation governing the warping torsion problem, for a constant cross-section, becomes [10]: 40 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE6: Warping Torsion Bar EC M d 4 ϕ d 4 − G T d 2 ϕ d 2 = m T , (9.5) where m T the distributed torque along the bar. The natural boundary conditions are: M ω = EC M d 2 ϕ d 2 , z = 0 or l (9.6) and −EC M d 3 ϕ d 3 + G T dϕ d = M T , z = 0 or l (9.7) where M T is the concentrated end torque and M ω the bimoment. Introducing λ, the so called decay factor, in the above equation, and a simplified notation for the derivatives of ϕ, we obtain: ϕ − λ 2 ϕ = m T EC M . (9.8) The solution of the warping torsion equation depends on the type of the torsional load and the kinematic boundary conditions, especially the amount of prevention of the warping. The complete solution system of Eq. 9.8, for the load type given in Fig. 9.3 (c), is thus: ϕ = C 1 λ 2 snh λ + C 2 λ 2 cosh λ + C 3 + C 4 − 1 2G T (m 0 + 1 3 m 1 ) 2 (9.9) ϕ = C 1 λ cosh λ + C 2 λ snh λ + C 3 − 1 2G T (2m 0 + m 1 ) (9.10) ϕ = C 1 snh λ + C 2 cosh λ − 1 G T (m 0 + m 1 ) (9.11) ϕ = C 1 λcosh λ + C 2 λsnh λ − m 1 G T (9.12) The values of the constants C 1 to C 4 can be derived with respect to the kinematic boundary conditions of the problem. For the case of warping-free sections, where C M = 0, the differential equation is shortened, leading to the St. Venant torsion problem. 9.3 Model and Results The properties of the analysed model, are defined in Table 9.1. The corresponding results are presented in Table 9.2. Figure 9.4 shows the deformed shape of the structure and the angle of twist. SOFiSTiK 2014 | VERiFiCATiON MANUAL 41 BE6: Warping Torsion Bar Table 9.1: Model Properties Material Properties Cross-sectional Properties Loading E = 217396.3331684 N/ mm 2 = 1 m m T = 1 Nmm/ mm G = 81386.6878 N/ mm 2 h = 80 mm ν = 0.33557673 t = 2 mm b = 40 mm C M = 0.323 × 10 8 mm 6 T = 431.979 mm 4 Table 9.2: Results Twist in x-direction Ref. [7] ϕ [mrd] 0.329659 0.329262 Figure 9.4: Deformed Stucture 9.4 Conclusion This example presents the warping torsion problem. The total torsional moment resisted by the cross- section is the sum of that due to pure torsion, which is always present, and that due to warping. It has been shown that the behaviour of the beam for warping is captured correctly. 9.5 Literature [6] S. Timoshenko. Strength of Materials, Part II, Advanced Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. [7] C-N. Chen. “The Warping Torsion of a Bar Model of the Differential Quadrature Element Method”. In: Computers and Structures 66.2-3 (1998), pp. 249–257. [8] F.P. Beer, E.R. Johnston, and J.T. DeWolf. Mechanics of Materials. 4th. McGraw-Hill, 2006. [9] P. Seaburg and C.J. Carter. Steel Design Guide Series 9: Torsional Analysis of Structural Steel Members. AISC. 2003. [10] C. Petersen. Stahlbau. 2nd. Vieweg, 1990. 42 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE7: Large Deflection of Cantilever Beams I 10 BE7: Large Deflection of Cantilever Beams I Overview Element Type(s): B3D, SH3D Analysis Type(s): STAT, GNL Procedure(s): LSTP Topic(s): Module(s): ASE Input file(s): beam elem.dat, quad elem.dat 10.1 Problem Description A cantilever beam is supported as shown in Fig. 10.1. The beam is subjected to a total vertical load, applied at the tip of the cantilever, which should cause the tip to deflect significantly. The determination of the non-dimensional tip deflections ratios are determined. | ←− −→| P Figure 10.1: Model Properties 10.2 Reference Solution The classical problem of deflection of a cantilever beam of linear elastic material, under the action of an external vertical concentrated load at the free end, is analysed (Fig. 10.2). The solution for large deflection of a cantilever beam cannot be obtained from elementary beam theory since basic assumptions are no longer valid. The elementary theory includes specific simplifications e.g. in the consideration of curvature derivatives, and provides no correction for the shortening of the moment arm as the loaded end of the beam deflects. For large finite loads, it gives deflections greater than the length of the beam [11]. P Δ δ L Figure 10.2: Problem Definition SOFiSTiK 2014 | VERiFiCATiON MANUAL 43 BE7: Large Deflection of Cantilever Beams I The mathematical treatment of the equilibrium of cantilever beams does not involve great difficulty. Nev- ertheless, unless small deflections are considered, an analytical solution does not exist, since for large deflections a differential equation with nonlinear terms must be solved. The problem is said to involve geometrical nonlinearity [12]. Therefore in order to account for this nonlinear term, third order theory is performed, where the equilibrium is established at the deformed configuration (geometrically nonlinear analysis). 10.3 Model and Results A circular pipe with cross-section of outer diameter 0.2 m and wall thickness 0.01 m is used, so that the beam is moderately slender. This type of problem becomes considerably more difficult numerically as the slenderness ratio increases [13]. The finite element model consists of twenty elements. The properties of the model are defined in Table 10.1. Table 10.1: Model Properties Material Properties Geometric Properties Loading E = 100 MP = 10 m P = 269.35 N D = 0.2 m t = 0.01 m As an alternative, the structure is analysed with quad plane elements with a cross-section of the same stiffness as the circular, in order to achieve the same results and compare the behaviour of the two types of elements. The quad cross-section has a width of 0.3mand a thickness of 0.10261m, and therefore the same moment of inertia = 2.701 m −5 as the one of the circular cross-section. Results for both models are presented in Table 10.2. Figure 10.3 shows the deflection of the beam for the two analysed models. Figure 10.4 presents the results, in terms of the motion of the tip of the cantilever, where they are compared to the exact solution for the inextensible beam, as given by Bisshopp and Drucker [11]. Figure 10.3: Deformed structure: a) Beam elements b) Quad elements 44 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE7: Large Deflection of Cantilever Beams I Table 10.2: Results Beam Quad δ[m] 8.113 8.107 Δ[m] 5.545 5.544 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 diplacement [m] load factor L − Δ δ 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 diplacement [m] load factor L − Δ δ Figure 10.4: Load - Deflection: (a). Beam elements (b). Quad elements SOFiSTiK 2014 | VERiFiCATiON MANUAL 45 BE7: Large Deflection of Cantilever Beams I 10.4 Conclusion This benchmark shows the classical problem of a cantilever beam undergoing large deformations under the action of a vertical load at the tip. Results are presented in terms of the motion of the tip of the cantilever where the accuracy of the solution is apparent. 10.5 Literature [11] K. E. Bisshopp and D. C. Drucker. “Large Deflection of Cantilever Beams”. In: Quarterly of Applied Mathematics 3 (1945), pp. 272–275. [12] T. Bel ´ endez, C. Neipp, and A. Bel ´ endez. “Large and Small Deflections of a Cantilever Beam”. In: European Journal of Physics 23.3 (2002), pp. 371–379. [13] Abaqus Benchmarks Manual 6.10. Dassault Systmes Simulia Corp. 2010. 46 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE8: Large Deflection of Cantilever Beams II 11 BE8: Large Deflection of Cantilever Beams II Overview Element Type(s): B3D, SH3D Analysis Type(s): STAT, GNL Procedure(s): LSTP Topic(s): Module(s): ASE Input file(s): moment beam.dat, moment quad.dat 11.1 Problem Description The cantilever beam of Benchmark Example No. 7 is analysed here for a moment load, as shown in Fig. 11.1, with both beam and quad plane elements. The accuracy of the elements is evaluated through the deformed shape of the beam retrieved by limit load iteration procedure. | ←− −→| P Figure 11.1: Model Properties 11.2 Reference Solution The classical problem of deflection of a cantilever beam of linear elastic material, is here extended for the case of a moment applied at the beam tip. The concentrated moment causes the beam to wind around itself, i.e. deflect upwards and bend towards the built-in end. The analytical solution can be derived from the fundamental Bernoulli-Euler theory, which states that the curvature of the beam at any point is proportional to the bending moment at that point [14]. For the case of pure bending, the beam will bend into a circular arc of curvature R R = E M , (11.1) and will wind n times around itself [13] ML E = 2πn, (11.2) where is the moment of inertia, E the Elasticity modulus and M the concentrated moment applied at the tip. SOFiSTiK 2014 | VERiFiCATiON MANUAL 47 BE8: Large Deflection of Cantilever Beams II 11.3 Model and Results The properties of the two models analysed are defined in Table 11.1. For the moment load, the deformed shape of the structure for quad elements at various increments throughout the steps, are shown in Fig. 11.2. According to the analytical solution and the moment load applied, the cantilever is expected to wind around itself n = 2. Table 11.1: Model Properties Material Properties Geometric Properties Geometric Properties Loading Beam elements Quad elements E = 100 MP = 10 m = 10 m M = 3.38478 kNm D = 0.2 m B = 0.3 m t = 0.01 m t = 0.10261 m Figure 11.2: Deformed Structure - Quad Elements Figure 11.3: Final Deformed Shape of Cantilever with Quad Elements Figure 11.4 presents the load - deflection curve for the horizontal and vertical direction for the two cases. From the final deformed shape of the beam (Fig. 11.3), it is evident that the cantilever achieves 48 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE8: Large Deflection of Cantilever Beams II n ≈ 2, which can also be observed at the second load-deflection curve where the vertical displacement becomes zero twice. Beam Elements Quad Elements Figure 11.4: Load - Deflection Curve 11.4 Conclusion This benchmark shows the classical problem of a cantilever beam undergoing large deformations under the action of a moment load applied at the tip. The accuracy of the deformation solution for the quad and beam elements is evident. 11.5 Literature [13] Abaqus Benchmarks Manual 6.10. Dassault Systmes Simulia Corp. 2010. [14] A. A. Becker. Background to Finite Element Analysis of Geometric Non-linearity Benchmarks. Tech. rep. NAFEMS, 1998. SOFiSTiK 2014 | VERiFiCATiON MANUAL 49 BE8: Large Deflection of Cantilever Beams II 50 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE9: Verification of Beam and Section Types I 12 BE9: Verification of Beam and Section Types I Overview Element Type(s): B3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): AQUA Input file(s): cross sections.dat 12.1 Problem Description In this Benchmark different cross-section types are analysed, in order to test the properties of each cross-section associated with their definition in AQUA module. Figure 12.1 shows the cross-sections. 0 . 1 0 0.10 C SC 0 . 0 1 0 . 1 0 0.10 C SC 0 . 1 0 0.10 C SC 0 . 1 0 0.10 C SC 0 . 1 0 0.10 C SC 0 . 1 0 0.10 C SC 0 . 1 0 0.10 C SC 0 . 1 0 0.20 C SC 0 . 1 0 0.20 C SC 0 . 1 0 0.10 C SC 0.10 C SC Figure 12.1: Cross-Section Types 12.2 Reference Solution The important values of a cross-section for the simple cases of bending and torsion are the moment of inertia and the torsional moment, respectively. The analytical solution for the moment of inertia y with respect to y axis is [5]: y = _ A z 2 dA, (12.1) SOFiSTiK 2014 | VERiFiCATiON MANUAL 51 BE9: Verification of Beam and Section Types I in which each element of area dA is multiplied by the square of its distance from the z-axis and the integration is extended over the cross-sectional area A of the beam (Fig. 12.2). The torsional moment T is more complicated to compute and depends on the cross-sections geometry. For circular cross- sections is: T = _ A r 2 dA, (12.2) For thick-walled non-circular cross-sections, it depends on the warping function. Tabulated formulas are given in all relevant handbooks for the most common geometries [15]. For closed thin-walled non-circular cross-sections T is [10]: T = 4A 2 m n =1 s t , (12.3) and for open thin-walled non-circular cross-sections is: T = 1 3 n =1 s t 3 , (12.4) where A m is the area enclosed from the center line of the wall (Fig. 12.2), and t , s the dimensions of the parts from which the cross-section consists of. For the specific case of an I-cross-section, another approximate formula can be utilised, as defined by Petersen [10]: T = 2 1 3 b t 3 _ 1 − 0.630 t b _ + 1 3 (h − 2t) s 3 + 2 α D 4 , (12.5) where s, t and D are described in Fig. 12.2 and α is extracted from the corresponding diagram, given in [10], w.r.t. the cross-section properties. For the same cross-section but according to Gensichen, T is accordingly computed as: T = 2 1 3 b t 3 _ 1 − 0.630 t b _ + 1 3 (h − 2t) s 3 + 0.29 s t _ _ _ s 2 _ 2 + t 2 t _ _ 4 (12.6) y r dA z A m h D t s b Figure 12.2: Cross-Sectional Properties 52 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE9: Verification of Beam and Section Types I 12.3 Model and Results The properties of the different cross-sections, analysed in this example, are defined in Table 12.1. The cross-sections types are modelled with various ways in AQUA and compared. For differentiation between them, the modelling type is specified next to the name of each cross-section. The results are presented in Table 12.2. Table 12.1: Cross-Sections Properties Material Properties Cross-sectional Properties E = 30 MP b = 100 mm ν = 0.3 h = 100 mm t = 10 mm D = 100 mm Table 12.2: Results y [cm 4 ] |e r | T [cm 4 ] |e r | Type SOF. Ref. [%] SOF. Ref. [%] square -srec 833.33 833.33 0.00 1400.00 1400.00 0.00 rectangular -srec 0.83 0.83 0.00 3.13 3.13 0.00 circul -scit 490.87 490.87 0.00 981.75 981.75 0.00 circul -tube 490.87 490.87 0.00 981.75 981.75 0.00 pipe -scit 289.81 289.81 0.00 579.62 579.62 0.00 pipe -tube 289.81 289.81 0.00 579.62 579.62 0.00 Tbeam -poly 180.00 180.00 0.00 6.45 6.33∗ 1.89 Tbeam -plat 181.37 182.82 0.79 6.50 6.50 0.00 Ibeam -poly 449.33 449.33 0.00 9.53 9.33∗ (12.4) 2.10 9.62 (12.6) 0.94 9.21 ( 12.5) 3.44 Ibeam -plat 465.75 467.42 0.36 9.67 9.67 0.00 Ibeam -weld 447.67 449.33 0.37 9.33 9.33 0.00 square box -poly 492.00 492.00 0.00 771.99 729.00∗ 5.90 square box -plat 486.00 487.50 0.31 741.00 729.00 1.65 square box open -plat 486.00 487.50 0.31 11.98 12.00 0.17 rectang. box -poly 898.67 898.67 0.00 2172.38 2088.64∗ 4.01 rectang. box -plat 891.00 889.17 0.21 2107.31 2088.64 0.89 SOFiSTiK 2014 | VERiFiCATiON MANUAL 53 BE9: Verification of Beam and Section Types I Table 12.2: (continued) y [cm 4 ] |e r | T [cm 4 ] |e r | Type SOF. Ref. [%] SOF. Ref. [%] C-beam -poly 2292.67 2292.67 0.00 13.99 12.67∗ 10.4 C-beam -plat 2286.33 2287.92 0.07 12.67 12.67 0.0 L-beam -poly 180.00 180.00 0.00 6.36 6.33∗ 0.36 L-beam -weld 179.25 180.00 0.42 6.33 6.33 0.0 L-beam -plat 178.62 179.41 0.44 6.33 6.33 0.0 L 100 10 (tabulated) 176.68 177.0 [16] 0.18 6.94 6.33 [17] 9.51 I 100 (tabulated) 170.21 171.0 [16] 0.46 1.46 1.60 [16] 8.64 170.3 [17] 0.06 1.511 [17] 3.26 UPE 100 (tabulated) 206.85 207.0 [16] 0.07 2.09 1.99 [16] 5.24 206.9 [17] 0.03 2.01 [17] 4.19 IPE 400 (tabulated) 23126.97 23130 [16] 0.01 51.01 51.40 [16] 0.76 23128 [17] 0.00 50.41 [17] 1.18 When evaluating the results of the torsional moment of inertia T , it has to be taken under consideration, that the presented reference solutions of Sect. 12.2, for all non-circular cross-sections, are approximate and various assumptions are involved, according every time to the adopted theory. For the case of the I-beam, it is observed in Table 12.2, that the absolute errors fluctuate between 3.44 % and 0.91 %, according to which approximate reference result are compared to. The reference values for the open sections I, L, C, T-beam, denoted with an asterisk, are computed with respect to the thin-walled theory reference solution (Eq. 12.4). Therefore for the calculated values with -POLY, which do not correspond to the thin-walled theory, deviations appear. If we now make a convergence study, for the case of the I-beam, decreasing the thickness of the cross-section and comparing it to the thin-walled reference solution, we will observe that the deviation is vanishing as we approach even thinner members. This is presented in Fig. 12.4 for an I-beam, where the absolute difference of the calculated from the reference value is depicted for decreasing thickness values. C SC C SC C SC C SC Figure 12.3: Rolled Steel Shapes For the case of open thin-walled non-circular cross-sections, modelled with -PLAT, we can observe that T matches exactly the reference solution. For closed thin-walled non-circular cross-sections though, some deviations arise. If we take a closer look at the case of the square box, at first glance it appears to be not accurate enough, since the calculated value is 741.00 cm 4 and the reference is 729.00 cm 4 . The difference between them is 741.00 - 729.00 = 12 cm 4 , which corresponds to the reference value 54 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE9: Verification of Beam and Section Types I of the open square box. This is due to the fact, that the reference solution for this type of sections given by Eq. 12.3, corresponds to the thin-walled theory and assumes a constant distribution of shear stresses over the thickness of the cross-section. However, SOFiSTiK assumes a generalised thin-walled theory, where the shear stresses due to torsion, are distributed linearly across the thickness, as shown in Fig. 12.5, and thus holds: T genersed thn−ed theory = T cosed,SOFSTK = T cosed,thn−ed theory + T open,thn−ed theory , (12.7) Eq. 12.7 is satisfied exactly for the square box cross-section and it can be visualised in Fig. 12.6 by the purple line for decreasing thicknesses, whereas the blue line denotes the deviation of the T genersed thn−ed theory from the T cosed,thn−ed theory . 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 0 2 · 10 −2 4 · 10 −2 6 · 10 −2 8 · 10 −2 0.1 0.12 0.14 0.16 0.18 0.2 Thickness [mm] __ T , r e ƒ − T , c c __ [ c m 4 ] Figure 12.4: Convergence of I-beam For the same cross-section, but now modelled with -POLY, it is evident that the difference from the reference solution is larger, reaching the value of 5.90 %, as presented by the green line. This is due to the fact that except from the difference in the stresses consideration, as explained above, the thin-walled assumption is also engaged. If we do a convergence study for this cross-section, and compared it to the one modelled with -PLAT, represented by the red line, we will observe that as the thickness decreases the deviation curves gradually coincide. For the case of the standard tabulated rolled steel shapes, depicted in Fig. 12.3, we can observe that the differences deviate between 0.46 to 0.00 % for the y and between 9.51 to 0.76 % for the T , according to the reference source. Thus, it is evident once more, that the calculation of the T is a complicated matter. SOFiSTiK 2014 | VERiFiCATiON MANUAL 55 BE9: Verification of Beam and Section Types I Open Closed Figure 12.5: Distribution of Stresses 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 Thickness [mm] D e v i a t i o n [ % ] -POLY w.r.t. Thin-walled Theory -POLY w.r.t. -PLAT -PLAT generalised w.r.t. Thin-walled Theory -PLAT thin-walled w.r.t. Thin-walled Theory Figure 12.6: Convergence of Square Box From the results in Table 12.1, we can see that for the definition of general cross-sections the use of -POLY option gives the exact values for y . For the definition of thin-walled cross-sections the use of -PLAT gives very good results for T whereas for the determination of y some deviations appear. This is due to the fact that in order for the cross-section to be connected for shear, some parts of the plates overlap at the connections giving an additional moment of inertia around the y-axis. This can be seen at Fig. 12.7 for the I beam. It can be avoided if the -PLAT option is used without overlapping of parts but in combination with -WELD in order to ensure the proper connection of the plates. This can be seen from the results for the I- and L-beam which are analysed for the three options -POLY, -PLAT, -PLAT and -WELD. 56 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE9: Verification of Beam and Section Types I C SC C SC C SC -PLAT -PLAT and -WELD -POLY Figure 12.7: Definition types of I-beam 12.4 Conclusion This example presents the different cross-sections and their properties according to their definition in AQUA. It has been shown that the properties of the cross-sections can be adequately captured irrele- vantly of their definition with small deviations from the exact solution. 12.5 Literature [5] S. Timoshenko. Strength of Materials, Part I, Elementary Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. [10] C. Petersen. Stahlbau. 2nd. Vieweg, 1990. [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. [16] M. Schneider-B¨ urger. Stahlbau-Profile. 24th. Verlag Stahleisen, 2004. [17] R. Kindmann, M. Kraus, and H. J. Niebuhr. Stahlbau Kompakt, Bemessungshilfen, Profiltabellen. Verlag Stahleisen, 2006. SOFiSTiK 2014 | VERiFiCATiON MANUAL 57 BE9: Verification of Beam and Section Types I 58 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE10: Verification of Beam and Section Types II 13 BE10: Verification of Beam and Section Types II Overview Element Type(s): B3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): ASE Input file(s): cross sections ii.dat 13.1 Problem Description The problem consists of a cantilever beam as shown in Fig. 13.1. For the first case analysed, a trans- verse load is applied at the end of the beam. For the second case, a moment is applied around the axis. The various cross-section types analysed in Benchmark Example 9 are used, in order to test the behaviour of the beam associated with each of the section definitions. P −→| | ←− M Figure 13.1: Problem Description 13.2 Reference Solution For a Bernoulli beam and a linear elastic material behaviour, the maximum deflection δ m of the can- tilever, under the action of a transverse load P, occurs at the tip and is [15]: δ m = PL 3 3E , (13.1) and the rotation ϕ z ϕ z = PL 2 2E . (13.2) For the case of the moment M, applied at the -axis the angle of twist ϕ is [10]: ϕ = ML G T , (13.3) where G is the shear modulus, E the flexural rigidity and T the torsional moment. SOFiSTiK 2014 | VERiFiCATiON MANUAL 59 BE10: Verification of Beam and Section Types II 13.3 Model and Results The properties of the model and the cross-sections analysed, are defined in Table 13.1. For all cross- section the shear deformation areas A y and A z are given equal to zero, in order to consider a Bernoulli beam formulation which doesn’t account for shear deformations. Table 13.1: Model Properties Material Properties Cross-sectional Properties Loading E = 30 MP L = 1 m P = 1 kN ν = 0.3 h = 100 mm M = 1 kNm t = 10 mm b = 100 mm D = 100 mm Table 13.2: Results Case 1 y [m] |e r | ϕ z [mrd] |e r | Type SOF. Ref. [%] SOF. Ref. [%] square -srec 1.333 1.333 0.00 2.000 2.000 0.00 rectangular -srec 1333.329 1333.333 0.00 1999.994 2000.000 0.00 circul -scit 2.264 2.264 0.00 3.395 3.395 0.00 circul -tube 2.264 2.264 0.00 3.395 3.395 0.00 pipe -scit 3.834 3.834 0.00 5.751 5.751 0.00 pipe -tube 3.834 3.834 0.00 5.751 5.751 0.00 Tbeam -poly 6.173 6.173 0.00 9.259 9.259 0.00 Tbeam -plat 6.126 6.078 0.80 9.189 9.116 0.80 Ibeam -poly 2.473 2.473 0.00 3.709 3.709 0.00 Ibeam -plat 2.386 2.377 0.36 3.578 3.566 0.36 Ibeam -weld 2.482 2.473 0.37 3.723 3.709 0.37 square box -poly 2.258 2.258 0.00 3.388 3.388 0.00 square box -plat 2.286 2.279 0.31 3.429 3.419 0.31 square box open -plat 2.286 2.279 0.31 3.429 3.419 0.31 rectang. box -poly 1.236 1.236 0.00 1.855 1.855 0.00 rectang. box -plat 1.247 1.250 0.21 1.871 1.874 0.21 C-beam -poly 0.485 0.485 0.00 0.727 0.727 0.00 60 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE10: Verification of Beam and Section Types II Table 13.2: (continued) y [m] |e r | ϕ z [mrd] |e r | Type SOF. Ref. [%] SOF. Ref. [%] C-beam -plat 0.486 0.486 0.07 0.729 0.728 0.07 L-beam -poly 6.173 6.173 0.00 9.259 9.259 0.00 L-beam -weld 6.199 6.173 0.42 9.298 9.259 0.42 L-beam -plat 6.221 6.193 0.44 9.331 9.290 0.44 The cross-sections types are modelled with various ways in AQUA as presented in Benchmark Example 9. The results are presented in Table 13.2 for the case of the transverse load P and in Table 13.3 for the case of the moment M. Table 13.3: Results Case 2 ϕ [mrd] |e r | Type SOF. Ref. [%] square -srec 6.190 6.190 0.00 rectangular -srec 2768.903 2768.903 0.00 circul -scit 8.828 8.828 0.00 circul -tube 8.828 8.828 0.00 pipe -scit 14.952 14.952 0.00 pipe -tube 14.952 14.952 0.00 Tbeam -poly 1342.974 1368.421 1.86 Tbeam -plat 1333.333 1333.333 0.00 Ibeam -poly 909.485 928.571 2.06 Ibeam -plat 896.552 896.552 0.00 Ibeam -weld 928.571 928.571 0.00 square box -poly 11.226 11.888 5.57 square box -plat 11.696 11.888 1.62 square box open -plat 723.428 722.222 0.17 rectang. box -poly 3.989 4.149 3.85 rectang. box -plat 4.113 4.149 0.89 C-beam -poly 619.612 684.211 9.44 C-beam -plat 684.210 684.211 0.00 L-beam -poly 1363.460 1368.421 0.36 SOFiSTiK 2014 | VERiFiCATiON MANUAL 61 BE10: Verification of Beam and Section Types II Table 13.3: (continued) ϕ [mrd] |e r | Type SOF. Ref. [%] L-beam -weld 1368.421 1368.421 0.00 L-beam -plat 1368.421 1368.421 0.00 From the above results, and with respect to the results of Benchmark Example 9, we can see that the differences are a direct influence of the calculations of the properties of the cross-sections according to their definition in AQUA, and are not associated to the beam formulation. This can also be verified, if instead of, e.g. the reference value for y REF , the calculated value is used y CALC in Eq. 13.1 . Then the error is eliminated for all the cross-sections types. 13.4 Conclusion This example presents the influence of the cross-sections types, for the case of a simple cantilever beam. It has been shown that the behaviour of the beam is accurately captured. 13.5 Literature [10] C. Petersen. Stahlbau. 2nd. Vieweg, 1990. [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. 62 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE11: Plastification of a Rectangular Beam 14 BE11: Plastification of a Rectangular Beam Overview Element Type(s): B3D, BF2D, SH3D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): Module(s): ASE, STAR2, TALPA Input file(s): beam star2.dat, fiber beam.dat, quad.dat 14.1 Problem Description The problem consists of a rectangular cantilever beam, loaded in pure bending as shown in Fig. 14.1. The model [18] is analysed for different load levels, including the capacity limit load, where the cross- section fully plastifies. The beam is modelled and analysed with different elements and modules. M yed | ←− −→| Figure 14.1: Problem Description 14.2 Reference Solution The model follows an elastic-perfectly-plastic stress-strain behaviour as shown in Fig. 14.2. Under this assumption, the beam remains elastic until the outermost fibers reach the yield stress. The correspond- ing limit load can be calculated as: M yed = σ yed bh 2 6 , (14.1) where σ yed is the yield stress, b and h the dimensions of the beam. The cross-section fully plastifies when the load reaches M = M t = 1.5×M yed , where all fibers of the beam are in condition of yielding [6]. σ yed −σ yed ε σ Figure 14.2: Stress-Strain Curve SOFiSTiK 2014 | VERiFiCATiON MANUAL 63 BE11: Plastification of a Rectangular Beam 14.3 Model and Results The properties of the model are defined in Table 14.1. A standard steel material is used and modified accordingly to account for the intended elastic-perfectly-plastic material behaviour. Table 14.1: Model Properties Material Properties Geometric Properties Loading E = 210000 MP L = 1 m M yed = 280 Nm ν = 0.3 h = 20 mm σ yed = 420 MP b = 10 mm The structure is modelled and analysed in various ways. For the first case the fiber beam is used (TALPA), where the cross-section is discretised into single fibers and directly integrates the continuum mechanical material reaction into beam theory, and physically nonlinear analysis is performed. For the second case the standard beam elements are used and the model is analysed with STAR2 where a nonlinear stress and strain evaluation determination is performed. For the third case, the quad elements are used and a nonlinear analysis is done with ASE. The results are presented in Table 14.2 for the three cases. Table 14.2: Results M/M yed Fiber Beam Standard Beam Quad Ref. σ [MP] σ [MP] σ σ eƒ ƒ 0.99 415.80 415.80 415.80 415.80 σ < 420.00 Fully Elastic 1.00 ≤ 420 ≤ 420 ≤ 420 ≤ 420 σ ≤ 420.00 First Yield 1.48 ≤ 420 ≤ 420 ≤ 430.9 ≤ 420 σ ≤ 420.00 Elastic-Plastic 1.50 Fully-Plastic Fully-Plastic Fully-Plastic σ = 420.00 No Convergence No Convergence Fully-Plastic 1.51 Fully-Plastic Fully-Plastic Fully-Plastic Fully-Plastic No Convergence No Convergence No Convergence No Convergence This benchmark is designed to test elastic-plastic material behaviour under uniaxial loading conditions. From the above results, it is evident that both beam element formulations adequately reproduce the intended behaviour. Fig. 14.3 shows the distribution of stresses for the case of the fiber beam with M/M yed = 0.99, 1.0 and 1.5. For the quad element, the stress appears to exceed the limit value of 420 MP. This is due to the fact that, as the plasticity involves at the cross-section, plastic strains also appear in the lateral direction. This causes a biaxial stress state, which is not neglected by the quad formulation, as shown in Fig. 14.4 for M/M yed = 1.0 and 1.48. A closer look at the list of results 64 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE11: Plastification of a Rectangular Beam though, reveals that the eƒ ƒ ecte stresses do not exceed the σ yed . 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 8 . 5 1 4 0 8 . 5 1 4 0 8 . 5 1 4 0 8 . 5 1 4 0 2 . 4 0 8 . 5 1 4 0 8 . 5 1 4 - 0 8 . 5 1 4 - 0 2 . 4 - 0 8 . 5 1 4 - 0 8 . 5 1 4 - 0 8 . 5 1 4 - 0 8 . 5 1 4 - 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 0 0 . 0 2 4 4 2 . 4 0 0 . 0 2 4 0 0 . 0 2 4 - 0 0 . 0 2 4 - 4 2 . 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - 0 0 . 0 2 4 - Figure 14.3: Fiber Beam Stress State Figure 14.4: Quad Stress State 14.4 Conclusion This example presents the pure bending of beams beyond their elastic limit for a non elastic material. It has been shown that the behaviour of the beam is accurately captured for all three modelling options. 14.5 Literature [6] S. Timoshenko. Strength of Materials, Part II, Advanced Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. [18] Verification Manual for the Mechanical APLD Application, Release 12.0. Ansys, Inc. 2009. SOFiSTiK 2014 | VERiFiCATiON MANUAL 65 BE11: Plastification of a Rectangular Beam 66 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE12: Cantilever in Torsion 15 BE12: Cantilever in Torsion Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): Topic(s): Module(s): ASE Input file(s): torsion.dat 15.1 Problem Description The problem consists of a cantilever beam as shown in Fig. 15.1. The tip of the cantilever is offsetted in y-direction by Δ y = / 200 = 2.5 cm, creating a geometrical imperfection. The beam is loaded with a transverse force P z and an axial force P . The imperfection acts as a lever arm for the loading, causing a torsional moment. The torsional moment at the support with respect to the local and global coordinate system is determined. P z P Δ y z y Figure 15.1: Problem Description 15.2 Reference Solution In order to account for the effect of the geometrical imperfection on the structure, second-order theory should be used, where the equilibrium is established at the deformed system. According to the equilib- rium of moments at the deformed system, with respect to the global -axis, the torsional moment at the support M gob is: M gob = P z ( y + Δ y ) − P y z , (15.1) whereas by the local -axis the torsional moment M oc is: M oc = P z y + P ( Δ y ) z , (15.2) where is the length of the beam, Δ y the initial geometrical imperfection and P is negative for compres- sion. SOFiSTiK 2014 | VERiFiCATiON MANUAL 67 BE12: Cantilever in Torsion 15.3 Model and Results The properties of the model [19] [20] are defined in Table 15.1. A standard steel material is used as well as a standard hot formed hollow section with properties according to DIN 59410, DIN EN 10210-2. A safety factor γ M = 1.1 is used, which according to DIN 18800-2 it is applied both to the yield strength and the stiffness. Furthermore, the self weight, the shear deformations and the warping modulus C M are neglected. At the support the warping is not constrained. Table 15.1: Model Properties Material Properties Geometric Properties Loading S 355 = 5 m P z = 10 kN γ M = 1.1 RRo/ SH 200 × 100 × 10 [15] P = 100 kN C M = 0 Δ y = 2.5 cm Table 15.2: Results y z M gob M oc P bck [cm] [cm] [kNcm] [kNcm] [kN] SOF. 3.208 10.204 57.07 26.97 163.7 Ref.[21] 3.20 10.2 57.0 26.9 164 The corresponding results are presented in Table 15.2. Figure 15.2 shows the deformed shape of the structure and the nodal displacements for the z and y direction. From the presented results, we can observe that the values of the moments are correctly computed. Here has to be noted that the reference results are according to [19], where they are computed with another finite element software, and not with respect to an analytical solution. [ y ] [ z ] Figure 15.2: Deformations [mm] 68 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE12: Cantilever in Torsion 15.4 Conclusion This example presents a case where torsion is induced to the system because of an initial geometrical imperfection. It has been shown that the behaviour of the beam is captured accurately. 15.5 Literature [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. [19] V. Gensichen and G. Lumpe. Zur Leistungsf ¨ ahigkeit, korrekten Anwendung und Kontrolle r ¨ aumlicher Stabwerksprogramme. Stahlbau Seminar 07. [20] V. Gensichen. Zur Leistungsf ¨ ahigkeit r ¨ aumlicher Stabwerksprogramme, Feldstudie in Zusamme- narbeit mit maßgebenden Programmherstellern. Stahlbau Seminar 07/08. [21] V. Gensichen and G. Lumpe. “Zur Leistungsf ¨ ahigkeit, korrekten Anwendung und Kontrolle von EDV-Programmen f ¨ ur die Berechnung r ¨ aumlicher Stabwerke im Stahlbau”. In: Stahlbau 77 (Teil 2) (2008). SOFiSTiK 2014 | VERiFiCATiON MANUAL 69 BE12: Cantilever in Torsion 70 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE13: Buckling of a Bar with Hinged Ends I 16 BE13: Buckling of a Bar with Hinged Ends I Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): STAB Topic(s): Module(s): ASE Input file(s): buckling bar.dat 16.1 Problem Description The problem consists of an axially loaded long slender bar of length with hinged ends, as shown in Fig. 16.1. Determine the critical buckling load. [18] P P y P y / 2 Figure 16.1: Problem Description 16.2 Reference Solution The problemof lateral buckling of bars is examined here. The case of a bar with hinged ends is very often encountered in practical applications and is called the ƒ ndment case of buckling of a prismatic bar. For the case of an axially compressed bar there is a certain critical value of the compressive force at which large lateral deflection may be produced by the slightest lateral load. For a prismatical bar with hinged ends (Fig. 16.1) this critical compressive force is [6]: P cr = π 2 E (β) 2 = π 2 E 2 , (16.1) SOFiSTiK 2014 | VERiFiCATiON MANUAL 71 BE13: Buckling of a Bar with Hinged Ends I where is the full length of the bar, E its flexural rigidity and β the effective length coefficient, whose value depends on the conditions of end support of the bar. For the fundamental case, β = 1. If the load P is less than its critical value the bar remains straight and undergoes only axial compression. This straight form of elastic equilibrium is stable, i.e., if a lateral force is applied and a small deflection is produced this deflection disappears when the lateral load is removed and the bar becomes straight again. By increasing P up to the critical load causes the column to be in a state of unstable equilibrium, which means, that the introduction of the slightest lateral force will cause the column to undergo large lateral deflection and eventually fail by buckling. 16.3 Model and Results Only the upper half of the bar is modelled because of symmetry (Fig. 16.1). The boundary conditions thus become free-fixed for the half symmetry model. A total of 20 elements are used to capture the buckling mode. The properties of the model are defined in Table 16.1. Table 16.1: Model Properties Material Properties Geometric Properties Loading E = 300 MP = 20 m P y = 1 kN h = 0.5 m P << 1 kN A = 0.25 m 2 = 5.20833 × 10 −3 m 4 β = 2, free-fixed ends Node 100 ux [mm] 1 4 .0 0 0 1 2 .0 0 0 1 0 .0 0 0 8 .0 0 0 6 .0 0 0 4 .0 0 0 2 .0 0 0 0 .0 0 0 0 .0 0 0 factor 30.0 20.0 10.0 0.0 Figure 16.2: Load-Deflection curve A small horizontal load at the top is necessary in order to induce an initial horizontal displacement. It should be sufficiently large to cause a nonlinear iteration, but it should not affect the result unintentionally. A buckling eigenvalue determination is performed where the critical load factor is calculated. The results 72 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE13: Buckling of a Bar with Hinged Ends I are presented in Table 16.2. Moreover, an ultimate limit load iteration is done and the produced Load- Deflection curve is shown in Fig. 16.2, as well as a part of the iteration summary. Table 16.2: Results SOF. Ref. P cr [kN] 38.553 38.553 16.4 Conclusion This example presents the buckling of slender bars. It has been shown that the buckling properties of the bar are accurately captured. 16.5 Literature [6] S. Timoshenko. Strength of Materials, Part II, Advanced Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. [18] Verification Manual for the Mechanical APLD Application, Release 12.0. Ansys, Inc. 2009. SOFiSTiK 2014 | VERiFiCATiON MANUAL 73 BE13: Buckling of a Bar with Hinged Ends I 74 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE14: Buckling of a Bar with Hinged Ends II 17 BE14: Buckling of a Bar with Hinged Ends II Overview Element Type(s): SH3D Analysis Type(s): STAT, GNL Procedure(s): STAB Topic(s): Module(s): ASE Input file(s): buckling bar quad.dat 17.1 Problem Description Benchmark Example 13 is tested here for QUAD plane elements. The problem consists of an axially loaded long slender bar of length with hinged ends, as shown in Fig. 17.1. Determine the critical buckling load [18]. P P z P/ 2 P/ 2 z / 2 Figure 17.1: Problem Description 17.2 Reference Solution The problem of lateral buckling of bars is presented at Benchmark Example 13. For a prismatic bar, the critical load is [6]: P cr = π 2 E (β) 2 . (17.1) From the above equation it is evident, that the critical load does not depend upon the strength of the material but only upon the dimensions of the structure and the modulus of elasticity of the material. Two equal slender axially compressed bars, will buckle at the same compressive force, if they consist of the SOFiSTiK 2014 | VERiFiCATiON MANUAL 75 BE14: Buckling of a Bar with Hinged Ends II same flexural rigidity and material with the same Young’s modulus. 17.3 Model and Results Only the upper half of the bar is modelled because of symmetry (Fig. 17.1). The boundary conditions thus become free-fixed for the half symmetry model. A total of 20 elements are used to capture the buckling mode. The properties of the model are defined in Table 17.1. Table 17.1: Model Properties Material Properties Geometric Properties Loading E = 300 MP = 20 m , h = 0.5 m P = 1 kN A = 0.25 m 2 P << 1 kN t = 0.5 m = 5.20833 × 10 −3 m 4 β = 2, free-fixed ends A buckling eigenvalue determination is performed where the critical load factor is calculated. For the different solver options for the eigenvalue calculation the results are presented in Table 17.2. The ref- erence value of the critical load for Benchmark Example 13 and 14 is calculated the same, since the properties of the two models are equivalent, as explained in Section 17.2. Table 17.2: Results Solver P cr [kN] Ref. |e r | [%] BUSI - Simultaneous vector iteration 38.524 38.553 0.0751 BURA - Method of Rayleigh 38.527 38.553 0.0675 BUCK - Fastest solver for current system 38.524 38.553 0.0751 17.4 Conclusion This example presents the buckling of slender bars. It has been shown that the buckling properties of the bar are accurately captured also with QUAD elements. 17.5 Literature [6] S. Timoshenko. Strength of Materials, Part II, Advanced Theory and Problems. 2nd. D. Van Nos- trand Co., Inc., 1940. [18] Verification Manual for the Mechanical APLD Application, Release 12.0. Ansys, Inc. 2009. 76 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE15: Flexural and Torsional Buckling 18 BE15: Flexural and Torsional Buckling Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): STAB Topic(s): Module(s): ASE Input file(s): flex tors buckling.dat 18.1 Problem Description The problem consists of a standard I-beam, subjected to a compressive load P and supported as shown in Fig. 18.1. The flexural and torsional buckling load is determined. P Figure 18.1: Problem Description 18.2 Reference Solution For the rolled steel profiles, such as PE 300, the torsional buckling is generally only decisive, when the buckling length for torsional buckling s θ is significantly larger than the one for the flexural buckling s z , and at the same time the slenderness ratio is low [19]. The analysed model fulfils the above prerequisites. The flexural buckling load is: P bck z = π 2 E z s 2 z γ M , (18.1) whereas the torsional buckling load is: SOFiSTiK 2014 | VERiFiCATiON MANUAL 77 BE15: Flexural and Torsional Buckling P bck θ = 1 2 m γ M _ G T + π 2 EC M s 2 θ _ , (18.2) where E z the flexural rigidity, C M the warping modulus, γ M a safety factor, G the shear modulus, T the torsional moment and M is the polar radius of gyration calculated as following: M = y + z A . (18.3) 18.3 Model and Results The properties of the model [19] are defined in Table 18.1. A standard steel material is used, as well as a standard rolled steel profile with properties according to DIN 1025-5. A safety factor γ M = 1.1 is used, which according to DIN 18800-2 it is applied both to the yield strength and the stiffness. Furthermore, the self weight and the shear deformations are neglected. At all the supports the warping is not constrained. Table 18.1: Model Properties Material Properties Geometric Properties Loading S 355 s θ = 5 or 6 m P = 600 kN γ M = 1.1 s z = 2.5 or 3 m C M = 125900 cm 6 PE 300 [15] y = 8360 cm 4 z = 604 cm 4 A = 53.81 cm 2 The corresponding results are presented in Table 18.2. Figure 18.2 shows the deformed shape of the structure for the first and second buckling eigenvalues. It is obvious that the first one corresponds to the torsional buckling while the second one to the flexural. Table 18.2: Results s θ = 5.0 [m] / s z = 2.5 [m] s θ = 6.0 [m] / s z = 3.0 [m] P bck z [kN] P bck θ [kN] P bck z [kN] P bck θ [kN] SOF. 1820.78 1462.56 1264.43 1288.51 Exact 1820.74 1462.56 1264.41 1288.51 Ref. [21] 1818 1459 1264 1285 78 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE15: Flexural and Torsional Buckling Figure 18.2: Buckling Eigenvalues 18.4 Conclusion This example presents the determination of torsional and flexural buckling loads. It has been shown that the behaviour of the beam is captured accurately. 18.5 Literature [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. [19] V. Gensichen and G. Lumpe. Zur Leistungsf ¨ ahigkeit, korrekten Anwendung und Kontrolle r ¨ aumlicher Stabwerksprogramme. Stahlbau Seminar 07. [21] V. Gensichen and G. Lumpe. “Zur Leistungsf ¨ ahigkeit, korrekten Anwendung und Kontrolle von EDV-Programmen f ¨ ur die Berechnung r ¨ aumlicher Stabwerke im Stahlbau”. In: Stahlbau 77 (Teil 2) (2008). SOFiSTiK 2014 | VERiFiCATiON MANUAL 79 BE15: Flexural and Torsional Buckling 80 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE16: Torsion due to Biaxial Bending 19 BE16: Torsion due to Biaxial Bending Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): Topic(s): Module(s): ASE Input file(s): torsion bending.dat 19.1 Problem Description The problem consists of a beam subjected to transverse load P z and a lateral load P y , as shown in Fig. 19.1. The effect of torsion due to biaxial bending is examined. P z P y z y Figure 19.1: Problem Description 19.2 Reference Solution For an I-beam subjected to biaxial bending, without the action of a normal force, it follows directly from the beam theory that a torsional moment will inevitably appear, even if the cross-section is double sym- metric, the load is centrically applied, and the beam is statically determined. In order to account for this effect, third order theory has to be utilised. 19.3 Model and Results The properties of the model [19] are defined in Table 19.1. A standard steel material is used as well as a cross-section with a standard rolled steel shape. A safety factor γ M = 1.1 is used, which according to DIN 18800-2 it is applied both to the yield strength and the stiffness. Furthermore, the self weight and the shear deformations are neglected. At the supports the warping is not constrained. SOFiSTiK 2014 | VERiFiCATiON MANUAL 81 BE16: Torsion due to Biaxial Bending Table 19.1: Model Properties Material Properties Geometric Properties Loading S 355 = 5 m P z = 20 kN γ M = 1.1 PE 300 [17] [22] P y = 4 kN C M = 125900 cm 6 The results are presented in Table 19.2. It has to be noted that the reference results are according to [19] and [21], where they are computed with another finite element software, and not with respect to an analytical solution. Table 19.2: Results max | | C M = 125900 [cm 6 ] Ref. [19] C M = 0 Ref. [19] ϕ [rd] 0.0316 0.0315 0.0321 0.0321 M [kNm] 0.183 0.185 0.185 0.189 M y [kNm] 24.88 24.9 24.85 24.9 M z [kNm] 5.57 5.6 5.71 5.7 19.4 Conclusion This example presents a case where torsion is induced to the system because of biaxial bending. It has been shown that the behaviour of the beam is captured accurately. 19.5 Literature [17] R. Kindmann, M. Kraus, and H. J. Niebuhr. Stahlbau Kompakt, Bemessungshilfen, Profiltabellen. Verlag Stahleisen, 2006. [19] V. Gensichen and G. Lumpe. Zur Leistungsf ¨ ahigkeit, korrekten Anwendung und Kontrolle r ¨ aumlicher Stabwerksprogramme. Stahlbau Seminar 07. [21] V. Gensichen and G. Lumpe. “Zur Leistungsf ¨ ahigkeit, korrekten Anwendung und Kontrolle von EDV-Programmen f ¨ ur die Berechnung r ¨ aumlicher Stabwerke im Stahlbau”. In: Stahlbau 77 (Teil 2) (2008). [22] R. Kindmann. “Neue Berechnungsformel f ¨ ur das IT von Walzprofilen und Berechnung der Schub- spannungen”. In: Stahlbau 75 (2006). 82 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE17: Lateral Torsional Buckling 20 BE17: Lateral Torsional Buckling Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): Topic(s): Module(s): ASE Input file(s): lateral torsional buckling.dat 20.1 Problem Description The problem consists of a single span beam with an initial geometrical imperfection at the middle, sub- jected to a uniformly distributed load q z , as shown in Fig. 20.1. The structure is examined for lateral torsional buckling. | ←− −→| y z Δy Total System Δy Equivalent System | ←− / 2 −→| q z Model z p = −20 cm Figure 20.1: Problem Description 20.2 Reference Solution The I-beam of Fig. 20.1 has an initial geometrical imperfection in the y-direction Δy = / 200 = 3.0 cm. Using the symmetry of the equivalent system the model can be reduced to half as shown at Fig. 20.1. Due to the bending moments, the load application on the upper flange of the beam (z p ) and the imper- fection, the beam is at risk for lateral torsional buckling. In order to account for this effect, third order theory has to be utilised. SOFiSTiK 2014 | VERiFiCATiON MANUAL 83 BE17: Lateral Torsional Buckling 20.3 Model and Results The properties of the model [19] are defined in Table 20.1. A standard steel material is used as well as a standard rolled steel profile with properties according to DIN 1025-5. A safety factor γ M = 1.1 is used, which according to DIN 18800-2 it is applied both to the yield strength and the stiffness. The loading is applied at the upper flange as shown in Fig. 20.1. Furthermore, the self weight and the shear deformations are neglected. At the supports the warping is not constrained. Table 20.1: Model Properties Material Properties Geometric Properties Loading S 355 , γ M = 1.1 = 6 m , Δy = 3 cm q z = 10 kN/ m C M = 490000 cm 6 PE 400 [15] z p = −20 cm The results are presented in Table 20.2. It is observed that second-order theory (TH. II) fails to capture the moments with respect to the z-axis, therefore third-order theory (TH. III) has to be used. It has to be noted that the reference results are according to [19], where they are computed with another finite element software, and not with respect to an analytical solution. Table 20.2: Results C M = 490000 [cm 6 ] Ref. [19] C M = 0 Ref. [19] TH. II TH. III TH. II TH. III y [cm] 0.094 0.082 0.089 0.184 0.158 0.172 z [cm] 0.422 0.425 0.424 0.470 0.479 0.475 ph [rd] 0.0167 0.0166 0.0167 0.0367 0.0363 0.0365 M [kNm] 0.439 0.437 0.438 0.510 0.504 0.508 M y [kNm] 45.0 45.0 45.0 45.0 45.0 45.0 M z [kNm] 0.001 0.747 0.752 0.001 1.627 1.641 M ω [kNm] 0.606 0.605 0.607 0.0 0.0 0.0 20.4 Conclusion This example examines the lateral torsional buckling of beams. It has been shown that the behaviour of the beam is captured accurately. 20.5 Literature [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. [19] V. Gensichen and G. Lumpe. Zur Leistungsf ¨ ahigkeit, korrekten Anwendung und Kontrolle r ¨ aumlicher Stabwerksprogramme. Stahlbau Seminar 07. 84 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE18: Three-storey Column under Large Compressive Force and Torsional Moment 21 BE18: Three-storey Column under Large Compres- sive Force and Torsional Moment Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): Topic(s): Module(s): ASE, DYNA Input file(s): three storey column.dat 21.1 Problem Description The problem consists of a three-storey column, subjected to a large compressive axial force N and a torsional moment M t at the middle, as shown in Fig. 21.1. The rotation and twisting as well as the torsional moments of the structure are determined. / 2 M t N / 2 / 2 Figure 21.1: Problem Description 21.2 Reference Solution A large axial compressive force is applied to the column of Fig. 21.1, in combination with a torsional moment at the middle, which can cause warping and potentially buckling of the structure. In order to account for this effect, second order theory has to be utilised. The total torsional moment M T is given as a sum of the different torsional parts, the primary, secondary and third respectively: M = M T = M T1 + M T2 + M T3 , (21.1) SOFiSTiK 2014 | VERiFiCATiON MANUAL 85 BE18: Three-storey Column under Large Compressive Force and Torsional Moment where M T1 = G T ϕ , (21.2) M T2 = − E C M ϕ , (21.3) M T3 = N 2 p ϕ , (21.4) and the warping moment M ω = − E C M ϕ . (21.5) where G is the shear modulus, T the torsional moment of inertia, p the polar radius of gyration and EC M the warping torsion stiffness. Introducing the above into Eq. 21.1 we have: _ G T + N 2 p _ ϕ − E C M ϕ = M T = M . (21.6) 21.3 Model and Results The properties of the model [23] are defined in Table 21.1. A standard steel material is used and an I-beam profile for the cross-section. A safety factor γ M = 1.1 is used, which according to DIN 18800-2 it is applied both to the yield strength and the stiffness. At the supports the warping is not constrained. The cross-sectional properties given in Table 21.1 are the values calculated by SOFiSTiK, matching the analytical solution, except from the torsional moment T and the warping modulus C M which are modified to match the values of the reference example. This modification is done only for the sake of comparison and it has to be noted that the reference results [23] are computed with another finite element software, and not with respect to an analytical solution. Table 21.1: Model Properties Material Properties Geometric Properties Loading γ M = 1.1 b = 180 mm N = 1712 kN = 6 m h = 400 mm M t = 272 kNcm S 355 t eb = 10 mm t ƒ nge = 14 mm C M = 506884 cm 6 y = 23071.6 cm 4 z = 1363.9 cm 4 T = 44.18 cm 4 86 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE18: Three-storey Column under Large Compressive Force and Torsional Moment The results are presented in Table 21.2 and Fig 21.2. The value of M T3 is not given in the Reference [23], but according to Eq. 21.4 is computed as −721 kNcm, which matches the calculated value by SOFiSTiK. If we now sum the torsional moment parts, it is observed that Eq. 21.1 is satisfied and that the total torsional moment at = 0 is 136 kNcm. Table 21.2: Results SOF. Ref.[23] ϕ [mrd] ( = / 2) 294.4 294 ϕ [mrd/ cm] ( = 0; = ) 1.5096 1.50965 M T1 [kNcm] ( = 0) 491 491 M T2 [kNcm] ( = 0) 366 366 M T3 [kNcm] ( = 0) −721 - M T [kNcm] ( = 0) 136 136 M ω [kNcm] 85638 85619 M T1 [kNm] M T2 [kNm] M T [kNm] Figure 21.2: Results [kNm] 21.4 Conclusion This example examines the torsional behaviour of the beam and the different parts involved in the cal- culation of the total torsional moment. The results are reproduced accurately. 21.5 Literature [23] V. Gensichen and G. Lumpe. Anmerkungen zur linearen und nichtlinearen Torsionstheorie im Stahlbau. Stahlbau Seminar 2012. SOFiSTiK 2014 | VERiFiCATiON MANUAL 87 BE18: Three-storey Column under Large Compressive Force and Torsional Moment 88 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE19: Two-span Beam with Warping Torsion and Compressive Force 22 BE19: Two-span Beam with Warping Torsion and Compressive Force Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): Topic(s): Module(s): ASE, DYNA Input file(s): two span beam.dat 22.1 Problem Description The problem consists of a two-span beam, subjected to a large compressive axial force N 2 at its right end node, as well as a torsional moment M t at the middle and an additional axial force N 1 in the middle of the right span, as shown in Fig. 22.1. The structure is examined for its torsional and warping behaviour. / 2 1090 1 2 3 4 / 4 / 4 N 1 N 2 M T Figure 22.1: Problem Description 22.2 Reference Solution While in first order theory, the axial force has no effect in the torsional deformations and moments, in second order torsional theory, the influence of the axial force in the rotation and twisting is considered. From the formulation of the equilibrium conditions at the twisted element, the torsional moment part M T3 results, which covers the contribution of the axial force in the total torsional moment. Therefore second order theory is utilised here, in order to account for the torsional effect of the axial force, as well as the warping torsion arising from the application of the torsional moment and the axial force at the intermediary nodes of the beam. The total torsional moment M T is given as a sum of the different torsional parts, the primary, secondary and third respectively: M = M T = M T1 + M T2 + M T3 , (22.1) where M T1 = G T ϕ , (22.2) M T2 = − E C M ϕ , (22.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 89 BE19: Two-span Beam with Warping Torsion and Compressive Force M T3 = N 2 p ϕ , (22.4) and the warping moment M ω = − E C M ϕ . (22.5) where G is the shear modulus, T the torsional moment of inertia, p the polar radius of gyration and EC M the warping torsion stiffness. Introducing the above into Eq. 22.1 we have: _ G T + N 2 p _ ϕ − E C M ϕ = M T = M . (22.6) 22.3 Model and Results The properties of the model [23] are defined in Table 22.1. A standard steel material is used and an I-beam profile for the cross-section. A safety factor γ M = 1.1 is used, which according to DIN 18800-2 it is applied both to the yield strength and the stiffness. At the supports the warping is not constrained. The cross-sectional properties, given in Table 22.1, are the values calculated by SOFiSTiK, matching the reference solution, except from the torsional moment T and the warping modulus C M , which are modified to match the values of the reference example. This modification is done only for the sake of comparison and it has to be noted that the reference results [23] are computed with another finite element software, and not with respect to an analytical solution. The results are presented in Table 22.2, 22.3 and Fig 22.2. The double result values given for some nodes, e.g. 309/308, indicate the value left and right of the node respectively, and the exact result lies in between. When ’—’ is used, it indicates a change in the moment diagram. Table 22.1: Model Properties Material Properties Geometric Properties Loading γ M = 1.1 b = 180 mm , h = 400 mm N 1 = 200 kN = 6 m t eb = 10 mm , t ƒ nge = 14 mm N 2 = 1600 kN S 355 y = 23071.6cm 4 , z = 1363.9cm 4 M t = 280 kNcm C M = 506900 cm 6 T = 45.00 cm 4 Table 22.2: Torsional Deformation Results Node 1 Node 2 SOF. Ref.[23] SOF. Ref.[23] ϕ [mrd] - - 294 294 ϕ [rd/ cm] 1.525 1.52 - - 90 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE19: Two-span Beam with Warping Torsion and Compressive Force Table 22.3: Torsional Moment Results Node M T [kNcm] M T1 [kNcm] M T2 [kNcm] M T3 [kNcm] M ω [kNm 2 ] 1 SOF. 121 505 382 −766 0 Ref. [23] 121 505 382 −766 0 1090 SOF. 121 363 309/ 308 −551/ − 550 5.35 Ref. [23] 121 363 308 −550 5.35 2 SOF. 121| − 159 −9 118| − 165 12/ 15 8.65 Ref. [23] 121| − 159 −9 117| − 163 14 8.65 3 SOF. −159 −364 −345| − 285 550|490 4.70 Ref. [23] −159 −363 −346| − 285 551|490 4.70 4 SOF. −159 −487 −328 656 0 Ref. [23] −159 −487 −328 656 0 M ω [kNm 2 ] M T1 [kNm] M T2 [kNm] M T [kNm] Figure 22.2: Results In reference [23], except from the second order theory, the example is also analysed with respect to geometrically nonlinear torsional theory which accounts additionally for the large torsional deformations. This is done by introducing an additional torsional moment part, the helix torsional moment M TH . The SOFiSTiK 2014 | VERiFiCATiON MANUAL 91 BE19: Two-span Beam with Warping Torsion and Compressive Force results of both analysis are compared, leading to the conclusion that second order theory lies almost always to the safe side. 22.4 Conclusion This example examines the torsional behaviour of the beam and the different parts involved in the cal- culation of the total torsional moment. The results are reproduced accurately. 22.5 Literature [23] V. Gensichen and G. Lumpe. Anmerkungen zur linearen und nichtlinearen Torsionstheorie im Stahlbau. Stahlbau Seminar 2012. 92 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE20: Passive Earth Pressure I 23 BE20: Passive Earth Pressure I Overview Element Type(s): C2D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): SOIL Module(s): TALPA Input file(s): passive earth pressure.dat 23.1 Problem Description The problem consists of a soil mass retained by a wall as shown in Fig. 23.1. The horizontal passive earth pressure is determined and is compared to the value obtained for the case of the soil mass externally forced to its limiting strength. B Soil H W a l l Figure 23.1: Problem Description 23.2 Reference Solution When a retaining wall is forced against a soil mass, lateral passive earth pressure is exerted from the soil to the wall. In order to describe the horizontal component of the pressure the soil will exert, an earth pressure coefficient K ph according to Coulomb theory is used: K ph = cos 2 (ϕ − α) _ 1 − _ sn(ϕ+δ p) · sn(ϕ+β) cos(α+δ p) · cos(α+β) _ 2 cos 2 α , (23.1) where the parameters α, ϕ, δ p and β are defined in Fig. 23.2. The wall friction angle is denoted by δ p and the soil friction angle by ϕ. The horizontal passive earth pressure resultant is [15]: E ph = 1 2 γ H 2 K ph . (23.2) In order to account for the development of irreversible strains in the soil, under the action of the passive load, a plasticity model has to be utilised. Whether plasticity occurs in a calculation, can be evaluated with a yield function ƒ , where the condition ƒ = 0 stands for the plastic yielding. This condition can be represented as a surface in principal stress space. In this Benchmark, the Mohr-Coulomb model is SOFiSTiK 2014 | VERiFiCATiON MANUAL 93 BE20: Passive Earth Pressure I adopted, which represents an elastic perfectly-plastic behaviour. A perfectly-plastic model corresponds to a fixed yield surface, i.e. a yield surface that is fully defined by model parameters and is not affected by plastic straining. Moreover, for stress state within the yield surface, the behaviour is purely elastic and all strains are reversible. Hence, the Mohr-Coulomb model requires the input of a total of five parameters, the Young’s modulus E and Poisson’s ratio ν for the definition of the elasticity, and three for the plasticity, the friction angle ϕ, the cohesion c and the dilatancy angle ψ. The dilatancy angle is involved in the plastic potential function and controls the evolution of plastic volumetric strain increments [24]. ƒ = σ 1 − 1 − snϕ 1 + snϕ · σ 3 − 2c cosϕ 1 + snϕ , (23.3) Q β H ϕ α δ p E p ϑ p Figure 23.2: Passive Earth Pressure by Coulomb The yield function for the Mohr-Coulomb model [24] is defined by Eq. 23.3, where σ 1 and σ 3 are the principal stresses, and its yield surface is shown in Fig. 23.3. −σ 1 −σ 3 −σ 2 Figure 23.3: Mohr-Coulomb Yield Surface in Principal Stress Space 23.3 Model and Results The properties of the model are defined in Table 23.1. The Mohr-Coulomb plasticity model is used for the modelling of the soil behaviour. The load is defined as a unit support displacement in the -direction and is increased gradually until a limit value. It is applied at node 405, which is kinematically coupled with the wall nodes as shown in Fig. 23.4, and therefore corresponds to a uniformly applied load at the wall nodes. Maximum displacement is recorded for each loading increment, and the curve of horizontal passive earth pressure-displacement (Fig. 23.5) is plotted against the reference solution according to Coulomb theory. 94 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE20: Passive Earth Pressure I 405 Figure 23.4: Finite Element Model Table 23.1: Model Properties Material Properties Geometric Properties Loading Wall Soil Wall Soil E = 30000 MP E = 300 MP B = 0.1 m B = 30 m W = 1 mm ν = 0.18 ν = 0.20 H = 0.8 m H = 6 m γ = 24 kN/ m 3 γ = 19 kN/ m 3 c = 1 kN/ m 2 ϕ = 38 ◦ ψ = 6 ◦ δ p = ϕ / 3 0 10 20 30 40 50 60 70 80 90 100 0 500 1,000 1,500 2,000 2,500 Wall displacement [mm] E p h [ k N / m ] Numerical evolution Theoretical value Figure 23.5: Horizontal Passive Earth Pressure-Displacement Curve SOFiSTiK 2014 | VERiFiCATiON MANUAL 95 BE20: Passive Earth Pressure I 23.4 Conclusion This example examines the horizontal passive earth pressure determination for a soil mass retained by a wall. The Mohr-Coulomb model for the definition of the soil material behaviour is adopted. It has been shown that the behaviour of the soil is captured accurately. 23.5 Literature [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. [24] AQUA Materials and Cross Sections. SOFiSTiK AG. 2012. 96 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE21: Passive Earth Pressure II 24 BE21: Passive Earth Pressure II Overview Element Type(s): C2D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): SOIL Module(s): TALPA Input file(s): passive earth pressure harden.dat 24.1 Problem Description The model of Benchmark 20 is here extended for the case of a soil material described by the hardening plasticity soil model. The problem consists of a soil mass retained by a wall as shown in Fig. 24.1. The horizontal passive earth pressure is determined and is compared to the value obtained for the case of the soil mass externally forced to its limiting strength. B Soil H W a l l Figure 24.1: Problem Description 24.2 Reference Solution When a retaining wall is forced against a soil mass, lateral passive earth pressure is exerted from the soil to the wall. In order to describe the horizontal component of the pressure the soil will exert, an earth pressure coefficient K ph according to Coulomb theory is used: K ph = cos 2 (ϕ − α) _ 1 − _ sn(ϕ+δ p) · sn(ϕ+β) cos(α+δ p) · cos(α+β) _ 2 cos 2 α , (24.1) where the parameters α, ϕ, δ p and β are defined in Fig. 24.2. The wall friction angle is denoted by δ p and the soil friction angle by ϕ. The horizontal passive earth pressure resultant is [15]: E ph = 1 2 γ H 2 K ph . (24.2) In order to account for the development of irreversible strains in the soil, under the action of the passive load, a plasticity model has to be used. In this Benchmark the hardening plasticity soil model is adopted, which is an extended elastoplastic material with an optimized hardening rule [24]. In contrast to the Mohr- Coulomb model (Be. 20), which is an elastic-perfectly-plastic model, the yield surface of a hardening SOFiSTiK 2014 | VERiFiCATiON MANUAL 97 BE21: Passive Earth Pressure II plasticity model is not fixed but it can expand due to plastic straining. Its hardening rule is based on a hyperbolic stress-strain relationship, derived from triaxial testing. Hardening is limited by the material ’s strength, represented by the classic Mohr-Coulomb failure criterion. Additionally, the model accounts for the stress dependent stiffness, it captures the loading state and can therefore account for the different stiffness in primary loading and un-/reloading paths. The important features of the model are [24]: • the deviatoric hardening based on hyperbolic stress-strain relationship: input parameter E 50,reƒ , R ƒ • the Mohr-Coulomb failure criterion: input parameter ϕ, c, ψ • the stress dependent stiffness: input parameter m, P reƒ • the loading dependent stiffness: input parameter μ, E r • the optional limitation of tensile stress: input parameter ƒ t • the modelling of the contractant behaviour and stiffness during primary compression (oedometric testing): input parameter E s,reƒ • the preservation of a realistic stress ratio: input parameter k 0 Q β H ϕ α δ p E p ϑ p Figure 24.2: Passive Earth Pressure by Coulomb The yield surface (Fig. 24.3) for the hardening plasticity model is bounded by the Mohr-Coulomb failure criterion, while the oedometric properties create a cap yield surface, closing the elastic region in the direction of the p-axis. −σ 1 −σ 3 −σ 2 Mohr-Coulomb yield surface p σ 1 − σ 3 M o h r - C o u l o m b f a i l u r e l i n e Figure 24.3: Yield Surface Properties 98 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE21: Passive Earth Pressure II 24.3 Model and Results The properties of the model are defined in Table 24.1. The hardening plasticity model (GRAN) is used for the modelling of the soil behaviour in order for a more realistic representation in comparison to the elastic-perfectly-plastic Mohr-Coulomb model (Benchmark 20). The load is defined as a unit support displacement in the -direction and is increased gradually until a limit value. It is applied at node 405, which is kinematically coupled with the wall nodes as shown in Fig. 24.4, and therefore corresponds to a uniformly applied load at the wall nodes. Maximum displacement is recorded for each loading increment, and the curve of horizontal passive earth pressure-displacement (Fig. 24.5) is plotted against the reference solution according to Coulomb theory. 405 Figure 24.4: Finite Element Model Table 24.1: Model Properties Material Properties Geometric Properties Loading Wall Soil Wall Soil E = 30000 MP E = 300 MP B = 0.1 m B = 30 m W = 1 mm ν = 0.18 μ = 0.20 H = 0.8 m H = 6 m γ = 24 kN/ m 3 γ = 19 kN/ m 3 c = 1 kN/ m 2 ϕ = 38 ◦ ψ = 6 ◦ E s,reƒ = 75 MP E 50,reƒ = 75 MP m = 0.55 R ƒ = 0.9 P reƒ = 0.1 MP δ p = ϕ / 3, γ boyncy = 9 kN/ m 3 From the comparison of the curves with respect to the two different plasticity models and the refer- ence solution, it can be observed that both approach the limit value accurately. Their basic difference lies on the accounting of the hardening effect, a more realistic approach, which corresponds to higher deformations for the limit value, as it can be observed by the hardening plasticity curve in Fig. 24.5. SOFiSTiK 2014 | VERiFiCATiON MANUAL 99 BE21: Passive Earth Pressure II 0 10 20 30 40 50 60 70 80 90 100 0 500 1,000 1,500 2,000 2,500 Wall displacement [mm] E p h [ k N / m ] Mohr-Coulomb plasticity model Hardening plasticity model Theoretical value Figure 24.5: Horizontal Passive Earth Pressure-Displacement Curve Hardening plasticity model Mohr-Coulomb plasticity model Figure 24.6: Nodal Displacement for End Load in y-direction 24.4 Conclusion This example examines the horizontal passive earth pressure determination for a soil mass retained by a wall. The hardening plasticity model for the definition of the soil material behaviour is adopted and compared to the Mohr-Coulomb model. It has been shown that the behaviour of the soil is captured accurately. 100 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE21: Passive Earth Pressure II 24.5 Literature [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. [24] AQUA Materials and Cross Sections. SOFiSTiK AG. 2012. SOFiSTiK 2014 | VERiFiCATiON MANUAL 101 BE21: Passive Earth Pressure II 102 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE22: Tunneling - Ground Reaction Line 25 BE22: Tunneling - Ground Reaction Line Overview Element Type(s): C2D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): SOIL Module(s): TALPA Input file(s): groundline hoek.dat 25.1 Problem Description This problem consists of a cylindrical hole in an infinite medium, subjected to a hydrostatic in-situ state, as shown in Fig. 25.1. The material is assumed to be linearly elastic-perfectly plastic with a failure surface defined by the Mohr-Coulomb criterion and with zero volume change during plastic flow. The calculation of the ground reaction line is performed and compared to the analytical solution according to Hoek [25] [26]. p o p o p Figure 25.1: Problem Description 25.2 Reference Solution The stability of deep underground excavations depends upon the strength of the rock mass surrounding the excavations and upon the stresses induced in this rock. These induced stresses are a function of the shape of the excavations and the in-situ stresses which existed before the creation of the excavations [25]. When tunnelling in rock, it should be examined how the rock mass, surrounding the tunnel, deforms and how the support system acts to control this deformation. In order to explore this effect, an analytical solution for a circular tunnel will be utilised, which is based on the assumption of a hydrostatic in-situ state. Furthermore, the surrounding rock mass is assumed to follow an elastic-perfectly-plastic material behaviour with zero volume change during plastic flow. Therefore the Mohr-Coulomb failure criterion is adopted, in order to model the progressive plastic failure of the rock mass surrounding the tunnel. The onset of plastic failure, is thus expressed as: SOFiSTiK 2014 | VERiFiCATiON MANUAL 103 BE22: Tunneling - Ground Reaction Line σ 1 = σ cm + kσ 3 , (25.1) where σ 1 is the axial stress where failure occurs, σ 3 the confining stress and σ cm the uniaxial compres- sive strength of the rock mass defined by: σ cm = 2c cosϕ 1 − snϕ . (25.2) The parameters c and ϕ correspond to the cohesion and angle of friction of the rock mass, respectively. The tunnel behaviour on the other hand, is evaluated in terms of the internal support pressure. A circular tunnel of radius r o subjected to hydrostatic stresses p o and a uniform internal support pressure p , as shown in Fig. 25.2, is assumed. p o p o p r p r Figure 25.2: Plastic zone surrounding a circular tunnel As a measure of failure, the critical support pressure p cr is defined: p cr = 2p o − σ cm 1 + k , (25.3) where k is the coefficient of passive earth pressure defined by: k = 1 + snϕ 1 − snϕ . (25.4) If the internal support pressure p is greater than p cr , the behaviour of the surrounding rock mass remains elastic and the inward elastic displacement of the tunnel wall is: e = r o (1 + ν) E (p o − p ), (25.5) where E is the Young’s modulus and ν the Poisson’s ratio. If p is less than p cr , failure occurs and the total inward radial displacement of the walls of the tunnel becomes: 104 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE22: Tunneling - Ground Reaction Line p = r o (1 + ν) E _ 2(1 + ν) (p o − p cr ) _ r p r o _ 2 − (1 − 2ν) (p o − p ) _ , (25.6) and the plastic zone around the tunnel forms with a radius r p defined by: r p = r o _ 2(p o (k − 1) + σ cm ) (1 + k) ((k − 1) p + σ cm ) _ 1 (k−1) (25.7) 25.3 Model and Results The properties of the model are defined in Table 25.1. The Mohr-Coulomb plasticity model is used for the modelling of the rock behaviour. The load is defined as a unit supporting pressure, uniform along the whole line of the circular hole, following the real curved geometry. The ground reaction line is calculated, which depicts the inward oriented deformation along the circumference of the opening that is to be expected in dependence of the acting support pressure. Figure 25.3: Finite Element Model Table 25.1: Model Properties Material Properties Geometric Properties Pressure Properties E = 5000000 kN/ m 2 r o = 3.3 m P o = 29700 kN/ m 2 ν = 0.2 P m = 7000 kN/ m 2 γ = 27 kN/ m 3 P cr = 8133.744 kN/ m 2 γ boyncy = 17 kN/ m 3 ϕ = 39 ◦ , ψ = 0 ◦ c = 3700 kN/ m 2 k = 4.395 SOFiSTiK 2014 | VERiFiCATiON MANUAL 105 BE22: Tunneling - Ground Reaction Line The uniaxial compressive stress of the rock mass σ cm is calculated at 15514.423 kN/ m 2 and the critical pressure p c r is 8133.744 kN/ m 2 . The ground reaction line is presented in Fig. 25.4, as the curve of the inward radial displacement over the acting support pressure. It can be observed that the calculated values are in agreement with the analytical solution according to Hoek. 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0 5,000 10,000 15,000 20,000 25,000 30,000 Displacement [m] S u p p o r t P r e s s u r e [ k N / m 2 ] Hoek elastic-plastic Linear Numerical Figure 25.4: Ground Reaction Line 25.4 Conclusion This example examines the tunnel deformation behaviour with respect to the acting support pressure. It has been shown that the behaviour of the tunnel in rock is captured accurately. 25.5 Literature [25] E. Hoek. Practical Rock Engineering. 2006. [26] E. Hoek, P.K. Kaiser, and W.F. Bawden. Support of Underground Excavations in Hard Rock. 1993. 106 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE23: Undamped Free Vibration of a SDOF System 26 BE23: Undamped Free Vibration of a SDOF System Overview Element Type(s): SPRI Analysis Type(s): DYN Procedure(s): TSTP Topic(s): Module(s): DYNA Input file(s): undamped sdof.dat 26.1 Problem Description This problem consists of an undamped linearly elastic SDOF system undergoing free vibrations, as shown in Fig. 26.1. The response of the system is determined and compared to the exact reference solution. (0) m k Figure 26.1: Problem Description 26.2 Reference Solution The essential physical properties of a linearly elastic structural system subjected to an external excitation or dynamic loading are its mass, stiffness and damping. In the simplest model of a SDOF system, as shown in Fig. 26.2 in its idealized form, these properties are concentrated in a single physical element. For this system the elastic resistance to displacement is provided by the spring of stiffness k, while the energy-loss mechanism by the damper c. The mass m is included in the rigid body, which is is able to move only in simple translation, and thus the single displacement coordinate (t) completely describes its position [27]. (t) p(t) m c k Figure 26.2: Problem Description SOFiSTiK 2014 | VERiFiCATiON MANUAL 107 BE23: Undamped Free Vibration of a SDOF System The motion of a linear SDOF system, subjected to an external force p(t) is governed by [27] [28]: m¨ + c˙ + k = p(t) (26.1) Setting p(t) = 0, gives the differential equation governing the free vibration of the system m¨ + c˙ + k = 0 (26.2) For a system without damping (c =0), Eq. 26.2 specialises to m¨ + k = 0 (26.3) Free vibration is initiated by disturbing the system from its static equilibrium position by imparting the mass some displacement (0) and/or velocity ( ˙ 0) at time zero. Subjected to these initial conditions, the solution to the homogeneous differential equation of motion is: (t) = (0) cos (ω n t) + ˙ (0) ω n sn(ω n t) (26.4) where ω n = _ _ _ k m (26.5) represents the natural circular frequency of vibration and ƒ the natural cyclic frequency of vibration ƒ n = ω n 2π (26.6) The period T represents the time required for the undamped system to complete one cycle of free vibration and is given by T n = 2π ω n = 1 ƒ n (26.7) 26.3 Model and Results The properties of the model are defined in Table 26.1. The system is initially disturbed from its static equilibrium position by a displacement of 20 mm and is then let to vibrate freely. Eq. 26.4 is plotted in Fig. 26.4, presenting that the system undergoes vibration motion about its undeformed ( = 0) position, and that this motion repeats itself every 2π/ ω n seconds. The exact solution is compared to the calculated time history of the displacement of the SDOF system for different time integration methods. The time step is taken equal to 0.02 sec corresponding to a dt/ T ratio of 1/ 50. 108 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE23: Undamped Free Vibration of a SDOF System Figure 26.3: Finite Element Model Table 26.1: Model Properties Model Properties Excitation Properties m = 1 t (0) = 20 mm k = 4π 2 kN/ m ˙ (0) = 0 T = 1 sec 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 −20 −15 −10 −5 0 5 10 15 20 Time [sec] D i s p l a c e m e n t [ m m ] Exact Newmark Wilson Hughes Alpha Figure 26.4: Undamped Free Vibration Response From the results presented in Table 26.2, we observe that the response computed by the examined integration schemes is in a good agreement with the exact solution. Table 26.2: Results Integration method Newmark Wilson Hughes Alpha Ref. m [mm] 19.949 19.963 19.956 20.000 SOFiSTiK 2014 | VERiFiCATiON MANUAL 109 BE23: Undamped Free Vibration of a SDOF System 26.4 Conclusion This example examines the response of a linear elastic undamped SDOF system undergoing free vibra- tion. It has been shown that the behaviour of the system is captured adequately. 26.5 Literature [27] R. W. Clough and J. Penzien. Dynamics of Structures. 3rd. Computers & Structures, Inc., 2003. [28] A. K. Chopra. Dynamics of Structures: Theory and Applications to Earthquake Engineering. Pren- tice Hall, 1995. 110 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE24: Free Vibration of a Under-critically Damped SDOF System 27 BE24: Free Vibration of a Under-critically Damped S- DOF System Overview Element Type(s): SPRI, DAMP Analysis Type(s): DYN Procedure(s): TSTP Topic(s): Module(s): DYNA Input file(s): damped sdof.dat 27.1 Problem Description This problem consists of an under-critically damped linearly elastic SDOF system undergoing free vi- brations, as shown in Fig. 27.1. The response of the system is determined and compared to the exact reference solution. (0) m c k Figure 27.1: Problem Description 27.2 Reference Solution The differential equation governing the free vibration of a linear elastic damped SDOF system, as shown in Fig. 27.1 is given by [27] [28]: m ¨ + c ˙ + k = 0 (27.1) where c is the linear viscous damping, k the linear spring stiffness and m the mass of the system. Dividing Eq. 27.1 by m gives ¨ + 2 ξ ω n ˙ + ω 2 n = 0 (27.2) where ω n = _ k/ m as defined in Benchmark 23 and ξ represents the damping ratio ξ = c 2 mω n = c c cr (27.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 111 BE24: Free Vibration of a Under-critically Damped SDOF System The parameter c cr is called the critical damping coefficient (Eq. 27.4), because it is the smallest value of c that inhibits oscillation completely. If c ¡ c cr or ξ ¡ 1 the system is said to be under-critically damped and thus oscillates about its equilibrium position with a progressively decreasing amplitude [28]. c cr = 2 mω n = 2 _ k m = 2k ω n (27.4) Free vibration is initiated by disturbing the system from its static equilibrium position by imparting the mass some displacement (0) and/or velocity ( ˙ 0) at time 0. Subjected to these initial conditions, the solution to the homogeneous differential equation of motion is: (t) = e −ξ ω n t _ (0) cos (ω D t) + _ ˙ (0) + ξ ω n (0) ω D _ sn(ω D t) _ (27.5) where ω n represents the natural frequency of damped vibration and T D the natural period of damped vibration given by ω n = ω n _ 1 − ξ 2 (27.6) T d = 2π ω D = T n _ 1 − ξ 2 (27.7) Figure 27.2: Effects of Damping on Free Vibration The damped system oscillates with a displacement amplitude decaying exponentially with every cycle of vibration, as shown in Fig. 27.2. The envelope curves ±ρe −ξ ω n t touch the displacement curve at points slightly to the right of its peak values, where 112 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE24: Free Vibration of a Under-critically Damped SDOF System ρ = _ _ _ (0) 2 + _ ˙ (0) + ξ ω n (0) ω D _ 2 (27.8) 27.3 Model and Results The properties of the model are defined in Table 27.1. The system is initially disturbed from its static equilibrium position by a displacement of 20 mm and is then let to vibrate freely. Eq. 27.5 is plotted in Fig. 27.3 and is compared to the calculated time history of the displacement of the SDOF system for different time integration methods. The time step is taken equal to 0.02 sec corresponding to a dt/ T ratio of 1/ 50. From the curves, it is obvious that the examined integration schemes are in a good agreement with the exact solution. The damping of the SDOF system is represented in two ways, either by the spring element with a damping value in axial direction or with the damping element. The results obtained are exactly the same for both case. This can be visualised in the result files for the case of the Newmark integration scheme. Table 27.1: Model Properties Model Properties Excitation Properties m = 1 t (0) = 20 mm k = 4π 2 kN/ m ( ˙ 0) = 0 T = 1 sec ξ = 5 % 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 −20 −15 −10 −5 0 5 10 15 20 ρe −ξω n t Time [sec] D i s p l a c e m e n t [ m m ] Exact Newmark Wilson Hughes Alpha Figure 27.3: Damped Free Vibration Response SOFiSTiK 2014 | VERiFiCATiON MANUAL 113 BE24: Free Vibration of a Under-critically Damped SDOF System 27.4 Conclusion This example examines the response of a linear elastic under-critically damped SDOF system undergo- ing free vibration. It has been shown that the behaviour of the system is captured adequately. 27.5 Literature [27] R. W. Clough and J. Penzien. Dynamics of Structures. 3rd. Computers & Structures, Inc., 2003. [28] A. K. Chopra. Dynamics of Structures: Theory and Applications to Earthquake Engineering. Pren- tice Hall, 1995. 114 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE25: Eigenvalue Analysis of a Beam Under Various End Constraints 28 BE25: Eigenvalue Analysis of a Beam Under Various End Constraints Overview Element Type(s): B3D Analysis Type(s): DYN Procedure(s): EIGE Topic(s): Module(s): DYNA Input file(s): eigenvalue analysis.dat 28.1 Problem Description This problem consists of a beam with various end constraints, as shown in Fig. 28.1. The eigenfrequen- cies of the the system are determined and compared to the exact reference solution for each case. Figure 28.1: Problem Description 28.2 Reference Solution The general formula to determine the eigenfrequency of a standard Bernoulli beam for a linear elastic material is given by [15] [29] ƒ = λ 2 2π _ _ _ E μ 4 (28.1) where E the flexural rigidity of the beam, the length, μ = γ ∗A/ g the mass allocation and λ a factor depending on the end constraints. The values of λ for various cases are given in Table 28.1. In this example, we analyse four different cases of a beam structure: 1. simple cantilever 2. cantilever with simply supported end 3. simply supported 4. both ends fixed SOFiSTiK 2014 | VERiFiCATiON MANUAL 115 BE25: Eigenvalue Analysis of a Beam Under Various End Constraints Table 28.1: Constraints Factor End Constraints λ λ = 1.875 λ = 3.926 λ = π λ = 4.73 28.3 Model and Results The properties of the model are defined in Table 28.2 and the resulted eigenfrequencies are given in Table 28.3. For the eigenvalue analysis a consistent mass matrix formulation is used as well as a Bernoulli beam. The finite element model for all examined cases consists of ten beam elements. Table 28.2: Model Properties Material Properties Geometric Properties E = 200 MP h = 1 cm, b = 1 cm, = 1 m γ = 25 kN/ m 3 A = 1 cm 2 , = 0.1 cm 4 , μ = 0.025 t/ m Table 28.3: Results Eigenfrequency SOF. [Hz] Ref. [Hz] simple cantilever 0.457 0.457 cantilever with simply supported end 2.004 2.003 simply supported 1.283 1.283 both ends fixed 2.907 2.907 28.4 Conclusion The purpose of this example is to test the eigenvalue capability of the program w.r.t. different options. It has been shown that the eigenfrequencies for all beam systems are calculated accurately. 28.5 Literature [15] K. Holschemacher. Entwurfs- und Berechnungstafeln f ¨ ur Bauingenieure. 3rd. Bauwerk, 2007. [29] S. Timoshenko. Vibration Problems in Engineering. 2nd. D. Van Nostrand Co., Inc., 1937. 116 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE26: Response of a SDOF System to Harmonic Excitation 29 BE26: Response of a SDOF System to Harmonic Ex- citation Overview Element Type(s): SPRI, DAMP Analysis Type(s): DYN Procedure(s): TSTP Topic(s): Module(s): DYNA Input file(s): harmonic damped.dat, harmonic undamped.dat 29.1 Problem Description This problem consists of an elastic SDOF system undergoing forced vibration, as shown in Fig. 29.1. The response of an undamped and damped system is determined and compared to the reference solu- tion. (t) p(t) m c k Figure 29.1: Problem Description 29.2 Reference Solution A harmonic force is p(t) = p o sn ω p t, where p o is the amplitude value of the force and its frequency ω p is called the exciting frequency. The differential equation governing the forced harmonic vibration of a damped system is given by [27] [28]: m ¨ + c ˙ + k = p o sn ω p t (29.1) m ¨ + k = p o sn ω p t (29.2) For undamped systems it simplifies to Eq. 29.2. Subjected also to initial conditions, (0) and ( ˙ 0), the total solution to Eq. 29.2 is: (t) = (0) cos ω n t + _ ˙ (0) ω n − p o k ω p / ω n 1 − (ω p / ω n ) 2 _ sn ω n t . _¸ . trnsent + p o k 1 1 − (ω p / ω n ) 2 sn ω p t . _¸ . stedystte (29.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 117 BE26: Response of a SDOF System to Harmonic Excitation Eq. 29.3 shows, that u(t) contains two distinct vibration components, first the term sn ω p t gives a vibration at the exciting frequency and second the terms sn ω n t and cos ω n t give a vibration at the natural frequency of the system. The first term is the steady state vibration, corresponding to the applied force and the latter is the transient vibration, depending on the initial conditions. It exists even if the initial conditions vanish, in which case it becomes (t) = p o k 1 1 − (ω p / ω n ) 2 _ sn ω p t − ω p ω n sn ω n t _ (29.4) For the case of a damped SDOF system, the total solution is given by (t) = e −ξω n t [A cos ω D t + B sn ω D t] . _¸ . trnsent + C sn ω p t + Dcos ω p t . _¸ . stedystte (29.5) The coefficients C and D are determined from the particular solution of the differential equation of motion (Eq. 29.1), whereas A and B are determined in terms of the initial conditions. For the special case of zero initial conditions, the coefficients are given by C = p o k 1 − (ω p / ω n ) 2 [1 − (ω p / ω n ) 2 ] 2 + [2ξ (ω p / ω n )] 2 (29.6) D = p o k −2ξ (ω p / ω n ) [1 − (ω p / ω n ) 2 ] 2 + [2ξ (ω p / ω n )] 2 (29.7) A = −D (29.8) B = A ξ − C (ω p / ω n ) _ 1 − ξ 2 (29.9) For the special case where the exciting frequency equals the natural frequency of the SDOF system, we observe the resonant response. For the undamped system, the steady state response amplitude tends towards infinity as we approach unity and the peak values build up linearly, as shown in Fig. 29.2. For the damped case though, they build up in accordance to ( st / 2ξ)e −ξω n t and towards a steady state level, as shown in Fig. 29.2. The static deformation st = p o / k, corresponds to the displacement which would be produced by the load p o if applied statically, and serves as a measure of amplitude. Undamped system Damped system t t Figure 29.2: Response to Resonant Loading for at-rest initial conditions 118 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE26: Response of a SDOF System to Harmonic Excitation 29.3 Model and Results The properties of the model are defined in Table 29.1. The system is excited by a harmonic sinusoidal force and undergoes a forced vibration with zero initial conditions. The cases of the elastic damped and undamped SDOF system with a frequency ratio ω p / ω n = 2 are examined and their responses are compared to the exact solutions presented in Section 29.2. The resonance response is also examined for both systems, as shown in Fig. 29.4. Table 29.1: Model Properties Model Properties Excitation Properties m = 1 t (0) = 0 k = 4π 2 kN/ m ( ˙ 0) = 0 T = 1 sec p 0 = 10 kN ξ = 2 % ω p = 2 ω n (a) Damped system (b) Undamped system Figure 29.3: Response to Harmonic Loading for at-rest initial conditions and ratio ω p / ω n = 2: (a) ξ = 2%, (b) ξ = 0 SOFiSTiK 2014 | VERiFiCATiON MANUAL 119 BE26: Response of a SDOF System to Harmonic Excitation (a) Damped system (b) Undamped system Figure 29.4: Response to Resonant Loading (ω p / ω n = 1) for at-rest initial conditions: (a) ξ = 2%, (b) ξ = 0 29.4 Conclusion The purpose of this example is to test the calculation of the response of a dynamic system in terms of a harmonic loading function. It has been shown that the behaviour of the system is captured adequately. 29.5 Literature [27] R. W. Clough and J. Penzien. Dynamics of Structures. 3rd. Computers & Structures, Inc., 2003. [28] A. K. Chopra. Dynamics of Structures: Theory and Applications to Earthquake Engineering. Pren- tice Hall, 1995. 120 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE27: Response of a SDOF System to Impulsive Loading 30 BE27: Response of a SDOF System to Impulsive Loading Overview Element Type(s): SPRI Analysis Type(s): DYN Procedure(s): TSTP Topic(s): Module(s): DYNA Input file(s): impulse sine wave.dat, impulse rectangular.dat 30.1 Problem Description This problem consists of an elastic undamped SDOF system undergoing forced vibration (Fig. 30.1) due to an impulsive loading as the one shown in Fig. 30.2. The response of the system is determined and compared to the exact reference solution. (t) p(t) m c k Figure 30.1: Problem Description p(t) t Figure 30.2: Arbitrary Impulsive Loading 30.2 Reference Solution Another special case of dynamic loading of the SDOF system is the impulsive load. Such a load consists of a single principal impulse of arbitrary form, as illustrated in Fig. 30.2, and generally is of relatively short duration. Damping has much less importance in controlling the maximum response of a structure to impulsive loads than for periodic or harmonic loads because the maximum response to a particular impulsive load will be reached in a very short time, before the damping forces can absorb much energy SOFiSTiK 2014 | VERiFiCATiON MANUAL 121 BE27: Response of a SDOF System to Impulsive Loading from the structure [27]. Therefore the undamped response to impulsive loads will be considered in this Benchmark. p(t) p 0 t t 1 (a) Half sine wave p(t) p 0 t t 1 (b) Rectangular Figure 30.3: Examined Impulse Loading The response to an impulse loading is always divided into two phases, the first corresponds to the forced vibration phase in the interval during which the load acts and the second corresponds to the free vibration phase which follows. Let us consider the case, where the structure is subjected to a single half sine wave loading as shown in Fig. 30.3(a). Assuming that the system starts from rest, the undamped response ratio time history R(t) = (t)/ (p 0 / k), is given by the simple harmonic load expression R(t) = 1 1 − β 2 _ sn ω p t − β sn ω n t _ (30.1) where β =ω p / ω n , p o is the amplitude value of the force and ω p its frequency. Introducing the non dimensional time parameter α = t/ t 1 so that ω p t = πα and ω n t = πα/ β, we can rewrite the equation accordingly R(α) = 1 1 − β 2 _ sn πα − β sn πα β _ 0 ≤ α ≤ 1 (30.2) where t 1 the duration of the impulse and β ≡ T/ 2t 1 . This equation is valid only for phase correspond- ing to 0 ≤ α ≤ 1. While it is very important to understand the complete time history behaviour as shown in Fig. 30.4, the engineer is usually only interested in the maximum value of response as represented by Points a, b, c, d, and e. If a maximum value occurs in Phase , the value of α at which it occurs can be determined by differentiating Eq. 30.2 with respect to α and equating to zero d R(α) d α = 0 (30.3) solving for α yields the α values for the maxima α = 2βn β + 1 n = 0, 1, 2, ... 0 ≤ α ≤ 1 (30.4) For phase where t ≥ t 1 and the free vibration occurs, the value of α is not necessary and the maximum response is given by 122 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE27: Response of a SDOF System to Impulsive Loading R m = _ −2β 1 − β 2 _ cos π 2β α ≥ 1 (30.5) Accordingly for the case of a rectangular impulse loading Fig. 30.3(b), the general response ratio solution for at rest initial conditions and for phase is given by R(α) = 1 − cos 2π t 1 T α 0 ≤ α ≤ 1 (30.6) The maximum response ratio R m is given again in terms of α and can be determined in the same manner by differentiating Eq. 30.6 with respect to α and equating to zero, yielding α = βn n = 0, 1, 2, ... 0 ≤ α ≤ 1 (30.7) For phase , the maximum response of the free vibrating system is given by R m = 2 sn π t 1 T α ≥ 1 (30.8) Special attention has to be given in the case of β = 1 where the expression of the response ratio becomes indeterminate and the L’ Hospital’s rule has to be utilised. R(t) t Figure 30.4: Response Ratios due to Half Sine Pulse 30.3 Model and Results In the expressions derived before, the maximum response produced in an undamped SDOF structure by each type of impulsive loading depends only on the ratio of the impulse duration to the natural period of the structure t 1 / T. Thus, it is useful to plot the maximum value of response ratio Rm as a function of t 1 / T for various forms of impulsive loading. Such plots are commonly known as displacement-response spectra and are derived here, for two forms of loading, a rectangular and a half sine wave impulse. SOFiSTiK 2014 | VERiFiCATiON MANUAL 123 BE27: Response of a SDOF System to Impulsive Loading Generally plots like these can be used to predict with adequate accuracy the maximum effect to be expected from a given type of impulsive loading acting on a simple structure. The properties of the model are defined in Table 30.1. The resulting figures are presented in Fig. 30.5. Table 30.1: Model Properties Model Properties Excitation Properties m = 1 t (0) = 0 k = 4π 2 kN/ m ( ˙ 0) = 0 T = 1 sec p 0 = 10 kN 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0 0.5 1 1.5 2 Ratio t 1 / T = Impulse duration/Period M a x i m u m r e s p o n s e r a t i o R m Exact Half-sine-wave impulse Exact Rectangular impulse Figure 30.5: Displacement - Response Spectra for Two Types of Impulse 30.4 Conclusion The purpose of this example is to test the calculation of the response of a dynamic system in terms of an impulsive loading. It has been shown that the behaviour of the system is captured adequately. 30.5 Literature [27] R. W. Clough and J. Penzien. Dynamics of Structures. 3rd. Computers & Structures, Inc., 2003. 124 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE28: Cylindrical Hole in an Infinite Elastic Medium 31 BE28: Cylindrical Hole in an Infinite Elastic Medium Overview Element Type(s): C2D Analysis Type(s): STAT Procedure(s): Topic(s): SOIL Module(s): TALPA Input file(s): hole elastic.dat 31.1 Problem Description This problem consists of a cylindrical hole in an infinite elastic medium subjected to a constant in-situ state, as shown in Fig. 31.1. The material is assumed to be isotropic and elastic. The stresses and the displacements are verified. p o p o Figure 31.1: Problem Description 31.2 Reference Solution The problem of calculating the displacements and stresses outside a circular hole in an infinite elastic medium, with a uniform stress state far from the hole, was first solved by the German engineer Kirsch in 1898. It is a rather important topic due to the fact that most of the holes drilled through rock are of circular section. The classical Kirsch solution can be used to find the radial and tangential displacement fields and stress distributions, for a cylindrical hole in an infinite isotropic elastic medium under plane strain conditions. The stresses σ r and σ θ for a point at polar coordinates (r, θ) outside the cylindrical opening of radius α are given by [30]: σ r = p 1 + p 2 2 _ 1 − α 2 r 2 _ + p 1 − p 2 2 _ 1 − 4α 2 r 2 + 3α 4 r 4 _ cos 2θ (31.1) σ θ = p 1 + p 2 2 _ 1 + α 2 r 2 _ − p 1 − p 2 2 _ 1 + 3α 4 r 4 _ cos 2θ (31.2) The radial outward displacement r , assuming conditions of plane strain, is given by: SOFiSTiK 2014 | VERiFiCATiON MANUAL 125 BE28: Cylindrical Hole in an Infinite Elastic Medium r = p 1 + p 2 4G α 2 r + p 1 − p 2 4G α 2 r _ 4(1 − ν) − α 2 r 2 _ cos 2θ (31.3) where G is the shear modulus, ν the Poisson ratio and p 1 , p 2 , θ, r are defined in Fig. 31.2 σ r σ θ θ r θ r p 1 p 2 Figure 31.2: Cyllindrical Hole in an Infinite Elastic Medium 31.3 Model and Results The properties of the model are defined in Table 31.1. The radius of the hole is 1 m and is assumed to be small compared to the length of the cylinder, therefore 2D plane strain conditions are in effect. A fixed external boundary is located 29.7 m from the hole center. The model is presented in Fig. 31.3. The stresses and displacements are calculated and verified with respect to the formulas provided in Section 31.2. Figure 31.3: Finite Element Model Table 31.1: Model Properties Material Properties Geometric Properties Pressure Properties E = 6777.9 MP α = 1 m P o = 30 MP ν = 0.21 r bondry = 29.7 m 126 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE28: Cylindrical Hole in an Infinite Elastic Medium 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 10 20 30 40 50 60 Radial distance from center [m] S t r e s s [ M P ] Exact σ θ Stress σ θ Exact σ r Stress σ r Figure 31.4: Radial and Tangential Stresses for Cylindrical Hole in Infinite Elastic Medium 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 1 2 3 4 5 6 Radial distance from center [m] R a d i a l d i s p l a c e m e n t [ m m ] Exact r Displacement r Figure 31.5: Radial Displacement for Cylindrical Hole in Infinite Elastic Medium SOFiSTiK 2014 | VERiFiCATiON MANUAL 127 BE28: Cylindrical Hole in an Infinite Elastic Medium Figure 31.6: Total Displacement Distribution Figures 31.4 and 31.5 show the radial and tangential stress and the radial displacement respectively, along a line, lying on the X-axis. This line (cut) can be visualised in Fig. 31.6, where the radial displace- ment distribution is illustrated. The results are in very good agreement with the reference solution. 31.4 Conclusion This example verifies the deformation and stresses behaviour of a cylindrical hole in an infinite elastic medium. It has been shown that the behaviour of the model is captured accurately. 31.5 Literature [30] J. C. Jaeger, N. G. W. Cook, and R. W. Zimmerman. Fundamentals of Rock Mechanics. 4th. Blackwell Publishing, 2007. 128 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE29: Cylindrical Hole in an Infinite Mohr-Coulomb Medium 32 BE29: Cylindrical Hole in an Infinite Mohr-Coulomb Medium Overview Element Type(s): C2D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): SOIL Module(s): TALPA Input file(s): hole mohr.dat 32.1 Problem Description This problem verifies stresses for the case of a cylindrical hole in an infinite elastic-plastic medium subjected to a constant in-situ state, as shown in Fig. 32.1. The material is assumed to be linearly elastic and perfectly plastic with a failure surface defined by the Mohr-Coulomb criterion. The stresses and the displacements are verified. P o P o P Figure 32.1: Problem Description 32.2 Reference Solution Consider a hollow cylinder with inner radius and outer radius r, under plane strain conditions, with a uniform pressure applied to its outer surface. If this pressure is slowly increased from 0 to some value P o , at first the cylinder will everywhere be in the elastic zone. As Po increases further, the yielding will start, the yielded zone will grow radially outward, and the cylinder will consist of an inner annular region that has yielded and an outer annulus that is still in its elastic state [30]. A specialised problem is now the calculation of the stresses outside a cylindrical hole in an infinite elastic-perfectly-plastic medium, here with a failure surface defined by the Mohr-Coulomb criterion. Assume that the rock mass is initially under hydrostatic stress P o and then a circular hole of radius is drilled into the rock, so that the stress at r = is reduced to some value P . The yield zone radius R o is given analytically by the theoretical model based on the solution of Salencon [31] SOFiSTiK 2014 | VERiFiCATiON MANUAL 129 BE29: Cylindrical Hole in an Infinite Mohr-Coulomb Medium R o = α _ 2 K p + 1 P o + q K p −1 P + q K p −1 _ 1 Kp−1 (32.1) where α is the radius of the hole, P o the initial in-situ stress, P the internal pressure and K p , q are given by K p = 1 + sn ϕ 1 − sn ϕ (32.2) q = 2c tn(45 + ϕ/ 2) (32.3) The parameters c and ϕ correspond to the cohesion and angle of friction of the medium respectively. For sufficiently small values of P o , where P o ¡ P holds, the medium will be in its elastic state, and the stresses will be given by [30] [32] σ r = P o − (P o − σ re ) _ R o r _ 2 (32.4) σ θ = P o + (P o − σ re ) _ R o r _ 2 (32.5) where r is the distance from the field point to the center of the hole and σ re is the radial stress at the elastic-plastic interface σ re = 1 K p + 1 (2P o − q) (32.6) For P o > P , the rock will fail within some annular region surrounding the borehole. The stresses in the yielded zone will be given by σ r = − q K p − 1 + _ P + q K p − 1 _ _ r α _ K p −1 (32.7) σ r = − q K p − 1 + K p _ P + q K p − 1 _ _ r α _ K p −1 (32.8) 32.3 Model and Results The properties of the model are defined in Table 32.1. The radius of the hole is 1 m and is assumed to be small compared to the length of the cylinder, therefore 2D plane strain conditions are in effect. A fixed external boundary is located 29.7 m from the hole center. The model is presented in Fig. 32.2. The stresses are calculated and verified with respect to the formulas provided in Section 32.2. 130 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE29: Cylindrical Hole in an Infinite Mohr-Coulomb Medium Table 32.1: Model Properties Material Properties Geometric Properties Pressure Properties E = 6777.9 MP α = 1 m P o = 30 MP ν = 0.21 r bondry = 29.7 m P = 0 or 1 MP Figure 32.2: Finite Element Model 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 10 20 30 40 50 Y i e l d z o n e r a d i u s Radial distance from center [m] S t r e s s [ M P ] Exact σ θ P = 0 P = 1 MP Exact σ r P = 0 P = 1 MP Figure 32.3: Radial and Tangential Stresses for Cylindrical Hole in Infinite Mohr-Coulomb Medium Figure 32.3 show the radial and tangential stress, along a line, lying on the X-axis. Results are presented for two cases, first with no internal pressure and second with P = 1 MP. The results in both cases are in very good agreement with the reference solution. SOFiSTiK 2014 | VERiFiCATiON MANUAL 131 BE29: Cylindrical Hole in an Infinite Mohr-Coulomb Medium 32.4 Conclusion This example verifies the stresses of a cylindrical hole in an infinite elastic-perfectly-plastic medium. It has been shown that the behaviour of the model is captured accurately. 32.5 Literature [30] J. C. Jaeger, N. G. W. Cook, and R. W. Zimmerman. Fundamentals of Rock Mechanics. 4th. Blackwell Publishing, 2007. [31] J. Salencon. “Contraction Quasi-Statique D’ une Cavite a Symetrie Spherique Ou Cylindrique Dans Un Milieu Elasto-Plastique”. In: Annales Des Ports Et Chaussees 4 (1969). [32] Phase 2 Stress Analysis Verification Manual Part I. Rocscience Inc. 2009. 132 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE30: Strip Loading on an Elastic Semi-Infinite Mass 33 BE30: Strip Loading on an Elastic Semi-Infinite Mass Overview Element Type(s): C2D Analysis Type(s): STAT Procedure(s): Topic(s): SOIL Module(s): TALPA Input file(s): strip load.dat 33.1 Problem Description This problem concerns the analysis of a strip loading on an elastic semi-infinite mass, as shown in Fig. 33.1. The material is assumed to be isotropic and elastic. The stresses are verified. Figure 33.1: Problem Description 33.2 Reference Solution The problem focuses on the calculation of the stresses due to a strip loading on an semi-infinite mass. The stresses under the surface are given by [33]: σ y = p π [α + sn α cos (α + 2δ)] (33.1) σ = p π [α − sn α cos (α + 2δ)] (33.2) and the principal stresses are σ 1 = p π [α + sn α] (33.3) σ 3 = p π [α − sn α] (33.4) where p, α, δ are described in Fig. 33.2 SOFiSTiK 2014 | VERiFiCATiON MANUAL 133 BE30: Strip Loading on an Elastic Semi-Infinite Mass α δ 2b p y Figure 33.2: Vertical Strip Loading on a Semi-Infinite Mass 33.3 Model and Results The properties of the model are defined in Table 33.1. The strip footing has a width of 2 m. The material is considered to be isotropic and elastic and plane strain conditions are in effect. For the analysis, boundary conditions are applied as shown in Fig. 33.3. The model is analysed with various dimensions in order to record the influence of the boundary in the results. The stresses are calculated and verified with respect to the formulas provided in Section 33.2. The results are printed for the case of a vertical line (cut) for = 0 where the stresses in and y coincide with the principal stresses. Figure 33.3: Finite Element Model Table 33.1: Model Properties Material Properties Geometric Properties Pressure Properties E = 20000 MP H = 25, 50, 100 m P = 1 MP / re ν = 0.2 B = 2 H 134 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE30: Strip Loading on an Elastic Semi-Infinite Mass 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 0 0.2 0.4 0.6 0.8 1 1.2 Vertical distance from (0,0) [m] S t r e s s [ M P ] Exact σ y Exact σ H = 25 m H = 50 m H = 100 m Figure 33.4: Comparison of Horizontal and Vertical Stresses Under the Strip Loading Figure 33.5: Vertical Stress Distribution for a Strip Loading on a Semi-Infinite Mass Fig. 33.4 shows the horizontal and vertical stress along the cutting line, for the analysed models with various dimensions. This line (cut) can be visualised in Fig. 33.5, where the contours of the vertical stress for the case of H = 50 m are illustrated. From the results of the stresses, it is evident that the vertical stresses are not influenced significantly from the dimensions of the model. On the contrary, for the horizontal stresses it is obvious, that as the boundary moves further away, its influence vanishes and the results are in very good agreement with the reference solution. SOFiSTiK 2014 | VERiFiCATiON MANUAL 135 BE30: Strip Loading on an Elastic Semi-Infinite Mass 33.4 Conclusion This example verifies the distribution of stresses of a semi-infinite mass under strip loading. It has been shown that the behaviour of the model is captured accurately. 33.5 Literature [33] H.G. Poulos and E.H. Davis. Elastic Solutions for Soil and Rock Mechanics. Centre for Geotech- nical Research, University of Sydney, 1991. 136 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE31: Snap-Through Behaviour of a Truss 34 BE31: Snap-Through Behaviour of a Truss Overview Element Type(s): TRUS Analysis Type(s): STAT, GNL Procedure(s): LSTP Topic(s): Module(s): ASE Input file(s): snap through.dat 34.1 Problem Description This problem is concerned with one of the fundamental geometric non-linearity (GNL) tests. A simple two-node truss, as shown in Fig. 34.1, is examined in terms of the limit load and snap-through behaviour. H P L Figure 34.1: Problem Description 34.2 Reference Solution In this problem a truss is pin-jointed to a rigid surface at one end and subjected to a transverse vertical point force at the other end as shown in Fig. 34.1. The loaded end is restrained to move only vertically and the truss in inclined with respect to the horizontal. This problem is effectively a symmetrical half of a two-bar structure and is utilised here in order to demonstrate the snap-through behaviour and limit points. The analytical solution assuming a shallow strut is given by [34] [14] P = EAH 3 2L 3 _ 2 H + 3 2 H 2 + 3 H 3 _ (34.1) where the parameters H and L are shown in Fig. 34.1. The reference solution is plotted in Fig. 34.2.The load-displacement curve rises until it reaches a limit point A. If we continue further, the next point will be B, where the bar is horizontal and the vertical load reduces to zero. Further increments cause the bar to deflect below the horizontal axis until the second limit point is reached at point C. Note that after point B, the load reverses its sign and acts upwards. After point C, the bar continues its motion downwards until it reaches point D, where the vertical load is zero. Note that under load-control approach, snap-through behaviour occurs after the first limit point A, where the bar suddenly jumps from point A to point E without any increase in the load. By switching from load- control to displacement-control, i.e. the displacement rather than the load is applied in small increments, the solution is able to progress beyond the limit point A, where further displacements cause the load to SOFiSTiK 2014 | VERiFiCATiON MANUAL 137 BE31: Snap-Through Behaviour of a Truss reduce as the bar reaches a horizontal position at B. Figure 34.2: Analytical Load-Displacement Curve 34.3 Model and Results The properties of the model are defined in Table 34.1. In the load-control approach, the load is applied in significantly small increments in order to be able to capture point A. In the displacement-control, the displacement increments are of 1 mm. The load-displacement curve for both approaches is presented in Fig. 34.3 and compared to the analytical solution. If we solve Eq. 34.1 with respect to the limit points, we observe that at point A the displacement is 10.57 mm and the corresponding critical loading is P cr = 9.6225 N. Table 34.1: Model Properties Material Properties Geometric Properties Loading E = 500 × 10 3 N/ mm 2 H = 25 mm P = 10 N ν = 0.4999 ≈ 0.5 L = 2500 mm Δ = 1 mm A = 100 mm 2 Fig. 34.3 shows that the load-control approach reaches the first limit point and suddenly snaps to the new equilibrium state, corresponding to point E in Fig. 34.2. The value of the load obtained before the snap-through occurs, corresponding to point A, is P = 9.62 N, with a displacement of 10.415 mm, which is in very good agreement with the theoretical critical value P cr . Furthermore, we can observe that the more suitable solution strategy, for such a simple system, for obtaining the load-deflection response is to adopt the displacement control approach, which clearly as shown in Fig. 34.3, has no difficulty with the local limit point at A and traces the complete equilibrium path. 138 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE31: Snap-Through Behaviour of a Truss Figure 34.3: Calculated Load-Displacement Curve 34.4 Conclusion This example verifies the determination of the limit load and snap-through behaviour of a simple truss. It has been shown that the geometric non-linear behaviour of the model is captured accurately. 34.5 Literature [14] A. A. Becker. Background to Finite Element Analysis of Geometric Non-linearity Benchmarks. Tech. rep. NAFEMS, 1998. [34] M. A. Crisfield. Non-linear Finite Element Analysis of Solids and Structures - Volume 1. John Wiley, 1991. SOFiSTiK 2014 | VERiFiCATiON MANUAL 139 BE31: Snap-Through Behaviour of a Truss 140 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE32: Thermal Extension of Structural Steel in case of Fire 35 BE32: Thermal Extension of Structural Steel in case of Fire Overview Element Type(s): BF2D, SH3D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): FIRE Module(s): TALPA, ASE Input file(s): thermal extension, quad 32.dat 35.1 Problem Description This benchmark is concerned with the validation of the structural analysis in case of fire with respect to the general calculation method according to DIN EN 1992-1-2. Therefore test case 4 is employed as presented in Annex CC of the standard DIN EN 1992-1-2/NA:2010-03 [35]. In this example the validation of the extension of structural steel, for the model of Fig. 35.1, at different constant temperature exposures is examined. h b Figure 35.1: Problem Description 35.2 Reference Solution The physical, mechanical and mathematical basics of engineering-based fire design programs, should be validated in terms of thermal, cross-sectional and system analysis. The aim of Annex CC [35] is, through a collection of test cases, to check their applicability for fire proof evaluation on real structures. For every example a parameter-dependent test matrix, for the relevant assessment criteria, is provided, where the computational accuracy of the program is examined. Results of existing analytical solutions or of approved programs are also provided, as well as the acceptable specified tolerances. 35.3 Model and Results The properties of the model are defined in Table 35.1. A fictional beam, as depicted in Fig. 35.1, with cross-sectional dimensions b / h = 100/100 mm and the length of 100 mm is examined. Different temperatures are assigned to the material S 355 of the cross-section. The analysis is performed with SOFiSTiK 2014 | VERiFiCATiON MANUAL 141 BE32: Thermal Extension of Structural Steel in case of Fire TALPA, where the FIBER beam element is utilised, as well as with ASE, where the QUAD element is tested. The computed and the reference results are presented in Table 35.2, Fig. 35.2 and Table 35.3 for the FIBER beam and QUAD element, respectively. Table 35.1: Model Properties Material Properties Geometric Properties Test Properties S 355 = 100 mm Initial Conditions: ƒ yk(20 ◦ C) = 355 N/ mm 2 h = 100 mm Θ = 20 ◦ C Stress-strain curve according to DIN EN 1993-1-2 b = 100 mm Homogeneous temperature component: Θ = 100, 300, 500, 600, 700, 900 ◦ C Table 35.2: Results for Thermal Elongation of Steel - FIBER Θ [ ◦ C] Ref. [35] SOF. |e r | [%] Tol. Δ [mm] Δ’ [mm] or e [mm] 100 0.09984 0.09984 0.000 mm for Θ ≤ 300 ◦ C 300 0.37184 0.37184 0.000 mm ± 0.05 mm 500 0.67584 0.67584 0.000 % 600 0.83984 0.83984 0.000 % for Θ > 300 ◦ C 700 1.01184 1.01184 0.000 % ± 1 % 900 1.18000 1.18000 0.000 % Table 35.3: Results for Thermal Elongation of Steel - QUAD Θ [ ◦ C] Ref. [35] SOF. |e r | [%] Tol. Δ [mm] Δ’ [mm] or e [mm] 100 0.09984 0.09984 0.000 mm for Θ ≤ 300 ◦ C 300 0.37184 0.37184 0.000 mm ± 0.05 mm 500 0.67584 0.67584 0.000 % 600 0.83984 0.83984 0.000 % for Θ > 300 ◦ C 700 1.01184 1.01184 0.000 % ± 1 % 900 1.18000 1.18000 0.000 % 142 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE32: Thermal Extension of Structural Steel in case of Fire Figure 35.2: Temperature Strains 35.4 Conclusion This example verifies the extension of structural steel at different constant temperature exposures. It has been shown that the calculation results are in excellent agreement with the reference results. 35.5 Literature [35] DIN EN 1991-1-2/NA: Eurocode 1: Actions on structures, Part 1-2/NA: Actions on structures ex- posed to fire. CEN. 2010. SOFiSTiK 2014 | VERiFiCATiON MANUAL 143 BE32: Thermal Extension of Structural Steel in case of Fire 144 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE33: Work Laws in case of Fire for Concrete and Structural Steel 36 BE33: Work Laws in case of Fire for Concrete and Structural Steel Overview Element Type(s): BF2D, SH3D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): FIRE Module(s): TALPA, ASE Input file(s): temperature compression.dat, quad 33.dat 36.1 Problem Description This benchmark is concerned with the validation of the structural analysis in case of fire with respect to the general calculation method according to DIN EN 1992-1-2. Therefore test case 5 is employed as presented in Annex CC of the standard DIN EN 1992-1-2/NA:2010-03 [35]. In this example the validation of the change in length of structural steel and concrete in compression, for the model of Fig. 36.1, at varying temperature and load capacity levels, is investigated. σ h b Figure 36.1: Problem Description 36.2 Reference Solution The aim of Annex CC [35] is to check the applicability of the programs for engineering-based fire de- sign on real structures. In this case the influence of the combination of increasing temperature and compressive loading with respect to the loading capacity of the structure is examined. 36.3 Model and Results The properties of the model are defined in Table 36.1. A fictional beam as depicted in Fig. 36.1 is examined here, for the case of structural steel S 355 and of concrete C 20/ 25, with cross-sectional dimensions b / h = 10 / 10 mm, = 100 mm and b / h = 31.6 / 31.6 mm, = 100 mm, respectively. SOFiSTiK 2014 | VERiFiCATiON MANUAL 145 BE33: Work Laws in case of Fire for Concrete and Structural Steel Different temperatures and load levels are investigated. The boundary conditions are set such that stability failure is ruled out. The analysis is performed with TALPA, where the FIBER beam element is utilised. The computed and the reference results are presented in Table 36.2 for structural steel and in Table 36.3 for concrete. Fig. 36.2 presents stress-strain curves for structural steel for different temperature levels. Table 36.1: Model Properties Material Properties Geometric Properties Test Properties Steel Concrete Steel Concrete S 355 C 20/ 25 = 100 mm = 100 mm Initial Conditions: ƒ yk = 355 MP ƒ ck = 20 MP h = 100 mm h = 31.6 mm Θ = 20 ◦ C Stress-strain: Stress-strain: b = 10 mm b = 31.6 mm Homog. temp.: DIN EN 1993-1-2 DIN EN 1992-1-2 20, 200, 400, 600, 800 ◦ C Loading: σ s(Θ) / ƒ yk(Θ) or σ c(Θ) / ƒ ck(Θ) = 0.2, 0.6, 0.9 Table 36.2: Results for Structural Steel - FIBER Θ [ ◦ C] Ref. [35] SOF. e r [%] Tol. σ s(Θ) / ƒ yk(Θ) Δ [mm] Δ’ [mm] [%] 20 0.2 0.034 0.034 0.560 0.6 0.101 0.101 −0.424 0.9 0.152 0.152 −0.094 200 0.2 −0.194 −0.194 −0.141 0.6 −0.119 −0.119 −0.119 0.9 0.159 0.156 1.794 400 0.2 −0.472 −0.472 0.097 0.6 −0.293 −0.294 −0.305 ± 3 % 0.9 0.451 0.449 0.524 600 0.2 −0.789 −0.789 0.053 0.6 −0.581 −0.581 −0.054 0.9 0.162 0.160 1.243 800 0.2 −1.059 −1.059 0.030 146 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE33: Work Laws in case of Fire for Concrete and Structural Steel Table 36.2: (continued) Θ [ ◦ C] Ref. [35] SOF. e r [%] Tol. σ s(Θ) / ƒ yk(Θ) Δ [mm] Δ’ [mm] [%] 0.6 −0.914 −0.914 −0.028 0.9 −0.170 −0.172 −1.163 Table 36.3: Results for Concrete - FIBER Θ [ ◦ C] Ref. [35] SOF. e r [%] Tolerance σ s(Θ) / ƒ yk(Θ) Δ [mm] Δ’ [mm] [%] 20 0.2 0.0334 0.0334 0.074 0.6 0.104 0.1036 0.428 0.9 0.176 0.1763 −0.173 200 0.2 −0.107 −0.1070 0.024 0.6 0.0474 0.0474 −0.035 0.9 0.2075 0.2075 0.014 400 0.2 −0.356 −0.3557 0.085 0.6 −0.075 −0.0750 0.016 ± 3 % 0.9 0.216 0.2160 −0.008 600 0.2 −0.685 −0.6850 −0.007 0.6 0.0167 0.0167 −0.182 0.9 0.744 0.7442 −0.033 800 0.2 −1.066 −1.0662 −0.023 0.6 −0.365 −0.3645 0.145 0.9 0.363 0.363 −0.010 Next step is the analysis of the same example with ASE where the QUAD element is now tested. The results are presented in Table 36.4 for structural steel and in Table 36.5 for concrete. Table 36.4: Results for Structural Steel - QUAD Θ [ ◦ C] Ref. [35] SOF. e r [%] Tolerance σ s(Θ) / ƒ yk(Θ) Δ [mm] Δ’ [mm] [%] 20 0.2 0.034 0.034 0.560 0.6 0.101 0.101 −0.424 0.9 0.152 0.152 −0.094 SOFiSTiK 2014 | VERiFiCATiON MANUAL 147 BE33: Work Laws in case of Fire for Concrete and Structural Steel Table 36.4: (continued) Θ [ ◦ C] Ref. [35] SOF. e r [%] Tolerance σ s(Θ) / ƒ yk(Θ) Δ [mm] Δ’ [mm] [%] 200 0.2 −0.194 −0.194 −0.173 0.6 −0.119 −0.119 −0.276 0.9 0.159 0.155 2.630 400 0.2 −0.472 −0.472 0.046 0.6 −0.293 −0.295 −0.669 ± 3 % 0.9 0.451 0.437 3.148 600 0.2 −0.789 −0.789 −0.022 0.6 −0.581 −0.585 −0.640 0.9 0.162 0.146 9.856 800 0.2 −1.059 −1.059 −0.029 0.6 −0.914 −0.917 −0.358 0.9 −0.170 −0.185 −8.765 Table 36.5: Results for Concrete - QUAD Θ [ ◦ C] Ref. [35] SOF. e r [%] Tolerance σ s(Θ) / ƒ yk(Θ) Δ [mm] Δ’ [mm] [%] 20 0.2 0.0334 0.0334 0.081 0.6 0.1040 0.1036 0.429 0.9 0.1760 0.1763 −0.173 200 0.2 −0.1070 −0.1047 2.137 0.6 0.0474 0.0475 −0.108 0.9 0.2075 0.2086 −0.527 400 0.2 −0.3560 −0.3557 0.082 0.6 −0.0750 −0.0750 0.014 ± 3 % 0.9 0.2160 0.2160 −0.008 600 0.2 −0.6850 −0.6851 −0.010 0.6 0.0167 0.0167 −0.197 0.9 0.7440 0.7442 −0.033 800 0.2 −1.0660 −1.0663 −0.026 0.6 −0.3650 −0.3645 0.145 0.9 0.3630 0.3630 −0.010 148 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE33: Work Laws in case of Fire for Concrete and Structural Steel 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 0 50 100 150 200 250 300 350 Strain [ ◦ / ◦◦ ] σ s ( Θ ) [ M P ] 800 ◦ C 600 ◦ C 400 ◦ C 200 ◦ C 20 ◦ C Figure 36.2: Steel Loading Strains 36.4 Conclusion This example verifies the change in length of structural steel and concrete at different temperature and load levels. It has been shown that the calculation results with TALPA and the FIBER beam element are in very good agreement with the reference results. For the case of the QUAD layer element the results present some deviation only for the structural steel and specifically for the case of a high stress level, reaching the 90% of the steel strength. 36.5 Literature [35] DIN EN 1991-1-2/NA: Eurocode 1: Actions on structures, Part 1-2/NA: Actions on structures ex- posed to fire. CEN. 2010. SOFiSTiK 2014 | VERiFiCATiON MANUAL 149 BE33: Work Laws in case of Fire for Concrete and Structural Steel 150 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE34: Ultimate Bearing Capacity of Concrete and Steel under Fire 37 BE34: Ultimate Bearing Capacity of Concrete and S- teel under Fire Overview Element Type(s): BF2D, SH3D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): FIRE Module(s): TALPA, ASE Input file(s): capacity.dat, quad 34.dat 37.1 Problem Description This benchmark is concerned with the validation of the structural analysis in case of fire with respect to the general calculation method according to DIN EN 1992-1-2. Therefore test case 6 is employed as presented in Annex CC of the standard DIN EN 1992-1-2/NA:2010-03 [35]. In this example the ultimate bearing capacity of structural steel and concrete in compression, for the model of Fig. 37.1, at varying temperature levels, is investigated. σ h b Figure 37.1: Problem Description 37.2 Reference Solution The aim of Annex CC [35] is to check the applicability of the programs for engineering-based fire design on real structures. In this case the influence of the combination of temperature and compressive loading, on the ultimate bearing capacity is examined. 37.3 Model and Results The properties of the model are defined in Table 37.1. A fictional beam as depicted in Fig. 37.1 is examined here, for the case of structural steel S 355 and of concrete C 20/ 25, with cross-sectional dimensions b / h = 10 / 10 mm, = 100 mm and b / h = 31.6 / 31.6 mm, = 100 mm, respec- SOFiSTiK 2014 | VERiFiCATiON MANUAL 151 BE34: Ultimate Bearing Capacity of Concrete and Steel under Fire tively. The boundary conditions are set such that stability failure is ruled out. The analysis is performed with TALPA, where the FIBER beam element is utilised. The computed and the reference results are presented in Table 37.2 for structural steel and in Table 37.3 for concrete. Table 37.1: Model Properties Material Properties Geometric Properties Test Properties Steel Concrete Steel Concrete S 355 C 20/ 25 = 100 mm = 100 mm Initial Conditions: ƒ yk = 355 MP ƒ ck = 20 MP h = 100 mm h = 31.6 mm Θ = 20 ◦ C Stress-strain: Stress-strain: b = 10 mm b = 31.6 mm Homog. temp.: DIN EN 1993-1-2 DIN EN 1992-1-2 20, 200, 400, 600, 800 ◦ C Table 37.2: Results for Structural Steel - FIBER beam Θ [ ◦ C] Ref. [35] SOF. e [kN] e r [%] Tol. N R,ƒ ,k [kN] N R,ƒ ,k ’ [kN] 20 −35.5 −35.5 0.000 0.000 200 −35.5 −35.5 0.000 0.000 ± 3 % 400 −35.5 −35.5 0.000 0.000 and 600 −16.7 −16.7 −0.015 0.090 ± 0.5 [kN] 800 −3.9 −3.9 0.005 −0.128 Table 37.3: Results for Concrete - FIBER beam Θ [ ◦ C] Ref. [35] SOF. e [kN] e r [%] Tol. N R,ƒ ,k [kN] N R,ƒ ,k ’ [kN] 20 −20.0 −20.0 −0.029 0.144 200 −19.0 −19.0 −0.027 0.144 ± 3 % 400 −15.0 −15.0 −0.022 0.144 and 600 −9.0 −9.0 −0.013 0.144 ± 0.5 [kN] 800 −3.0 −3.0 −0.004 0.144 Next step is the analysis of the same example with ASE where the QUAD element is now tested. The results are presented in Table 4 for structural steel and in Table 5 for concrete. 152 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE34: Ultimate Bearing Capacity of Concrete and Steel under Fire Table 37.4: Results for Structural Steel - QUAD Θ [ ◦ C] Ref. [35] SOF. e [kN] e r [%] Tol. N R,ƒ ,k [kN] N R,ƒ ,k ’ [kN] 20 −35.5 −35.5 0.000 0.000 200 −35.5 −35.5 0.000 0.000 ± 3 % 400 −35.5 −35.5 0.000 0.000 and 600 −16.7 −16.7 −0.015 0.090 ± 0.5 [kN] 800 −3.9 −3.9 0.005 −0.128 Table 37.5: Results for Concrete - QUAD Θ [ ◦ C] Ref. [35] SOF. e [kN] e r [%] Tol. N R,ƒ ,k [kN] N R,ƒ ,k ’ [kN] 20 −20.0 −20.0 −0.029 0.144 200 −19.0 −19.0 −0.037 0.193 ± 3 % 400 −15.0 −15.0 −0.023 0.156 and 600 −9.0 −9.0 −0.013 0.150 ± 0.5 [kN] 800 −3.0 −3.0 −0.015 0.489 37.4 Conclusion This example verifies the influence of compressive loading on the ultimate bearing capacity under differ- ent temperature levels. It has been shown that the calculation results are in very good agreement with the reference results for both the QUAD layer element and the FIBER beam element. 37.5 Literature [35] DIN EN 1991-1-2/NA: Eurocode 1: Actions on structures, Part 1-2/NA: Actions on structures ex- posed to fire. CEN. 2010. SOFiSTiK 2014 | VERiFiCATiON MANUAL 153 BE34: Ultimate Bearing Capacity of Concrete and Steel under Fire 154 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE35: Calculation of Restraining Forces in Steel Members in case of Fire 38 BE35: Calculation of Restraining Forces in Steel Members in case of Fire Overview Element Type(s): BF2D, SH3D Analysis Type(s): STAT, MNL Procedure(s): LSTP Topic(s): FIRE Module(s): TALPA, ASE Input file(s): restraining forces.dat, quad 35.dat 38.1 Problem Description This benchmark is concerned with the validation of the structural analysis in case of fire with respect to the general calculation method according to DIN EN 1992-1-2. Therefore test case 7 is employed as presented in Annex CC of the standard DIN EN 1992-1-2/NA:2010-03 [35]. In this example the restrain- ing forces developed in an immovable steel member due to temperature exposure are investigated for the model of Fig. 38.1. h Θ o Θ Figure 38.1: Problem Description 38.2 Reference Solution The aim of Annex CC [35] is to check the applicability of the programs for engineering-based fire design on real structures. In this case the influence of temperature exposure on the development of restraining forces in steel is investigated. To illustrate the development of the restraining forces, consider a steel bar fixed at both ends and exposed to fire. As the bar is heated it tries to expand. However, the fixture prevents expansion in the longitudinal direction. Thus, the fixture exerts restraining forces on the bar. Since the bar is prevented from longitudinal expansion, it is possible to expand in the other directions. 38.3 Model and Results The properties of the model are defined in Table 38.1. A beam with cross-sectional dimensions b/ h = 100/ 100 mm, = 1000 mm and fixed at both ends, as depicted in Fig. 38.1, is examined here. The material of the cross-section is structural steel with a fictive yield strength of ƒ yk(20 ◦ C) = 650 N/ mm 2 and thermo-mechanical properties according to EN 1993-1-2. The model is exposed to different tem- peratures. In the first case the same temperature is assigned across the cross-section height, whereas SOFiSTiK 2014 | VERiFiCATiON MANUAL 155 BE35: Calculation of Restraining Forces in Steel Members in case of Fire in the second case, the temperature difference of the upper and lower fiber is 200 ◦ C. The analysis is performed with TALPA, where the FIBER beam element is utilised. The computed and the reference results are presented in Table 38.2. Table 38.1: Model Properties Material Properties Geometric Properties Test Properties ƒ yk(20 ◦ C) = 650 N/ mm 2 = 1000 mm Case 1 E (20 ◦ C) = 210000 N/ mm 2 h = 100 mm Θ o = 120 ◦ , C Θ = 120 ◦ C Stress-strain curve b = 100 mm Case 2 according to DIN EN 1993-1-2 Θ o = 20 ◦ , C Θ = 220 ◦ C Table 38.2: Results for Structural Steel - FIBER beam Temperature Load Ref. [35] SOF. |e r | [%] Tol. Θ [ ◦ C] X X’ [%] 120/ 120 N Z [kN] −2585.0 −2584.8 0.006 ± 1 M Z [kNm] 0.0 0.0 0.000 ± 1 σ Z [N/mm 2 ] −258.5 −258.5 0.006 ± 5 20/ 220 N Z [kN] −2511.0 −2503.9 0.282 ± 1 M Z [kNm] −40.3 −40.2 0.249 ± 1 σ Z [N/mm 2 ] −479.0 −479.0 0.000 ± 5 Next step is the analysis of the same example with ASE where the QUAD element is now tested. The results are presented in Table 38.3 for both temperature loads. Table 38.3: Results for Structural Steel - QUAD Temperature Load Ref. [35] SOF. |e r | [%] Tol. Θ [ ◦ C] X X’ [%] 120/ 120 N Z [kN] −2585.0 −2589.8 0.186 ± 1 M Z [kNm] 0.0 0.0 0.000 ± 1 σ Z [N/mm 2 ] −258.5 −258.98 0.186 ± 5 20/ 220 N Z [kN] −2511.0 −2523.8 0.510 ± 1 M Z [kNm] −40.3 −40.73 1.067 ± 1 σ Z [N/mm 2 ] −479.0 −480.23 0.257 ± 5 156 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE35: Calculation of Restraining Forces in Steel Members in case of Fire For the quad element, the results appear to deviate from the reference solution. This is due to the fact that, as the plasticity involves at the cross-section, plastic strains appear also in the lateral direction. This causes a biaxial stress state (σ = 0), which is not neglected by the quad formulation, as shown in Fig. 38.2, and has an effect on both the stresses and moments in the y direction. Figure 38.2: Nonlinear Stresses for Temperature 220 ◦ C at Bottom Quad Layer 38.4 Conclusion This example verifies the development of restraining forces in steel due to temperature exposure. It has been shown that the calculation results are in very good agreement with the reference results for both the QUAD layer element and the FIBER beam element. 38.5 Literature [35] DIN EN 1991-1-2/NA: Eurocode 1: Actions on structures, Part 1-2/NA: Actions on structures ex- posed to fire. CEN. 2010. SOFiSTiK 2014 | VERiFiCATiON MANUAL 157 BE35: Calculation of Restraining Forces in Steel Members in case of Fire 158 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE36: Pushover Analysis: Performance Point Calculation by ATC-40 Procedure 39 BE36: Pushover Analysis: Performance Point Calcu- lation by ATC-40 Procedure Overview Element Type(s): Analysis Type(s): Procedure(s): Topic(s): EQKE Module(s): SOFiLOAD Input file(s): pushover-pp-atc.dat 39.1 Problem Description The following example is intended to verify the ATC-40 procedure for the calculation of the performance point (illustrated schematically in Fig. 39.1), as implemented in SOFiSTiK. The elastic demand and capacity diagrams are assumed to be know. S d S dp S S p El. Demand Diagram Performance Point Capacity Diagram Demand Diagram Figure 39.1: Determination of the performance point PP (S dp , S p ) 39.2 Reference Solution The reference solution is provided in [36], 8.3.3.3 ”Performance Point Calculation by Capacity Spectrum Method - Procedure A”. Assuming that the elastic demand diagram (5% elastic response spectrum in ADRS format 1 ) and the capacity diagram are known, it is possible to determine the performance point PP (S dp , S p ) (Fig. 39.1). The procedure comprises of a series of trial calculations (trial performance points PP t (S dp,t , S p,t )), in which the equivalent inelastic single degree of freedom system (SDOF), represented by the capacity diagram, is transformed to an equivalent elastic SDOF system whose response in form of the perfor- mance point PP is then calculated from the reduced elastic response spectrum (demand diagram). The computation stops when the performance point PP is within a tolerance of a trial performance point PP t . The ATC-40 Procedure A is a semi-analytical procedure since it involves graphical bilinear idealization of the capacity diagram. Detailed description of this step-by-step procedure can be found in [36]. 1 ADRS = Spectral Acceleration S - Spectral Displacement S d format SOFiSTiK 2014 | VERiFiCATiON MANUAL 159 BE36: Pushover Analysis: Performance Point Calculation by ATC-40 Procedure 39.3 Model and Results In order to verify the analysis procedure for the determination of the performance point, a test case has been set up in such a way that it comprises of a SDOF with a unit mass and a non-linear spring element. It is obvious that for such an element the quantities governing the transformation from the original system to the equivalent inelastic SDOF system must be equal to one, i.e. ϕ cnod = 1 ; = 1 ; m = 1 , (39.1) where ϕ cnod is the eigenvector value at control node, is the modal participation factor and m is the generalized modal mass. Writing now the equations which govern the conversion of the pushover curve to capacity diagram, we obtain [37] S d = cnod ϕ cnod · = cnod , (39.2a) S = V b 2 · m = V b , (39.2b) where V b is the base shear and cnod is the control node displacement. Since the original system is a SDOF system, V b and cnod are nothing else but the force in spring P and the displacement of the unit mass , respectively. It follows further that the force-displacement work law assigned to the spring element corresponds to the capacity diagram in ADRS format, with the force P and displacement equal to S and S d , respectively. The capacity diagram used in the reference example is defined by four points, whose coordinates are listed in the Table 39.1. According to the analysis above, these points can be used to define the force- displacement work law P − of the non-linear spring element (Fig.39.2). Table 39.1: Model Properties [36] Capacity Diagram Elastic Demand Point _ S d [mm], S [m/ s 2 ] _ UBC 5% Elastic Response Spectrum. A ( 48.77, 2.49) Seismic Zone 4, ZEN = 0.40. B ( 71.37, 3.03) No near-fault effects. C ( 96.01, 3.39) Soil Profile: D (199.14, 3.73) - Type S B : C A = 0.40, C V = 0.40 - Type S D : C A = 0.44, C V = 0.64 160 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE36: Pushover Analysis: Performance Point Calculation by ATC-40 Procedure D C B A O u [mm] 2 0 0 . 0 1 5 0 . 0 1 0 0 . 0 5 0 . 0 0 . 0 0 . 0 P [kN] 3.0 2.0 1.0 0.0 0.0 Figure 39.2: Force-displacement work law of the non-linear spring The elastic demand is an UBC 5% damped elastic response spectrum, whose properties are summa- rized in Table 39.1. Two soil profile types are considered - soil profile type S B and S D . The outcome of the analysis is shown in Figures 39.3 and 39.4. Capacity UBC,Soil-B Demand, ξ-eff = 9.36% PY PP Ty Tp Tb = 0.1 Tc = 0.4 T = 0.5 T = 1.0 T = 1.5 T = 2.0 T = 4.0 SPL 1 SPL 2 SPL 3 SPL4 Sd [mm] 2 0 0 . 0 0 0 1 5 0 . 0 0 0 1 0 0 . 0 0 0 5 0 . 0 0 0 0 . 0 0 0 Sa [m/sec2] 10.00 5.00 0.00 Figure 39.3: Capacity-Demand-Diagram (Soil Profile Type S B ) Capacity UBC,Soil-D Demand, ξ-eff = 14.59% PY PP Ty Tp Tb = 0.1 Tc = 0.6 Td = 3.0 T = 0.5 T = 1.0 T = 1.5 T = 2.0 T = 4.0 SPL 1 SPL 2 SPL 3 SPL4 Sd [mm] 2 0 0 . 0 0 0 1 5 0 . 0 0 0 1 0 0 . 0 0 0 5 0 . 0 0 0 0 . 0 0 0 Sa [m/sec2] 10.00 5.00 0.00 Figure 39.4: Capacity-Demand-Diagram (Soil Profile Type S D ) SOFiSTiK 2014 | VERiFiCATiON MANUAL 161 BE36: Pushover Analysis: Performance Point Calculation by ATC-40 Procedure The results of the SOFiSTiK calculation and the comparison with the reference solution are summarized in Table 39.2. Table 39.2: Results ξ eƒ ƒ SR SR S dy S y S dp S p Soil type [%] [−] [−] [mm] [m/ s 2 ] [mm] [m/ s 2 ] SOF. 9.4 0.80 0.84 51.30 2.62 85.04 3.23 S B Ref. [36] 9.2 0.80 0.85 53.34 2.65 83.36 3.24 |e| [%] 2.2 0.0 1.2 3.8 1.1 2.0 0.3 SOF. 14.6 0.65 0.73 59.86 3.06 149.34 3.57 S D Ref. [36] 14.2 0.66 0.74 58.42 3.04 149.86 3.63 |e| [%] 2.8 1.5 1.4 2.5 0.7 0.3 1.7 ξ eƒ ƒ effective viscous damping of the equivalent linear SDOF system SR , SR spectral reduction factors in constant acceleration and constant velocity range of spectrum S dy , S y spectral displacement and spectral acceleration at yielding point S dp , S p spectral displacement and spectral acceleration at performance point The results are in excellent agreement with the reference solution. Small differences can mainly be attributed to the approximate nature of the graphical procedure for the bilinear idealization of the capacity used in the reference solution, while the procedure implemented in SOFiLOAD is refrained from such approximation and computes the hysteretic energy directly from the area underneath the capacity curve and the coordinates of the performance point [37]. Apart from that, the performance point displacement tolerance used in SOFiLOAD is lower than the one used in the reference solution (1% compared to 5%). 39.4 Conclusion Excellent agreement between the reference and the results computed by SOFiSTiK verifies that the procedure for the calculation of the performance point according to ATC-40 is adequately implemented. 39.5 Literature [36] ATC-40. Seismic Evaluation and Retrofit of Concrete Buildings. Applied Technology Council. Red- wood City, CA, 1996. [37] SOFiLOAD Manual: Loadgenerator for Finite Elements and Frameworks. Version 2014.1. SOFiSTiK AG. Oberschleißheim, Germany, 2013. 162 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory 40 BE37: BeamCalculation of Varying Cross-Section ac- cording to Second Order Theory Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): STAB Topic(s): Module(s): ASE, STAR2, DYNA Input file(s): beam th2.dat 40.1 Problem Description The problem consists of a column of continuously varying cross-section, subjected to a large compres- sive force in combination with imperfections as well as horizontal and temperature loads, as shown in Fig. 40.1. The forces and deflections of the structure, calculated according to second order theory, are determined. h t 0 ψ 0 ∗ A E q E H H ∗ ΔT q A V ∗ ∗ s b Figure 40.1: Problem Description 40.2 Reference Solution This example attempts to give a complete description of the forces and the deflections of a beam with varying cross-section calculated with second order theory. As a reference solution, a general formulation concept is adopted, where through the application of series functions, uniform formulas can be derived to describe the beam behaviour of varying cross-section. In this concept, the cross-section properties can vary according to a polynomial of arbitrary degree, the normal force, with respect to second order theory, is assumed constant, the imperfections or predeformations as well as the temperature loads are taken into account and the deformations due to moments and normal forces are treated. Further SOFiSTiK 2014 | VERiFiCATiON MANUAL 163 BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory information on the reference solution can be found in Rubin (1991) [38]. 40.3 Model and Results The general properties of the model [38] are defined in Table 40.1 and the cross-sections in Table 40.2. A general linear material is used and a linearly varying, thin-walled I-beam profile for the cross-section. The shear deformations are neglected. A safety factor of 1.35 is used for the dead weight, giving a total normal force of N = −0.5 − 0.0203 = −0.5203 MN. An imperfection of linear distribution with maximum value of 60 mm at node E is applied, as well as one of quadratic distribution with maximum value of −48 mm at the middle. The temperature load is given as a difference of temperature of 25 ◦ C, between the left and the right side of the beam. The height of the cross-section is taken as the height of the web only. Second order theory is utilised and the structure is analysed both with ASE and STAR2. Table 40.1: Model Properties Model Properties Loading E = 21 MN/ cm 2 , γ g = 1.35 V = 500 kN ψ 0 = 1/ 200, 0 = −48 mm q E = 6 kN/ m, q A = 10 kN/ m α T = 1.2 × 10 −5 1/ ◦ K ΔT = T rght − T eƒ t = −25 ◦ C = 12 m, ∗ = 4 m H = 20 kN, H ∗ = 10 kN Table 40.2: Cross-sectional Properties Position Web [mm] Flange [mm] Area [cm 2 ] y [cm 4 ] h s b t E 200 12 194 20 101.6 8560 ∗ 300 12 260.7 20 140.27 26160.3 A 500 12 394 20 217.6 111000 M [kNm] [mm] N [kN] ϕ [mrd] [mm] Figure 40.2: Results with Twenty Four Beam Elements calculated by ASE 164 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory The results are presented in Table 40.3, where they are compared to the reference solution according to Rubin (1991) [38]. Fig. 40.2 shows the forces and deflections of the structure as they have been calculated by ASE with twenty four beam elements. Table 40.3: Results Number of ϕ E E E M A N A N E N K Elements [mrd] [mm] [mm] [MNm] [MN] [MN] [MN] - Ref. [38] 67.70 423.50 −1.9050 −1.10 −0.5203 −0.50 −1.882 3 ASE 63.8819 402.8733 −1.8937 −1.0816 −0.5202 −0.50 −1.9556 STAR2 82.1398 508.5219 −1.9288 −1.1352 −0.5203 −0.50 - 6 ASE 65.8787 413.7793 −1.9018 −1.0871 −0.5203 −0.50 −1.8993 STAR2 70.0395 437.2366 −1.9108 −1.0990 −0.5203 −0.50 - 24 ASE 66.5406 417.2853 −1.9045 −1.0888 −0.5203 −0.50 −1.8827 STAR2 66.7933 418.7001 −1.9050 −1.0895 −0.5203 −0.50 - 40.4 Conclusion This example examines the behaviour of a tapered beam, treated with second order theory. The results, calculated both with ASE and STAR2, converge to the same solution as the number of elements increas- es. Their deviation arises from the fact that ASE uses an exponential interpolation based on area and inertia as well as numerical integration of the stiffness, while STAR2 uses the geometric mean value of the stiffness. The first is slightly too stiff, the latter is too soft and therefore resulting on the safe side. With a total of twenty four beam elements, the results are reproduced adequately. However, the obtained solution deviates from the reference. The reason for that is the fact, that for second order effects, Rubin has taken an unfavourable constant normal force of −520.3 kN for the whole column. If that effect is accounted for, the results obtained with twenty four elements are: Table 40.4: Results with Constant Normal Force Number of ϕ E E M A Elements [mrd] [mm] [MNm] - Ref. [38] 67.70 423.50 −1.100 24 ASE 67.657 423.27 −1.0995 STAR2 67.918 424.73 −1.1002 In the case where the example is calculated with DYNA, where a constant normal force of −520.3kN is considered as a primary load case, leading to linearised second order theory and therefore satis- fying Rubin’s assumption, the results converge to the reference solution. The results, calculated with DYNA and twenty four elements, are presented in Table 40.5. Furthermore, different single loadings are examined and the results are given in Table 40.6. SOFiSTiK 2014 | VERiFiCATiON MANUAL 165 BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory Table 40.5: Results with DYNA Number of ϕ E E E M A N A N E Elements [mrd] [mm] [mm] [MNm] [MN] [MN] - Ref. [38] 67.70 423.50 −1.9050 −1.10 −0.5203 −0.50 24 DYNA 67.6569 423.2766 −1.9045 −1.0994 −0.5203 −0.50 Table 40.6: Results with DYNA for Combination of Constant Normal Force and Single Loadings Load ϕ E E M A Case [mrd] [mm] [MNm] H ∗ , H 23.6779 148.3151 −0.3972 q 22.5896 163.3797 −0.6130 ΔT 14.8218 77.1333 −0.0401 ψ 0 2.6481 15.9532 −0.0395 0 3.9194 18.4953 −0.0096 67.6569 423.2766 −1.0994 40.5 Literature [38] H. Rubin. “Ein einheitliches, geschlossenes Konzept zur Berechnung von St ¨ aben mit stetig ver ¨ andlichem Querschnitt nach Theorie I. und II. Ordnung”. In: Bauingenieur 66 (1991), pp. 465– 477. 166 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE38: Calculation of Slope Stability by Phi-C Reduction 41 BE38: Calculation of Slope Stability by Phi-C Reduc- tion Overview Element Type(s): C2D Analysis Type(s): STAT, MNL Procedure(s): LSTP, PHIC Topic(s): SOIL Module(s): TALPA Input file(s): slope stability.dat 41.1 Problem Description In this benchmark the stability of an embankment, as shown in Fig. 41.1, is calculated by means of a ph − c reduction. The factor of safety and its corresponding slip surface are verified. h 1 h 2 sope 1 2 3 Figure 41.1: Problem Description 41.2 Reference Solution The classical problem of slope stability analysis involves the investigation of the equilibrium of a mass of soil bounded below by an assumed potential slip surface and above by the surface of the slope. Forces and moments, tending to cause instability of the mass, are compared to those tending to resist instabil- ity. Most procedures assume a two-dimensional cross-section and plane strain conditions for analysis. Successive assumptions are made regarding the potential slip surface until the most critical surface, i.e. lowest factor of safety, is found. If the shear resistance of the soil along the slip surface exceeds that necessary to provide equilibrium, the mass is stable. If the shear resistance is insufficient, the mass is unstable. The stability of the mass depends on its weight, the external forces acting on it, the shear strengths and pore water pressures along the slip surface, and the strength of any internal reinforcement crossing potential slip surfaces. The factor of safety is defined with respect to the shear strength of the soil as the ratio of the available shear strength to the shear strength required for equilibrium [39]: FS = be sher strength eqbrmsher stress (41.1) SOFiSTiK 2014 | VERiFiCATiON MANUAL 167 BE38: Calculation of Slope Stability by Phi-C Reduction The stability analysis according to Fellenius, i.e. ph − c reduction, is based on the investigation of the material’s shear strength in the limit state of the system, i.e. the shear strength that leads to failure of the system. In this method, the forces on the sides of the slice are neglected and the forces that operate are only the weight, the normal stress and shear stress on the base of the slice [40]. On the other hand, the Bishop’s Method is based on the assumption that the interslice forces are horizontal and the slip surface is circular. Forces are summed in the vertical direction. The resulting equilibrium equation is combined with the Mohr-Coulomb equation and the definition of the factor of safety to determine the forces on the base of the slice [39]. In SOFiSTiK, the safety factors according to ph −c reduction are defined as the ratio between available shear strength and shear strength in the limit state of the system [41]: η ϕ = tn ϕ np tn ϕ t (41.2) η c = c np c t (41.3) where c is the cohesion and ϕ the friction angle. The ph − c reduction stability analysis is based on an incremental reduction of the shear strength adopting a synchronized increase of the safety factors η = η ph = η c . The reached safety η at system failure represents the computational safety against stability failure. 41.3 Model and Results The properties of the model [42] are presented in Table 41.1. The embankment has a slope of 1 : 2, a height of 4.5 m and a width of 9.0 m. The initial stresses are generated using gravity loading. Then the embankment is subjected to the ph − c reduction. Plane strain conditions are assumed. Table 41.1: Model Properties Material Properties Geometric Properties E = 2600 kN/ m 2 , ν = 0.3 h 1 = 6.5 m γ = 16 kN/ m 3 h 2 = 2.0 m ϕ = 20 ◦ , ψ = 20 ◦ 1 = 2 = 3 = 3.0 m c = 5 kN/ m 2 sope = 9.0 m 168 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE38: Calculation of Slope Stability by Phi-C Reduction FS = 1.56 Figure 41.2: Bishop’s circular slip surface result FS = 1.55 Figure 41.3: Results for ph − c Reduction Figure 41.3 presents the nodal displacement vector distribution for the factor of safety obtained with the ph − c reduction analysis. The reference safety factor is given by Bishop’s slip circle method as 1.56 [42]. The calculated factor of safety is in very good agreement with the reference solution. 41.4 Conclusion This example verifies the stability of a soil mass and the determination of the factor of safety. The results, obtained with the ph − c reduction method, are compared to the widely acclaimed Bishop’ method and is shown that the behaviour of the model is captured accurately. 41.5 Literature [39] USACE Engineering and Design: Slope Stability. USACE. 2003. [40] A. Verruitz. Grondmechanica. Delft University of Technology. 2001. [41] TALPA Manual: Statics of Plane or Axissymmetric Geomechanical Structures. Version 27.01. SOFiSTiK AG. Oberschleißheim, Germany, 2012. [42] PLAXIS Validation & Verification. Plaxis. 2011. SOFiSTiK 2014 | VERiFiCATiON MANUAL 169 BE38: Calculation of Slope Stability by Phi-C Reduction 170 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE39: Natural Frequencies of a Rectangular Plate 42 BE39: Natural Frequencies of a Rectangular Plate Overview Element Type(s): SH3D Analysis Type(s): DYN Procedure(s): EIGE Topic(s): Module(s): DYNA Input file(s): freq plate.dat 42.1 Problem Description This problem consists of a rectangular plate which is simply supported on all four sides, as shown in Fig. 42.1. The eigenfrequencies of the system are determined and compared to the exact reference solution for each case. b Figure 42.1: Problem Description 42.2 Reference Solution The general formula to determine the eigenfrequencies of a simply-supported thin plate, consisting of a linear elastic homogeneous and isotropic material is given by [29], [43] ƒ m,n = λ m,n 2 2π _ _ _ gD γh (42.1) where λ m,n 2 = π 2 _ m 2 2 + n 2 b 2 _ (42.2) SOFiSTiK 2014 | VERiFiCATiON MANUAL 171 BE39: Natural Frequencies of a Rectangular Plate and D is the flexural rigidity of the plate D = Eh 3 12(1 − ν 2 ) (42.3) Combining the above equations gives ƒ m,n = π 2 2 _ m 2 + n 2 2 b 2 _ _ _ _ g γh Eh 3 12(1 − ν 2 ) (42.4) where , b the dimensions of the plate, h the thickness and γh/ g the mass of the plate per unit area. The values of λ m,n 2 for the first five combinations of m, n are given in Table 42.1 for a simply-supported plate. Table 42.1: Dimensionless parameter λ m,n 2 m n λ m,n 2 Mode number 1 1 32.08 1 2 1 61.69 2 1 2 98.70 3 3 1 111.03 4 2 2 128.30 5 42.3 Model and Results The properties of the model are defined in Table 42.2 and the resulted eigenfrequencies are given in Table 42.3. The corresponding eigenforms are presented in Fig. 42.2. Table 42.2: Model Properties Material Properties Geometric Properties E = 30000 MP = 4.5 m γ = 80 kN/ m 3 b = 3.0 m ν = 0.3 h = 0.02 m Table 42.3: Results Eigenfrequency Number SOF. [Hz] Ref. [Hz] |e r | [%] 1 2.941 2.955 0.476 2 5.623 5.682 1.047 3 9.200 9.091 1.197 172 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE39: Natural Frequencies of a Rectangular Plate Table 42.3: (continued) Eigenfrequency Number SOF. [Hz] Ref. [Hz] |e r | [%] 4 10.206 10.228 0.214 5 11.706 11.819 0.954 Mode 1 Mode 2 Mode 3 Mode 4 Mode 5 Figure 42.2: Eigenforms 42.4 Conclusion The purpose of this example is to verify the eigenvalue determination of plate structures modelled with plane elements. It has been shown that the eigenfrequencies for a simply-supported thin rectangular plate are calculated accurately. 42.5 Literature [29] S. Timoshenko. Vibration Problems in Engineering. 2nd. D. Van Nostrand Co., Inc., 1937. [43] Schneider. Bautabellen f ¨ ur Ingenieure. 19th. Werner Verlag, 2010. SOFiSTiK 2014 | VERiFiCATiON MANUAL 173 BE39: Natural Frequencies of a Rectangular Plate 174 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE40: Portal Frame 43 BE40: Portal Frame Overview Element Type(s): B3D Analysis Type(s): STAT, GNL Procedure(s): Topic(s): Module(s): ASE Input file(s): frame.dat 43.1 Problem Description The problem consists of a rigid rectangular frame, with an imperfection at the columns, subjected to a uniform distributed load q across the span and to various single loads, as shown in Fig. 43.1. For the linear case, the structure is subjected to the uniform load only, whereas for the nonlinear case, all defined loads including the imperfection are considered. The response of the structure is determined and compared to the analytical solution. h c c b ψ 0 ψ 0 H 1 2 q F F δ Figure 43.1: Problem Description 43.2 Reference Solution For the linear case, where only the distributed load is considered, the moments M are determined in terms of the shear force H as follows: SOFiSTiK 2014 | VERiFiCATiON MANUAL 175 BE40: Portal Frame H 1 = H 2 = q 2 4h(k + 2) (43.1) M 1 = M 2 = Hh 3 (43.2) M 3 = M 4 = M 1 − H 1 h (43.3) where k = b h / c . For the nonlinear case, in order to account for the effect of the normal force and the imperfections on the determination of the resulting forces and moments, second order theory has to be used. The moments at nodes 1 − 4 are determined in dependency of the column characteristic ratio ε = c _ N/ E c , giving the influence of the normal force N = F + q/ 2 with respect to the column properties, length c and bending stiffness E c . Further information on the analytical formulas can be found in Schneider [43]. 43.3 Model and Results The properties of the model are defined in Table 43.1. The frame has an initial geometrical imperfection at the columns of linear distribution ψ 0 = 1/ 200, with a maximum value of 25 mm at nodes 3 and 4. The normal force N, used to determine ε, is calculated to be 430 kN at the columns and the ratio ε = 1.639. For the linear case the results are presented in Table 43.3 and they are compared to the analytical solution calculated from the formulas presented in Section 43.2. For the nonlinear case, the results are presented in Table 43.2 and they are compared to the reference example provided in Schneider [43]. Table 43.1: Model Properties Material Properties Geometric Properties Loading E c = 6000 kNm 2 = 6 m q = 10 kN/ m E b = 4000 kNm 2 h = 5 m H = 20 kN ψ 0 = 1/ 200 F = 400 kN Table 43.2: Nonlinear Case Results Ref. [43] SOF. M 1 [kNm] 38.2 38.62 M 2 [kNm] 22.5 22.52 M 3 [kNm] 58.1 58.02 M 4 [kNm] 58.8 58.79 δ [mm] 65.3 65.44 176 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE40: Portal Frame Nonlinear Case Linear Case Figure 43.2: Bending Moments Table 43.3: Linear Case Results Ref. [Sect.43.2] SOF. H 1 = H 2 [kN] 5.54 5.52 M 1 = M 2 [kNm] 9.23 9.18 M 3 = M 4 [kNm] 18.46 18.43 Nonlinear Case Linear Case Figure 43.3: Deformed Shape 43.4 Conclusion This example examines a rigid frame under different loading conditions. It has been shown that the behaviour of the structure is captured accurately for both the linear and the nonlinear analysis. 43.5 Literature [43] Schneider. Bautabellen f ¨ ur Ingenieure. 19th. Werner Verlag, 2010. SOFiSTiK 2014 | VERiFiCATiON MANUAL 177 BE40: Portal Frame 178 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE41: Linear Pinched Cylinder 44 BE41: Linear Pinched Cylinder Overview Element Type(s): C3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): ASE Input file(s): cylinder.dat 44.1 Problem Description The problem consists of a thin cylinder shell with rigid end diaphragms, which is loaded in its middle by two oppositely directed radially point loads, as shown in Fig 44.1. The maximum deflection at the center of the cylinder, under the point loads, is determined and verified for refined meshes [44]. 1 . 0 1 . 0 p p Figure 44.1: Problem Description 44.2 Reference Solution There is a convergent numerical solution of = 1.8248·10 −5 for the radial displacement at the loaded points, as given by Belytschko [45]. This problem is one of the most severe tests for both inextensional bending and complex membrane states of stress [46] . SOFiSTiK 2014 | VERiFiCATiON MANUAL 179 BE41: Linear Pinched Cylinder 44.3 Model and Results The properties of the model are defined in Table 44.1. The geometric parameters and the material are all dimensionless. The compressive point load p = 1.0 is applied radially and in opposite directions at the middle nodes of the cylinder, as shown in Fig. 44.1. Using symmetry, only one-eighth of the cylinder needs to be modeled, as shown in Fig. 44.2. For the simplified model only one fourth of the load p∗ is applied at the the upper middle node, as it can be visualised in Fig. 44.2. The end of the cylinder is supported by a rigid diaphragm [47], while at the two edges of the cylinder, parallel to the - and y- axis, symmetry support conditions are employed. In the plane of middle of the cylinder, the displacements in the longitudial direction, as well as the rotations around - and y- axis are fixed. The example allows the verification of the calculation of thin shells with increasingly refined regular meshes. Table 44.1: Model Properties Material Properties Geometric Properties Loading E = 3.0 · 10 6 MP L = 600, = 300 p = 1.0 μ = 0.30 r = 300 p∗ = 0.25 t = 3 p∗ r Figure 44.2: FEM model Table 44.2: Normalised Point-Load Displacement / with Mesh Refinement Element/Side Conforming Element Non-Conforming Element 4 0.4525 0.5917 8 0.8214 0.9056 16 0.9701 1.0082 180 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE41: Linear Pinched Cylinder The results are presented in Fig 44.3 and Table 44.2, where they are compared to the analytical solution as presented in Section 44.2. Two element formulations are considered. The first one, represented by the red curve, corresponds to the 4-node regular conforming element whereas the second, represented by the purple curve corresponds to the non-conforming element with six functions, offering a substantial improvement of the results. 0 2 4 6 8 10 12 14 16 18 0 0.2 0.4 0.6 0.8 1 1.2 Number of Elements N o r m a l i s e d D i s p l a c e m e n t / Analytical Solution Conforming Element Non-Conforming Element Figure 44.3: Convergence Diagram Figure 44.4: Deformed Shape 44.4 Conclusion The example allows the verification of the calculation of thin shells. For increasing refined meshes, the calculated result for both types of elements convergence fast to the predetermined analytical solu- SOFiSTiK 2014 | VERiFiCATiON MANUAL 181 BE41: Linear Pinched Cylinder tion. The advantage of the utilisation of the non-conforming element is evident, since it is in excellent agreement with the analytical solution for a refined mesh. 44.5 Literature [44] VDI 6201 Beispiel: Softwaregest ¨ utze Tragwerksberechnung - Beispiel Zylinderschale mit starren Endscheiben, Kategorie 1: Mechanische Grundlagen. Verein Deutscher Ingenieure e. V. [45] T. Belytschko et al. “Stress Projection for Membrane and Shear Locking in Shell Finite Elements”. In: Computer Methods in Applied Mechanics and Engineering 53(1-3) (1985), pp. 221–258. [46] T. Rabczuk, P. M. A. Areias, and T. Belytschko. “A meshfree thin shell method for non-linear dynam- ic fracture”. In: International Journal for Numerical Methods in Engineering 72(5) (2007), pp. 524– 548. [47] P. Krysl and T. Belytschko. “Analysis of thin shells by the element-free Galerkin method”. In: Inter- national Journal for Solids and Structures 33(20-22) (1996), pp. 3057–3080. 182 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE42: Thick Circular Plate 45 BE42: Thick Circular Plate Overview Element Type(s): C3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): ASE Input file(s): thick plate.dat 45.1 Problem Description The problem consists of a circular plate with a constant area load, as shown in Fig. 45.1. The system is modelled as a plane problem and the deflection in the middle of the plate is determined for various thicknesses [48]. p Figure 45.1: Problem Description 45.2 Reference Solution Depending on the various thicknesses of the plate, the maximum deflection in the middle of the plate can be obtained as = B + S , where B is the dislacement due to bending and S is the displacement due to shear strains, determined as follows [49]: B = p · r 4 64 · K (5 + μ) (5 + μ) (45.1) K = E · h 3 12(1 − μ 2 ) (45.2) SOFiSTiK 2014 | VERiFiCATiON MANUAL 183 BE42: Thick Circular Plate S = 1.2 · p · r 2 4 · G· h (45.3) where p is the load ordinate, r the radius, E the elasticity modulus, h the plate thickness, μ the Poissons ratio and G the shear modulus. The maximum bending moment at the middle of the plate is independent of the plate thickness and corresponds for the specific load case to M = M y = p · r 2 16 · (3 + μ) = 4928.125 [kNm/ m] (45.4) 45.3 Model and Results The properties of the model are defined in Table 45.1. The plate is modelled as a plane system with three degrees of freedom, z , ϕ , ϕ y , per node and z hinged at the edge, as shown in Fig. 45.1. The weight of the system is not considered. A constant area load p = 1000 kN/ m 2 is applied, as shown in Fig. 45.1. The system is modelled with 1680 quadrilateral elements, as presented in Fig. 45.2, and a linear analysis is performed for increasing thicknesses. The results are presented in Table 45.2 where they are compared to the analytical solution calculated from the formulas presented in Section 45.2 and the influence of the varying thickness is assesed. Table 45.1: Model Properties Material Properties Geometric Properties Loading E = 3000 kN/ cm 2 h = 0.5 − 2.5 m p = 1000 kN/ m 2 G = 1300 kN/ cm 2 r = 5 m μ = 0.154 D = 10 m y x Figure 45.2: FEM model The maximum bending moment is calculated at the middle of the plate, as M = M y = 4932.244 184 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE42: Thick Circular Plate [kNm/ m] with a deviation of 0.08 %. Table 45.2: Results h [m] h/ D Analytical z [mm] SOF. z [mm] |e r | [%] 0.50 0.05 137.413 137.440 0.02 1.00 0.10 17.609 17.618 0.05 1.50 0.15 5.431 5.437 0.11 2.00 0.20 2.418 2.421 0.14 2.50 0.25 1.321 1.324 0.23 Figure 45.3: Displacements 45.4 Conclusion The example allows the verification of the calculation of thick plates. It has been shown, that the calcu- lated results are in very good agreement with the analytical solution even for thicker plates. 45.5 Literature [48] VDI 6201 Beispiel: Softwaregest ¨ utze Tragwerksberechnung - Beispiel Dicke Platte, Kategorie 1: Mechanische Grundlagen. Verein Deutscher Ingenieure e. V. [49] F. U. Mathiak. Ebene Fl ¨ achentragwerke Teil II, Grundlagen der Plattentheorie. Hochschule Neubrandenburg. 2011. SOFiSTiK 2014 | VERiFiCATiON MANUAL 185 BE42: Thick Circular Plate 186 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE43: Panel with Circular Hole 46 BE43: Panel with Circular Hole Overview Element Type(s): C3D Analysis Type(s): STAT Procedure(s): Topic(s): Module(s): ASE Input file(s): vdi 3 panel.dat 46.1 Problem Description The problem consists of a rectangular panel with a circular hole in its middle, loaded by a constant linear load p on the vertical edges, as shown in Fig. 46.1. The system is modelled as a plane stress problem and the maximum stress at the edge of the hole is determined and verified for various meshes [50]. p p t L/ 2 h h d L/ 2 D A A y Figure 46.1: Problem Description 46.2 Reference Solution The maximum stress σ A,,m resulting from a load p, at the edge of the hole can be determined at points A and A across a vertical cut, visualised in Fig. 46.1, as follows [51] [52]: σ A,,m = K t · σ ,nom (46.1) where P = p · D = 1000 [kN] (46.2) σ ,nom = P t · (D− d) = 33.33 [N/ mm 2 ] (46.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 187 BE43: Panel with Circular Hole K t = 3.000 − 3.140 · (d/ D) + 3.667 · (d/ D) 2 − 1.527 · (d/ D) 3 , (0 < d/ D < 1) (46.4) 46.3 Model and Results The properties of the model are defined in Table 46.1. Plane stress conditions are assumed, with two degrees of freedom, , y , per node, and a line load p = 200.0 kN/ m is applied at both vertical ends. The length of the panel is considered to be large enough in order to avoid any disturbances in the area of the hole, due to the loaded ends. Due to symmetry conditions only one fourth of the panel is modelled. Table 46.1: Model Properties Material Properties Geometric Properties Loading E = 2.1 · 10 5 MP L = 15.00 m p = 200.0 kN/ m ν = 0.30 D = 5.00 m, d = 2.00 m h = 1.50 m , t = 0.01 m [390] [310] [168] [44] (a) Structured Meshing [804] [424] [140] [44] (b) Unstructured Meshing Figure 46.2: FEM Models 188 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE43: Panel with Circular Hole Four manually structured meshes, with refinement around the hole area, are considered, shown in Fig. 46.2(a), with increasing number of quadrilateral elements and the convergence behaviour is evaluated. For the sake of comparison, unstructured meshes, shown in Fig. 46.2(b), are also considered. The number of degrees of freedom for every mesh is given in the red brackets. The results are presented in Fig 46.3 where they are compared to the analytical solution calculated from the formulas presented in Section 46.2. For the case of structured meshing two element formulations are considered. The first one, represented by the red curve, corresponds to the 4-node regular conforming element whereas the second, represented by the purple curve corresponds to the non-conforming element with six functions. The blue curve represents the unstructured meshing. 0 100 200 300 400 500 600 700 800 900 40 45 50 55 60 65 70 75 Number of Degrees of Freedom n DOF σ A , , m [ N / m m 2 ] Analytical: 74.43 [ N/ mm 2 ] Structured Mesh - Conforming Element Structured Mesh - Non-conforming Element Unstructured Mesh - Non-Conforming Element Figure 46.3: Convergence Diagram 74.75 16.54 Figure 46.4: Maximum Stresses σ ,m The regular 4-node element is characterised through a bilinear accretion of the displacements and rotations. This element is called conforming, because the displacements and the rotations between elements do not have any jumps. The results at the gravity centre of the element represent the actual internal force variation fairly well, while the results at the corners are relatively useless, especially the ones at the edges or at the corners of a region. On the other hand the non-conforming elements, are based one the idea of describing more stress states through additional functions that their value is zero SOFiSTiK 2014 | VERiFiCATiON MANUAL 189 BE43: Panel with Circular Hole at all nodes. As a rule, these functions lead to a substantial improvement of the results, however, they violate the continuity of displacements between elements and thus they are called non-conforming. 46.4 Conclusion The example allows the verification of the calculation of plane stress problems and the convergence be- haviour of quadrilateral elements. For both types of elements, the calculated results convergence rather fast to the predetermined precise analytical solution, within acceptable tolerance range. Furthermore, it is evident that the unstructured mesh, which is a more often choice in practice, gives results which are in very good agreement with the analytical solution. 46.5 Literature [50] VDI 6201 Beispiel: Softwaregest ¨ utze Tragwerksberechnung - Beispiel Scheibe mit kreisf ¨ ormigem Loch - Konvergenztest f ¨ ur Scheibenelemente, Kategorie 1: Mechanische Grundlagen. Verein Deutscher Ingenieure e. V. [51] C. Petersen. Stahlbau. Grundlagen der Berechnung und baulichen Ausbildung von Stahlbauten. Vieweg, 1997. [52] W. D. Pilkey. Formulaes for Stress, Strain and Structural Matrices. Wileys & Sons, 1994. 190 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE44: Undrained Elastic Soil Layer Subjected to Strip Loading 47 BE44: Undrained Elastic Soil Layer Subjected to Strip Loading Overview Element Type(s): C2D Analysis Type(s): STAT Procedure(s): Topic(s): SOIL Module(s): TALPA Input file(s): soil layer el undr.dat 47.1 Problem Description The example concerns the behavior of the rectangular soil layer subjected to an uniform strip loading of intensity p acting on the surface. Base of the soil is rigidly fixed while the sides are laterally constrained. Geometry, load and boundary conditions are depicted in Fig. 47.1. The soil material is elastic, isotropic and saturated with water. Two soil conditions have been analyzed - drained and undrained. The drained and undrained displacements and stresses obtained by the finite element method are compared with the analytical solution. 4 h = 2 p A B C Figure 47.1: Problem Definition 47.2 Reference Solution The analytical solution to the problem obtained using a Fourier series analysis is provided in [53]. 47.3 Model and Results Elastic, isotropic soil under drained and undrained conditions has been analyzed. Material, geometry and loading properties are summarized in Table 47.1. The undrained soil condition is considered with the help of the method based on the undrained effective stress (σ ) analysis using effective material parameters. G and ν are effective soil parameters, while B represents the Skempton’s B-parameter. Self-weight is not taken into consideration. SOFiSTiK 2014 | VERiFiCATiON MANUAL 191 BE44: Undrained Elastic Soil Layer Subjected to Strip Loading Table 47.1: Model Properties Material Geometry Loading G, ν = 0.3 p B = 0.998 h = 4 ρ = 0.0 kg/ m 3 Finite element mesh of the model is shown in Fig. 47.2. Mesh is regular and consist of quadrilateral finite elements. Figure 47.2: Finite Element Model The drained and undrained vertical displacement of the surface nodes along the A−B line are compared with the analytical solution from [53] and depicted in Fig. 47.3. 0 1 2 3 4 −0.1 0 0.1 0.2 0.3 / G· y p·h analytical, drained fem, drained analytical, undrained fem, undrained Figure 47.3: Vertical displacement y of the surface The drained and undrained horizontal and vertical total stresses (σ = σ + p e ) in the nodes along the vertical A−C line have been computed and compared with the analytical ones, as show in Figures 47.4a and 47.4b. 192 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE44: Undrained Elastic Soil Layer Subjected to Strip Loading −1 −0.8 −0.6 −0.4 −0.2 0 0 0.2 0.4 0.6 0.8 1 σ / p y/ h (a) Horizontal stress σ −1 −0.9 −0.8 −0.7 −0.6 0 0.2 0.4 0.6 0.8 1 σ y / p y/ h (b) Vertical stress σ y Figure 47.4: Stresses beneath footing center Pore excess pressure (p e ) distribution for the undrained condition along the center line (A − C) is shown in Fig. 47.5. −1 −0.8 −0.6 −0.4 0 0.2 0.4 0.6 0.8 1 p e / p y/ h Figure 47.5: Excess pore pressure p e beneath footing centre 47.4 Conclusion This example verifies that the drained and undrained displacements and stresses obtained by the finite element method are in a good agreement with the analytical solution. 47.5 Literature [53] J.R. Booker, J.P. Carter, and J.C. Small. “An efficient method of analysis for the drained and undrained behaviour of an elastic soil”. In: International Journal of Solids and Structures 12.8 (1976), pp. 589 –599. SOFiSTiK 2014 | VERiFiCATiON MANUAL 193 BE44: Undrained Elastic Soil Layer Subjected to Strip Loading 194 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE45: One-Dimensional Soil Consolidation 48 BE45: One-Dimensional Soil Consolidation Overview Element Type(s): C2D Analysis Type(s): STAT Procedure(s): Topic(s): SOIL Module(s): TALPA Input file(s): soil 1d consolidation.dat 48.1 Problem Description In the following example a one-dimensional consolidation problem has been analyzed. The soil layer is subjected to an uniform loading of the intensity p 0 acting on the surface. Base of the soil is rigidly fixed while the sides are laterally constrained. Only the soil surface is allowed to drain. Geometry, load and boundary conditions are depicted in Fig. 48.1. The soil material is elastic, isotropic and saturated with water. The surface settlements and pore excess pressures for the two extreme cases (time zero and time infinity) of the consolidation process are compared to the analytical solution. y z h p 0 G, ν , ν , ρ Figure 48.1: Problem Definition 48.2 Reference Solution The analytical solution to the problem was given by Terzaghi in 1925 [54]. The solution assumes that the soil is saturated with water, the soil and water are non-deformable, the volume change takes place only on the account of the water drainage and the Darcy’s filtration law applies. Then the differential equation of the one-dimensional process of consolidation for the excess water pressure p e can be written as [55]: ∂p e ∂t = c ∂ 2 p e ∂z 2 , (48.1) where: c = k · E s / γ coefficient of consolidation, SOFiSTiK 2014 | VERiFiCATiON MANUAL 195 BE45: One-Dimensional Soil Consolidation E s stiffness modulus, k coefficient of permeability, γ unit weight of water, h soil thickness, z = h − y elevation. Taking into account the initial and boundary conditions for the problem illustrated by Fig. 48.1 t = 0 and 0 ≤ z < h ⇒ p e = p 0 , (48.2a) 0 ≤ t ≤ ∞ and z = 0 ⇒ ∂p e ∂z = 0, (48.2b) 0 ≤ t ≤ ∞ and z = h ⇒ p e = 0, (48.2c) t = ∞ and 0 ≤ z ≤ h ⇒ p e = 0, (48.2d) the Eq. 48.1 can be analytically solved for p e as a function of the time t and the elevation z = h − y p e (t, z) p 0 = 4 π · ∞ j=0 1 2j + 1 · sin _ (2j + 1) π 2 z h _ · e −(2j+1) 2 π 2 / 4·T (48.3) where: p 0 surface pressure, T = c / h 2 · t time factor. With the known change of excess pore pressure with respect to time, the settlement due to consolidation s(t) at time t can be determined s(t) = p 0 h E s · _ _ 1 − 8 π 2 · ∞ j=0 1 (2j + 1) 2 · e −(2j+1) 2 π 2 / 4·T _ _ . (48.4) For the time infinity, the excess pore pressures will completely dissipate (see Eq. 48.2d) and the final settlements due to consolidation s ∞ will be s ∞ = s(t = ∞) = p 0 h E s . (48.5) 48.3 Model and Results Elastic, isotropic soil under undrained and drained conditions has been analyzed. Material, geometry and loading properties are summarized in Table 48.1. The undrained soil condition is considered with the help of the method based on the undrained effective stress (σ ) analysis using effective material parameters. G and ν are effective soil parameters, while B represents the Skempton’s B-parameter. Self-weight is not taken into consideration. 196 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE45: One-Dimensional Soil Consolidation Table 48.1: Model Properties Material Geometry Loading G, ν = 0.0 h p 0 B = 0.998 = h ρ = 0.0 kg/ m 3 Finite element mesh of the model is shown in Fig. 48.2. Mesh is regular and consist of quadrilateral finite elements. Figure 48.2: Finite Element Model The results are summarized in the Table 48.2. Final settlement of the surface of the soil due to consol- idation s ∞ is compared to the analytical solution given by Eq. 48.5. The excess water pressures p e for the time zero (T = 0, undrained) and time infinity (T = ∞, drained) are compared to the analytical solutions from Eqs. 48.2a and 48.2d. Table 48.2: Results T = 0 |e| T = ∞ |e| SOF. Ref. [−] SOF. Ref. [−] s(T ) · E s / (p 0 h) [−] – 1.0 1.0 0.0 p e (T )/ p 0 [−] 0.994 1.000 0.006 0.000 0.000 0.000 48.4 Conclusion The example verifies that the settlements and excess pore pressures for initial (t = 0) and finial (t = ∞) time of the consolidation process obtained by the finite element method are in a good agreement with the analytical solution. SOFiSTiK 2014 | VERiFiCATiON MANUAL 197 BE45: One-Dimensional Soil Consolidation 48.5 Literature [54] K. Terzaghi. Erdbaumechanik auf bodenphysikalischer Grundlage. Leipzig (usw.): F. Deuticke, 1925. [55] K. Terzaghi. Theoretical Soil Mechanics. Wiley, 1948. 198 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE46: Material Nonlinear Analysis of Reinforced Concrete Beam 49 BE46: Material Nonlinear Analysis of Reinforced Con- crete Beam Overview Element Type(s): B3D, SH3D Analysis Type(s): STAT, MNL Procedure(s): Topic(s): Module(s): STAR2, ASE Input file(s): nonl rein conc.dat 49.1 Problem Description The problem consists of a single span beam of reinforced concrete, subjected to a single load P in the middle of the span, as shown in Fig. 49.1. The material nonlinear behaviour of the beam is examined and compared to test results. P b h Figure 49.1: Problem Description 49.2 Reference Solution Materially nonlinear analysis is utilised more and more for the structural design in concrete construction. It is often overlooked that for such analysis, both in-depth knowledge of the computational algorithms as well as the behavior of the concrete in cracked condition, are required. The following simple example will serve for verification of material nonlinear calculations of reinforced concrete beams. It will also highlight the unavoidable variations in practice. Therefore, the individual test results are given below and not only the mean values. The load-displacement curves of seven identical concrete beams, which were prefabricated almost at the same time and under the same controlled conditions, are graphed below. As a reference solution, these load-displacement curves of the test beams are used. 49.3 Model and Results The properties of the model [56] are defined in Table 49.1. The simply supported beam is shown in Fig. 49.1, as well as the dimensions and the reinforcement of the beams. The total length of the span is tot = 3.0 m. The square rectangular cross-section with edge lengths of 20 cm is reinforced by four longitudinal bars of 10 mm and stirrups of 6-15 cm. For this example the stirrups are not influential and can be neglected. The load is applied at the midspan and the beam is loaded to failure. Self weight is accounted for. The material properties of the concrete, B 35 or equivalently a C 35, were determined on a total of twelve cylinders 150/ 300, and are given in Table 49.1. The concrete cover of the longitudinal reinforcement is c , = 2.4 cm. The reinforcing steel is a BST 500 S, following a SOFiSTiK 2014 | VERiFiCATiON MANUAL 199 BE46: Material Nonlinear Analysis of Reinforced Concrete Beam stress-strain law, as shown in Fig. 49.2. The results are presented in Fig. 49.4. The deflection in the middlespan is recorded and plotted versus the load. The expectance is for the numerical calculations to fall into the gray shaded area, which bounds the curves of the seven tests beams. Of particular importance, are the onset of cracking, the slope after the completion of the cracking and by the yielding of the reinforcement, as well as the limit load. SOFiSTiK results are presented by the three additional curves included in the original figure. For beam elements: (a) yellow color with triangles for concrete C 35, (b) red color with circles for concrete B 35. For quad elements: (c) black color with squares for concrete C 35. 0 100 200 300 400 500 600 700 0 10 20 30 40 50 60 0. . . .. . . 00 0 0 1 07 213 0 2 00 395 9 3 04 517 0 4 04 539 4 5 00 551 7 6 10 560 9 7 10 566 2 10 07 574 9 20 01 589 4 40 09 604 1 50 06 605 6 55 05 604 4 58 19 512 2 S t r e s s N / m m ² Strain mm/m . . . . . . . . . . . . . . . . . . . . . . . . . . . . Strain mm/m Stress N/mm² Figure 49.2: Stress-Strain Curve for Reinforcing Steel σ-m σ-u o/oo - 3 . 5 0 - 1 . 9 3 0 . 0 0.00 -60.00 -40.00 -20.00 0.00 MPa o/oo 0 . 0 0.00 -60.00 -40.00 -20.00 0.00 MPa C 35 B 35 σ-m σ-u σ-r - 3 . 5 0 - 2 . 4 0 - 2 . 0 0 - 1 . 2 0 Figure 49.3: Stress-Strain Curve for Concrete Table 49.1: Model Properties Material Properties Geometric Properties Loading Concrete Steel b = h = 20.0 cm P = 1 kN B 35 or C 35 BST 500S = 3.0 m until failure ρ = 2320 kg/ m 3 c , = 2.4 cm ƒ cm = 54.0 MN/ m 2 4 bars 10 mm 200 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE46: Material Nonlinear Analysis of Reinforced Concrete Beam Table 49.1: (continued) Material Properties Geometric Properties Loading E = 28000 MN/ m 2 10 20 30 40 50 60 70 80 90 100 25 20 15 10 5 0 L o a d i n k N Deflection in mm Figure 49.4: Load-Displacement Curves 49.4 Conclusion This example examines the material nonlinear analysis of reinforced concrete beams. It has been shown that the behaviour is captured accurately. 49.5 Literature [56] VDI 6201 Beispiel: Softwaregest ¨ utze Tragwerksberechnung - Beispiel Stofflich nichtlineare Berechnung von Stahlbetonbalken, Kategorie 1: Mechanische Grundlagen. Verein Deutscher In- genieure e. V. SOFiSTiK 2014 | VERiFiCATiON MANUAL 201 BE46: Material Nonlinear Analysis of Reinforced Concrete Beam 202 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE47: Pushover Analysis: SAC LA9 Building 50 BE47: Pushover Analysis: SAC LA9 Building Overview Element Type(s): B3D Analysis Type(s): MNL Procedure(s): EIGE Topic(s): EQKE Module(s): ASE, SOFiLOAD Input file(s): pushover sac la9.dat 50.1 Problem Description In this example a pushover analysis of a moment resisting frame structure is performed. The pushover curve is identified and compared to the reference solution, as described in Chopra [57]. Figure 50.1: Problem Description 50.2 Reference Solution In this Benchmark the interest is focused in the retrieval of the pushover curve. The steps involved in this process are described schematically in Figure 50.2. Important is the definition of the pushover lateral load case pattern. The pushover analysis is performed by subjecting the structure to this monotonically increasing load pattern of lateral forces. Here the first three eigenmodes of the structure will be used. Choosing the characteristic force and displacement of the structure, a so called pushover curve of the multi-degree-of-freedom (MDOF) system can be obtained. The force, here denoted as V b , is usually base-shear, while the displacement is a displacement of the characteristic point on the structure cnod , also called the roof displacement and the control node displacement. SOFiSTiK 2014 | VERiFiCATiON MANUAL 203 BE47: Pushover Analysis: SAC LA9 Building AQUA + SOFIMSH Mat. / Hinges + System SOFILOAD Push Load ASE Push Analysis SOFILOAD Pushover Curve cnod V b V b cnod Figure 50.2: Pushover curve determination workflow 50.3 Model and Results The properties of the model are presented in Table 50.1 and Figure 50.3. The model utilised in this Benchmark consists of the benchmark structure for the SAC project, as has been described by Gupta and Krawinkler [58], Chopra and Goel [57] and FEMA [59]. “The 9-story structure, was designed by Brandow & Johnston Associates for the SAC2 Phase II Steel Project. Although not actually constructed, this structure meets seismic code and represents typical medium-rise buildings designed for the Los Angeles, California, region. The building is square in plan and rises nine floors above ground in elevation. The bays are 9.15 m on center, in both directions, with five bays each in the north-south (N-S) and east-west (E-W) directions. The buildings lateral load- resisting system is composed of steel perimeter moment-resisting frames (MRFS) with simple (simple hinged connection) framing on the farthest south E-W frame. The columns are steel wide-flange sec- tions. The levels of the 9-story building are numbered with respect to the ground level, with the ninth level being the roof. The building has a basement level, denoted B-1. The column lines employ two-tier construction, i.e., monolithic column pieces are connected every two levels beginning with the first level. Column splices, which are seismic (tension) splices to carry bending and uplift forces, are located on the first, third, fifth, and seventh levels at h s = 1.83 m above the center-line of the beam to column joint. The column bases are modeled as pinned and secured to the ground (B-1). Concrete foundation walls and surrounding soil are assumed to restrain the structure at the ground level from horizontal displace- ment. The floor system is composed of steel wide-flange beams in acting composite action with the floor slab. Each frame resists one half of the seismic mass associated with the entire structure. The seismic mass of the structure is due to various components of the structure. The model is based on centerline dimensions of the bare frame in which beams and columns extend from centerline to centerline. The strength, dimension, and shear distortion of panel zones are neglected.” [57] “Shear deformations in beam and column elements are neglected. Plastic zones in beams and columns are modeled as point hinges. The hysteretic behavior at plastic hinge locations is described by a bilinear moment-rotation diagram. All elements have 3% strain hardening. Expected rather than nominal yield strength values are used (49.2 ks for A 36 steel and 57.6 ks for A 50 steel). Viscous damping 2% is used in first mode and at T = 0.2 sec.” [59] 204 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE47: Pushover Analysis: SAC LA9 Building Table 50.1: Model Properties Material Geometry A 50 = 9.15 m A 36 h b = 3.65 m, h g = 5.49 m h ƒ = 3.96 m, h s = 1.83 m N 8 × h ƒ h g h b 5 × B-1 Ground 1st 2nd 3rd 4th 5th 6th 7th 8th 9th Figure 50.3: Model Description (a) Mode 1 (b) Mode 2 (c) Mode 3 Figure 50.4: Eigenmodes Table 50.2: First three natural-vibration periods Periods Ref. [57] SOF. T 1 2.27 2.26 T 2 0.85 0.85 T 3 0.49 0.49 SOFiSTiK 2014 | VERiFiCATiON MANUAL 205 BE47: Pushover Analysis: SAC LA9 Building 0 20 40 60 80 0 0.2 0.4 0.6 0.8 1 ·10 4 Roof Displacement [cm] B a s e S h e a r [ k N ] Chopra [57] SOFiSTiK (a) Mode 1 Pushover Curve 0 5 10 15 20 25 0 0.2 0.4 0.6 0.8 1 ·10 4 Roof Displacement [cm] B a s e S h e a r [ k N ] Chopra [57] SOFiSTiK (b) Mode 2 Pushover Curve 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 ·10 4 Roof Displacement [cm] B a s e S h e a r [ k N ] Chopra [57] SOFiSTiK (c) Mode 3 Pushover Curve Figure 50.5: Pushover Curves The first three vibration modes and periods of the building for linearly elastic vibration are shown in Figure 50.4. The vibration periods are 2.26, 0.85, and 0.49 sec, respectively. The force distributions of these first three modes are used in the pushover analysis in order to retrieve the pushover curves. The pushover curves for the first three eigenmodes, are presented in Figures 50.5. The hinge formation distribution for each pushover analysis, corresponding to approximatelly the last load case depicted in 206 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE47: Pushover Analysis: SAC LA9 Building each pushover curve, is presented in Figures 50.6. -2.23 2.19 -2.19 2.19 -2.19 -2.17 2.16 2.14 -2.08 2.00 2.00 -2.00 1.99 -1.99 -1.98 -1.95 1.95 1.84 1.84 -1.84 -1.84 1.84 -1.83 1.80 -1.71 -1.65 1.65 -1.65 1.65 1.62 -1.62 1.59 -1.50 -1.35 -1.19 -0.989 -0.982 0.966 0.965 -0.931 0 . 6 2 5 0 . 6 1 2 0 . 6 0 9 0 . 4 8 0 0 . 2 9 7 -0.291 -0.267 -0.172 -0.153 0.130 -0.127 0.117 0.0742 -0.0676 (a) Mode 1 -1.80 1.76 -1.75 1.75 -1.75 1.75 -1.74 1.71 -1.50 -1.38 1.38 1.38 -1.37 -1.36 1.34 1.34 -1.22 1.07 1.04 -1.02 -1.02 1.01 -1.01 -0.892 -0.571 -0.132 0 . 1 1 1 0 . 1 0 3 0 . 0 8 9 9 -0.0815 0.0217 (b) Mode 2 -0.942 0.856 -0.920 0.853 0.834 0.830 0.754 0.727 (c) Mode 3 Figure 50.6: Hinge distribution 50.4 Conclusion This example adresses the determination of the pushover curve for a benchmark structure. It has been shown that the results obtained are in a good agreement with the reference given by Chopra [57]. 50.5 Literature [57] A.K. Chopra and R. K. Goel. A Modal Pushover Analysis Procedure to Estimate Seismic De- mands for Buildings: Theory and Preliminary Evaluation. Tech. rep. PEER Report 2001/03. Pacific Earthquake Engineering Research Center - University of California Berkeley, 2001. [58] A. Gupta and H. Krawinkler. Seismic Demands for Performance Evaluation of Steel Moment Re- sisting Frame Structures. Tech. rep. Report No. 132. The John A. Blume Earthquake Engineering Center, 1999. [59] Prepared for the SAC Joint Venture Partnership by Helmut Krawinkler. State of the Art Report on Systems Performance of Steel Moment Frames Subject to Earthquake Ground Shaking. Tech. rep. FEMA-355C. Federal Emergency Management Agency (FEMA), 2000. SOFiSTiK 2014 | VERiFiCATiON MANUAL 207 BE47: Pushover Analysis: SAC LA9 Building 208 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE48: Triaxial Consolidated Undrained (CU) Test 51 BE48: Triaxial Consolidated Undrained (CU) Test Overview Element Type(s): CAXI Analysis Type(s): MNL Procedure(s): LSTP Topic(s): SOIL Module(s): TALPA Input file(s): triaxial cu test.dat, triaxial cu test 200.dat 51.1 Problem Description In this example a consolidated undrained (CU) triaxial test on a loose Hostun-RF sand is simulated. The specimen is subjected to different levels of triaxial confining stresses and the results are compared to those of the experimental tests and numerical simulations, as described in Wehnert [60]. D H σ 3 σ 1 Figure 51.1: Problem Description 51.2 Reference Solution In this example two soil models are utilised, the Mohr-Coulomb (MC) and the Hardening Soil (HS) model. Further details on these two models can be found in Benchmarks 20 and 21. The choice of the appropriate model for the soil is of a significant importance. For example, MC mod- el can significantly overestimate the undrained shear strength for a normally consolidated soil. More advanced models can provide better estimate for the undrained strength than the MC model. In par- ticular, the HS model is able to represent the change of the excess pore water pressure occurring under undrained shear loading conditions, providing more realistic effective stress paths and values for undrained shear strength. However, the results of the analysis with the Hardening Soil model are very sensitive to the used model parameters and the choice of the dilatancy model. Therefore, in this example for the HS model different dilatancy formulations are tested and their influence on the result examined. A well-established stress dilatancy theory is described by Rowe [61], where the so-called mobilized SOFiSTiK 2014 | VERiFiCATiON MANUAL 209 BE48: Triaxial Consolidated Undrained (CU) Test dilatancy angle ψ m is defined as sinψ m = sinφ m − sinφ cs 1 − sinφ m sinφ cs (51.1) Therein, the critical state friction angle φ cs marks the transition between contractive (small stress ratios with φ m < φ cs ) and dilatant (higher stress ratios with φ m > φ cs ) plastic flow. The mobilized friction angle φ m in Equation 51.1 is computed according to sinφ m = σ 1 − σ 3 2c · cot φ − σ 1 − σ 3 (51.2) At failure, when φ m ≡ φ, also the dilatancy angle reaches its final value ψ m ≡ ψ. Accordingly, from Equation 51.1 the critical state friction angle can be derived as sinφ cs = sinφ − sinψ 1 − sinφsinψ (51.3) It has been recognized that in some cases the Rowe’s model for dilatancy angles (Eq. 51.1) can over- estimate the contractive behavior of the soil at low mobilized friction angles, φ m < φ cs . As a remedy, several researchers have developed modified formulations based on the original Rowe’s model. Some of these models which are implemented in SOFiSTiK are described below. One of the models which does not require additional input parameters is the model according to Soreide [62] which modifies the Rowe’s formulation by using the scaling factor sinφ m / sinφ sinψ m = sinφ m − sinφ cs 1 − sinφ m sinφ cs · sinφ m sinφ . (51.4) Wehnert [60] proposed a model based on a lower cut-off value ψ 0 for the modification of the Rowe’s formulation from Eq. 51.1 at low mobilized friction angles sinψ m = _ _ _ _ _ sinψ 0 ; 0 < ψ m ≤ ψ Roe m sinφ m − sinφ cs 1 − sinφ m sinφ cs ; ψ Roe m < ψ m ≤ ψ . (51.5) This dilatancy model obviously requires a specification of an additional parameter, ψ 0 . 210 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE48: Triaxial Consolidated Undrained (CU) Test 5 15 25 φ = 35 -20 -10 0 ψ = 10 20 ψ 0 φ cs φ m [ ◦ ] ψ m [ ◦ ] Rowe Soreide Wehnert, ψ 0 = −3 ◦ Constant Figure 51.2: Comparison of models for mobilized dilatancy angle ψ m implemented in SOFiSTiK for φ = 35 ◦ and ψ = 10 ◦ 51.3 Model and Results The properties of the model are presented in Table 51.1. Two material models are considered: the Mohr- Coulomb and the Hardening Soil, which is combined with the different dilatancy models as described by the formulations presented in Section 51.2. For the model according to Wehnert (Eq. 51.5) additional parameter, dilatancy ψ 0 at low stress ratios, is used. The undrained calculation is conducted in the form of effective stresses with effective shear parameters (c , φ ) and stiffness parameters. Skempton’s parameter B ≈ 0.9832 (corresponding undrained Poisson’s ratio is ν = 0.495) is considered to describe the incompressibility of the pore water and saturated soil [60]. The analysis is carried out using an axisymmetric model. Two confining stress levels are considered, σ c = 200 and 300 kP. The undrained triaxial test on loose Hostun-RF sand is used as a reference. More information about the Hostun-RF sand can be found in Wehnert [60]. Table 51.1: Model Properties Material Geometry Loading E = 60.0 MN/ m 2 E s,reƒ = 16.0 MN/ m 2 H = 0.09 m Phase I: ν r = 0.25 E 50,reƒ = 12.0 MN/ m 2 D = 0.036 m σ 1 = σ 3 = σ c = γ = 0.0 MN/ m 3 m = 0.75 = 200, 300 kP c = 0.01 kN/ m 2 R ƒ = 0.9 Phase II: φ = 34 ◦ K 0 = 0.44 σ 3 = σ c = 200, 300 kP ψ = 2 ◦ B = 0.9832 σ 1 = σ > σ c ψ 0 = −4 ◦ The results, as calculated by SOFiSTiK, are presented in Figures 51.3 - 51.9 (MC, HS-Rowe, HS-Cons, HS-Soreide and HS-Wehnert). Figures 51.3 - 51.8, also include the results of the numerical simulations and of the experimental tests from Wehnert [60] (Wehnert, Exp. 1 and Exp. 2). On a p − q diagram, apart from the effective stress paths (ESP), the total stress paths (TSP) as well as the Mohr-Coulomb SOFiSTiK 2014 | VERiFiCATiON MANUAL 211 BE48: Triaxial Consolidated Undrained (CU) Test failure condition (MC failure) based on the used shear parameters, c and φ , are displayed. First the numerical simulation results by Wehnert [60] are compared to the results from the laboratory tests (Exp. 1 and Exp. 2). Although the oedometer and the drained triaxial tests (see also Benchmark 49) show good agreement with the results from the laboratory tests, the results from the undrained triaxial tests show deviation from the experimental results (see Figs. 51.3 - 51.8) 1 . The difference comes mainly as a result of the used dilatancy model (Eq. 51.5) and the choice of the model parameters, i.e. the peak dilatancy angle ψ and the lower cut-off dilatancy angle ψ 0 . Comparing the results of the development of the deviatoric stress q and the excess pore water pressure p e between the experiment and the calculation, one can notice a considerable difference, both for the confining stress level of 200 kP as well as for the level of 300 kP (Figs. 51.4, 51.5, 51.7 and 51.8). As explained in [60], the test sample with confining stress of 200 kP behaves significantly more dilatant than the sample with the confining stress of 300 kP. Since only one material model has been used to model the soil, only one peak dilatancy angle can be used to represent the dilatancy effects of both test cases. This peak dilatancy angle of ψ = 2 ◦ represents therefore a compromise, leading to a underestimation of the results for a test with a smaller confining stress level and to overestimation of the results with larger confining stress level at higher mobilized friction angels. Further differences arise from the chosen dilatancy model and the used lower cut-off dilatancy angle ψ 0 = −4 ◦2 . Due to the presence of the negative mobilized dilatancy angle (ψ m < 0) at low stress levels, the soil has the tendency to decrease its volume (contraction) under increase of the deviatoric stress q (shear). However, since the soil is under undrained conditions, the volumetric strains cannot develop, and as a result the excess pore pressure increases under shear. The increase of the excess pore pressure means that the effective stresses will reduce (ESP lines curve to the left in the p− q plot, Figs. 51.3 and 51.6). With the increase of the stress level, the contractive behavior turns to dilatant, meaning that the negative rate of excess pore pressures (pore water under-pressure) will arise, excess pore pressures decrease and hence the effective stresses increase. This transition from contractant to dilatant behavior occurs when the mobilized friction angle φ m becomes larger than the phase transition angle φ ƒ which is approximately equal to the critical state friction angle φ cs (see Fig. 51.2). As further noted by Wehnert [60], due to the fact that mobilized dilatancy angle at low stress levels is slightly heigher and kept constant (ψ m = ψ 0 for 0 ≤ ψ m ≤ ψ Roe m , Eq. 51.5), the pore water under-pressures are overestimated. Next the SOFiSTiK results obtained using the same soil model and dilatancy formulation as in [60] (HS- Wehnert) can be compared with the reference numerical simulation results (Wehnert). They show good agreement. Finally, in other to illustrate the effect that the chosen dilatancy model can have on the results of the undrained soil, the results of the computation using the hardening soil model with different dilatancy formulations from Section 51.2 are included. 1 Note also that the experimental test results for different samples of the same soil deviate significantly from each other. 2 The used value ψ 0 = −4 ◦ is much higher than the values obtained from experimental tests, which range from −13 ◦ to −21 ◦ . The reason for choosing this higher value is due to the fact that the experimental test used to obtain the dilatancy parameters involve not only shear but also some normal stress application to the test samples [60]. 212 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE48: Triaxial Consolidated Undrained (CU) Test 51.3.1 Hostun-RF Sand, σ c = 200 kN/ m 2 0 50 100 150 200 250 300 0 50 100 150 200 250 300 350 400 p, p [kN/ m 2 ] q [ k N / m 2 ] MC failure TSP ESP, MC ESP, HS-Rowe ESP, HS-Cons ESP, HS-Soreide ESP, HS-Wehnert ESP, Wehnert [60] ESP, Exp. 1 [60] ESP, Exp. 2 [60] Figure 51.3: Effective stress path curve (q-p) 0 5 10 15 0 50 100 150 200 250 300 350 400 ϵ 1 [%] q [ k N / m 2 ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp.1 [60] Exp.2 [60] Figure 51.4: Deviatoric stress - axial strain curve (q-ϵ 1 ) SOFiSTiK 2014 | VERiFiCATiON MANUAL 213 BE48: Triaxial Consolidated Undrained (CU) Test 0 5 10 15 0 50 100 150 200 ϵ 1 [%] p e [ k N / m 2 ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp. 1 [60] Exp. 2 [60] Figure 51.5: Excess porewater pressure - axial strain curve (p e -ϵ 1 ) 51.3.2 Hostun-RF Sand, σ c = 300 kN/ m 2 0 50 100 150 200 250 300 350 400 450 0 50 100 150 200 250 300 350 400 p, p [kN/ m 2 ] q [ k N / m 2 ] MC failure TSP ESP, MC ESP, HS-Rowe ESP, HS-Cons ESP, HS-Soreide ESP, HS-Wehnert ESP, Wehnert [60] ESP, Exp. 1 [60] Figure 51.6: Effective stress path curve (q-p) 214 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE48: Triaxial Consolidated Undrained (CU) Test 0 5 10 15 20 0 50 100 150 200 250 300 350 400 450 ϵ 1 [%] q [ k N / m 2 ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp.1 [60] Figure 51.7: Deviatoric stress - axial strain curve (q-ϵ 1 ) 0 5 10 15 0 50 100 150 200 250 300 ϵ 1 [%] p e [ k N / m 2 ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp. 1 [60] Figure 51.8: Excess porewater pressure - axial strain curve (p e -ϵ 1 ) SOFiSTiK 2014 | VERiFiCATiON MANUAL 215 BE48: Triaxial Consolidated Undrained (CU) Test 0 5 10 15 20 25 30 35 −30 −20 −10 0 φ m [ ◦ ] ψ m [ ◦ ] HS-Rowe HS-Cons HS-Soreide HS-Wehnert Figure 51.9: Mobilised dilatancy angle - friction angle curve (ψ m -φ m ) 51.4 Conclusion This example concerning the consolidated undrained triaxial test of a loose sand soil verifies that the Hardening Soil material model in combination with an appropriate choice of model parameters and dilatancy model is able to capture important behavior characteristics of the undrained soil. The numerical results are in a good agreement with the reference solution provided by Wehnert [60]. 51.5 Literature [60] M. Wehnert. Ein Beistrag zur dreainerten und undrainerten Analyse in der Geotechnik. Institut f ¨ ur Geotechnik, Universit ¨ at Stuttgart: P. A. Vermeer, 2006. [61] P.W. Rowe. “The stress-dilatancy relation for static equilibrium of an assembly of particles in con- tact”. In: Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sci- ences 269.1339 (1962), pp. 500–527. [62] O. K. Soreide. “Mixed hardening models for frictional soils”. PhD thesis. NTNU Norges teknisk- naturvitenskapelige universitet, 2003. 216 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE49: Triaxial Drained Test 52 BE49: Triaxial Drained Test Overview Element Type(s): CAXI Analysis Type(s): MNL Procedure(s): LSTP Topic(s): SOIL Module(s): TALPA Input file(s): triaxial d test.dat, triaxial d test 100.dat 52.1 Problem Description In this example a drained (D) triaxial test on a loose Hostun-RF sand is simulated. The specimen is subjected to different levels of triaxial confining stresses and the results are compared to those of the experimental tests and numerical simulations, as described in Wehnert [60]. D H σ 3 σ 1 Figure 52.1: Problem Description 52.2 Reference Solution In this example, the same triaxial test described in Benchmark 48 is examined, but for the case of a drained sample. Two soil models are utilised, the Mohr-Coulomb (MC) and the Hardening Soil (HS) model with different dilatancy configurations. Further details on the material models can be found in Benchmarks 48. 52.3 Model and Results The properties of the model are presented in Table 52.1. Two material models are considered: the Mohr-Coulomb and the Hardening Soil, which is combined with the different dilatancy configurations as described by the formulations presented in Section 52.2 in Benchmark 48. The analysis is carried out using an axisymmetric model. Two confining stress levels are considered, SOFiSTiK 2014 | VERiFiCATiON MANUAL 217 BE49: Triaxial Drained Test σ c = 100 and 300 kP. The drained triaxial test on loose Hostun-RF sand is used as a reference. More information about Hostun-RF sand can be found in Wehnert [60] and Benchmark 48. Table 52.1: Model Properties Material Geometry Loading E = 60.0 MN/ m 2 E s,reƒ = 16.0 MN/ m 2 H = 0.09 m Phase I: ν r = 0.25 E 50,reƒ = 12.0 MN/ m 2 D = 0.036 m σ 1 = σ 3 = σ c = γ = 0.0 MN/ m 3 m = 0.75 = 100, 300 kP c = 0.01 kN/ m 2 R ƒ = 0.9 Phase II: φ = 34 ◦ K 0 = 0.44 σ 3 = σ c = 100, 300 kP ψ = 2 ◦ B = 0.9832 σ 1 = σ > σ c ψ 0 = −4 ◦ The results, as calculated by SOFiSTiK, are presented in Figures 52.2 - 52.8 (MC, HS-Rowe, HS-Cons, HS-Soreide and HS-Wehnert). Figures 52.2 - 52.7, also include the results of the numerical simulations and of the experimental tests from Wehnert [60] (Wehnert, Exp. 1 and Exp. 2). On a p − q diagram the Mohr-Coulomb failure condition (MC failure) based on the used shear parameters, c and φ , is also displayed. If we first analyse the reference curves from Wehnert [60], we will notice, that the agreement between the numerical simulation and the experimental tests is quite good. Comparing the SOFiSTiK results for the HS model with the dilatancy model acc. to Wehnert (HS- Wehnert) with the reference numerical results from Wehnert [60], we can notice that the stress paths p-q are captured exactly for both σ c -stress levels. Accordingly, the deviatoric stress q versus the axial strain ε 1 curve fits very well to the reference results. For the case of the strain curves some deviation in results is identified and it seems that the Soreide dilatancy model shows better agreement with the simulation results from Wehnert. 218 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE49: Triaxial Drained Test 52.3.1 Hostun-RF Sand, σ c = 100 kN/ m 2 0 100 200 300 400 500 600 0 100 200 300 400 500 600 p, p [kN/ m 2 ] q [ k N / m 2 ] MC failure MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1 Figure 52.2: Effective stress path curve (q-p) 0 5 10 15 20 0 50 100 150 200 250 300 350 400 ϵ 1 [%] q [ k N / m 2 ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp.1 Figure 52.3: Deviatoric stress - axial strain curve (q-ϵ 1 ) SOFiSTiK 2014 | VERiFiCATiON MANUAL 219 BE49: Triaxial Drained Test 0 5 10 15 20 0.000 0.200 0.400 0.600 0.800 1.000 1.200 1.400 1.600 1.800 2.000 ϵ 1 [%] ϵ [ % ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1 Figure 52.4: Volumetric strain - axial strain curve (ϵ -ϵ 1 ) 52.3.2 Hostun-RF Sand, σ c = 300 kN/ m 2 0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700 800 p, p [kN/ m 2 ] q [ k N / m 2 ] MC failure MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1 Figure 52.5: Effective stress path curve (q-p) 220 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE49: Triaxial Drained Test 0 5 10 15 20 0 100 200 300 400 500 600 700 800 ϵ 1 [%] q [ k N / m 2 ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp.1 Figure 52.6: Deviatoric stress - axial strain curve (q-ϵ 1 ) 0 5 10 15 20 0.000 0.200 0.400 0.600 0.800 1.000 1.200 1.400 1.600 1.800 ϵ 1 [%] ϵ [ % ] MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1 Figure 52.7: Volumetric strain - axial strain curve (ϵ -ϵ 1 ) SOFiSTiK 2014 | VERiFiCATiON MANUAL 221 BE49: Triaxial Drained Test 0 5 10 15 20 25 30 35 −30 −20 −10 0 φ m [ ◦ ] ψ m [ ◦ ] HS-Rowe HS-Cons HS-Soreide HS-Wehnert Figure 52.8: Mobilised dilatancy angle - friction angle curve (ψ m -φ m ) 52.4 Conclusion This example, concerning the triaxial test of a loose consolidated undrained sand soil, verifies that the results obtained by the Hardening Soil material model with a cut-off in the dilatancy are in a good agreement with the solution given by Wehnert [60]. 52.5 Literature [60] M. Wehnert. Ein Beistrag zur dreainerten und undrainerten Analyse in der Geotechnik. Institut f ¨ ur Geotechnik, Universit ¨ at Stuttgart: P. A. Vermeer, 2006. 222 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE50: A Circular Cavity Embedded in a Full-Plane Under Impulse Pressure 53 BE50: A Circular Cavity Embedded in a Full-Plane Un- der Impulse Pressure Overview Element Type(s): C2D Analysis Type(s): DYN Procedure(s): Topic(s): SOIL Module(s): DYNA Input file(s): sbfem 2d cric cavity.dat 53.1 Problem Description This example addresses a circular cavity with radius r 0 embedded in a full-plane subjected to a radial pressure p(t) (Fig. 53.1). The full-plane is assumed to be elastic, homogeneous, isotropic, without material damping which stretches to infinity and it is modeled with the help of the Scaled Boundary Finite Elements (SBFE). Plane-strain condition is considered. Load is in a form of a triangular impulse and applied on the cavity wall (Fig. 53.1b). Radial displacement response of the cavity wall has been computed and compared to the analytical solution. r 0 p(t) (a) Circular cavity embedded in a full- plane ¯ t = t · c s / r 0 3 1.5 0 p( ¯ t) p 0 (b) Pressure load Figure 53.1: Problem Definition 53.2 Reference Solution This problem is essentially a one dimensional problem which has an analytical solution [63]. The force- displacement relationship for this problem in frequency domain is given by P(ω) = S ∞ (ω) · r (ω) , (53.1) where ω = 2πƒ represents the circular frequency, P(ω) is the total force applied on the cavity wall, r (ω) is the radial displacement and S ∞ (ω) is the dynamic-stiffness coefficient. SOFiSTiK 2014 | VERiFiCATiON MANUAL 223 BE50: A Circular Cavity Embedded in a Full-Plane Under Impulse Pressure The dynamic-stiffness coefficient for this particular problem has an analytical expression and it reads S ∞ ( 0 ) = 2πG 0 1 − 2ν · _ _ 2(1 − ν) λ − 1 − F λ − 2ν + 2(1 − ν) H (2) F+1 (λ 0 ) H (2) F (λ 0 ) 0 _ _ , (53.2) where G 0 shear modulus, ν Poisson’s ratio, ρ mass density, c s = _ G/ ρ shear wave velocity, c p = c s _ (2 − 2ν)/ (1 − 2n) P-wave velocity, 0 = ωr 0 / c p dimensionless frequency, λ = 2/ (2 − α) coefficient, α non-homogeneity parameter of elasticity (α = 0 for the homo- geneous case), H (2) k the second kind Hankel function of the order k, F = _ _ (λ − 1) 2 − λ 2 ν(α + 1) − 1 1 − ν order of the Hankel function. The static-stiffness coefficient K ∞ is used to non-dimensionlize displacement response K ∞ = 2πG 0 1 − 2ν · _ α(1 − ν) − 2ν + _ (α(1 − ν) − 2ν) 2 + 4 − 8ν _ . (53.3) The radial displacement response in frequency domain r (ω) is obtained by first making the Fourier transformation of the total triangular impulse load P(ω) (Fig. 53.1b) and then dividing it with the dynamic- stiffness coefficient S ∞ (ω) (Eq. 53.1). Finally the displacement response is transformed in the time domain ( r (t)) using the inverse Fourier transformation. 53.3 Model and Results Material, geometry and loading properties of the model are summarized in the Table 53.1. The plane- strain model of the full-pane is assumed to be elastic, homogeneous (α = 0) and isotropic. Table 53.1: Model Properties Material Geometry Loading Integration parameters c s , ρ, ν = 0.3 r 0 P(t) = 2πr 0 p(t) Δt = 0.04 · r 0 / c p G 0 = ρc 2 s P 0 = 2πr 0 p 0 M, N, θ = 1.4 Load and the finite element model of the structure are depicted in Fig. 53.2. The structure is comprised 224 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE50: A Circular Cavity Embedded in a Full-Plane Under Impulse Pressure solely of the 2-node line scaled boundary finite elements and the load is applied directly to the nodes of the boundary. Figure 53.2: Finite Element Model The transient radial displacement response of the cavity wall r (t) has been computed using the S- caled Boundary Finite Element Method (SBFEM) in the time domain. The integration of the governing equations of the SBFEM is performed using the original discretization scheme (const ) [63][64] and the extrapolation scheme from [65] based on the parameters M, N and θ 1 . The results in dimensionless form are plotted in Fig. 53.3 together with the analytical solution. The numerical results show excellent agreement with the analytical solution for all three cases. 0 2 4 6 8 10 12 14 16 18 20 −0.2 0 0.2 0.4 0.6 0.8 1 t · c p / r 0 r · K ∞ / P 0 analytic constant M = 40, N = 10 M = 20, N = 15 Figure 53.3: Radial displacement response 53.4 Conclusion The example verifies the accuracy of the SBFEM method in modeling unbounded domain problems. The integration scheme for the solution of the governing equations of the SBFEM in time domain based on the work from [65] provides the solution with high computational efficiency and little loss of accuracy compared to the original method from [64]. 1 For the full description of the scheme based on the extrapolation parameter θ and the meaning of the integration parameters M, N and θ consult [65]. SOFiSTiK 2014 | VERiFiCATiON MANUAL 225 BE50: A Circular Cavity Embedded in a Full-Plane Under Impulse Pressure 53.5 Literature [63] M.H. Bazyar. “Dynamic Soil-Structure Interaction Analysis Using the Scaled Boundary Finite- Element Method”. PhD thesis. Sydney, Australia: The University of New South Wales, School of Civil and Environmental Engineering, 2007. [64] J.P. Wolf and C. Song. Finite-Element Modelling of Unbounded Media. Chichester, UK: John Wiley and Sons, 1996. [65] B. Radmanovi ´ c and C. Katz. “A High Performance Scaled Boundary Finite Element Method”. In: IOP Conference Series: Materials Science and Engineering 10 (2010). 226 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure 54 BE51: Pushover Analysis: Performance Point Calcu- lation by EC8 Procedure Overview Element Type(s): Analysis Type(s): Procedure(s): Topic(s): EQKE Module(s): SOFiLOAD Input file(s): pushover-pp-ec8.dat 54.1 Problem Description The following example is intended to verify the Eurocode 8 (EC8) procedure for the calculation of the performance point (illustrated schematically in Fig. 54.1), as implemented in SOFiSTiK. The elastic demand and capacity diagrams are assumed to be know. S d S dp S S p El. Demand Diagram Performance Point Capacity Diagram Demand Diagram Figure 54.1: Determination of the performance point PP (S dp , S p ) 54.2 Reference Solution The reference solution is provided in [66]. Assuming that the elastic demand diagram (5% elastic response spectrum in ADRS format 1 ) and the capacity diagram are known, it is possible to determine the performance point PP (S dp , S p ) (Fig. 54.1). The procedure comprises of a series of trial calculations (trial performance points PP t (S dp,t , S p,t )), in which the equivalent inelastic single degree of freedom system (SDOF), represented by the capac- ity diagram, is idealized with the equivalent inelastic SDOF system with a bi-linear force-deformation relationship. The response in form of the performance point PP is then calculated from the inelastic response spectrum (demand diagram). The computation stops when the performance point PP is within a tolerance of a trial performance point PP t . Detailed description of this procedure can be found in [67], [68], [66] and [37]. In the reference example [66] the bi-linear idealization of the capacity is assumed to be independent of 1 ADRS = Spectral Acceleration S - Spectral Displacement S d format SOFiSTiK 2014 | VERiFiCATiON MANUAL 227 BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure the performance point and it is performed at the beginning of the analysis. This eliminates the need for the iterations and the solution of the problem can be obtained in a single calculation step. S d S μ = 1 T c T ∗ = T y PE S de S e PP S dp S p PY S dy μ > 1 (a) Short period range, T ∗ < T C S d S μ = 1 T c T ∗ = T y PE S dp = S de S e PP S p S y PY S dy μ > 1 (b) Medium and long range, T ∗ > T C Figure 54.2: Determination of the performance point PP for the equivalent SDOF system Hence in this example it is assumed that the bi-linear idealization of the capacity diagram is already known, which means that the point PY (S dy , S y ) is given. The procedure to calculated the performance point is illustrated in Fig. 54.2 and can be summarized as follows [37]: 1. Determine the period of the idealized system T ∗ = T y from the known PY (S dy , S y ): T ∗ = T y = 2π · _ _ _ S dy S y ; (54.1) 2. Calculate the elastic spectral response PE (S de , S e ) of the idealized equivalent SDOF system with the period T ∗ = T y from the given 5%-damped elastic response spectrum (Fig. 54.2); 3. Calculate the yield strength reduction factor R y : R y = S e S y ; (54.2) 4. Calculate ductility μ: μ = _ _ _ (R y − 1) · T C T ∗ + 1 for T ∗ < T C R y for T ∗ ≥ T C ; (54.3) 5. Determine the performance point PP (S dp , S p ) from the inelastic design spectrum: S dp = μ · S dy = μ · S de R y , (54.4a) S p = S e (T ∗ ) R y . (54.4b) 228 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure 54.3 Model and Results In order to verify the analysis procedure for the determination of the performance point, a test case has been set up in such a way that it comprises of a SDOF with a unit mass and a non-linear spring element. It is obvious that for such an element the quantities governing the transformation from the original system to the equivalent inelastic SDOF system must be equal to one, i.e. ϕ cnod = 1 ; = 1 ; m = 1 , (54.5) where ϕ cnod is the eigenvector value at control node, is the modal participation factor and m is the generalized modal mass. Writing now the equations which govern the conversion of the pushover curve to capacity diagram, we obtain [37] S d = cnod ϕ cnod · = cnod , (54.6a) S = V b 2 · m = V b , (54.6b) where V b is the base shear and cnod is the control node displacement. Since the original system is a SDOF system, V b and cnod are nothing else but the force in spring P and the displacement of the unit mass , respectively. It follows further that the force-displacement work law assigned to the spring element corresponds to the capacity diagram in ADRS format, with the force P and displacement equal to S and S d , respectively. The bi-linear idealization of the capacity diagram used in the reference example is defined by two points, whose coordinates are listed in the Table 54.1 2 . According to the analysis above, these points can be used to define the force- displacement work law P − of the non-linear spring element (Fig. 54.3). Table 54.1: Model Properties [66] Capacity Diagram Elastic Demand Point _ S d [mm], S [m/ s 2 ] _ 5%-Damped Elastic Response Spectrum A (61, 3.83) g = {0.60g, 0.30g, 0.16g} B (∞, 3.83) S A = 1.0, S B = 2.5, k 1 = 1.0 T B = 0.15s, T C = 0.60s, T D = 3.00s 2 Not that the point A is nothing else but the point PY (S dy , S y ). SOFiSTiK 2014 | VERiFiCATiON MANUAL 229 BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure P B A O u [mm] 2 0 0 .0 1 5 0 .0 1 0 0 .0 5 0 .0 0 .0 0 .0 P[kN] 3.0 2.0 1.0 0.0 0.0 Figure 54.3: Force-displacement work law of the non-linear spring The elastic demand is a 5%-damped elastic response spectrum, whose properties are summarized in Table 54.1. Three levels of peak ground acceleration g have been taken into an account. The shape of the spectrum and the meaning of the parameters specified in Table 54.1 are shown in Figure 54.4. S(T) T S A S B T B T C T D 0 ≤ T ≤ T B : S = S A + T T B · (S B − S A ) T B ≤ T ≤ T C : S = S B T C ≤ T ≤ T D : S = S B · _ T C T _ k 1 Figure 54.4: 5%-Damped Elastic Response Spectrum (El. Demand Diagram) The outcome of the analysis is shown in Figures 54.5 to 54.7. Capacity El. Demand, 0.60g Demand, μ = 2.91 PY PP Ty Tp Tb = 0.2 Tc = 0.6 Td = 3.0 T = 0.5 T = 1.0 T = 1.5 T = 2.0 T = 4.0 SPL 1 SPL 2 Sd [mm] 2 0 0 . 0 0 0 1 5 0 . 0 0 0 1 0 0 . 0 0 0 5 0 . 0 0 0 0 . 0 0 0 Sa [m/sec2] 15.00 10.00 5.00 0.00 Figure 54.5: Capacity-Demand-Diagram ( g = 0.60g) 230 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure Capacity El. Demand, 0.30g Demand, μ = 1.45 PY PP Ty Tp Tb = 0.2 Tc = 0.6 Td = 3.0 T = 0.5 T = 1.0 T = 1.5 T = 2.0 T = 4.0 SPL 1 SPL 2 Sd [mm] 2 0 0 . 0 0 0 1 5 0 . 0 0 0 1 0 0 . 0 0 0 5 0 . 0 0 0 0 . 0 0 0 Sa [m/sec2] 8.00 6.00 4.00 2.00 0.00 Figure 54.6: Capacity-Demand-Diagram ( g = 0.30g) Capacity El. Demand, 0.15g Demand, μ = 1.00 PY PP Ty Tp Tb = 0.2 Tc = 0.6 Td = 3.0 T = 0.5 T = 1.0 T = 1.5 T = 2.0 T = 4.0 SPL 1 SPL 2 Sd [mm] 2 0 0 . 0 0 0 1 5 0 . 0 0 0 1 0 0 . 0 0 0 5 0 . 0 0 0 0 . 0 0 0 Sa [m/sec2] 4.00 2.00 0.00 Figure 54.7: Capacity-Demand-Diagram ( g = 0.15g) The results of the SOFiSTiK calculation and the comparison with the reference solution are summarized in Table 54.2. Table 54.2: Results g μ R yp T y S dy S dp S p [g] [−] [−] [s] [mm] [mm] [m/ s 2 ] SOF. 2.9 2.9 0.79 61 177 3.83 0.60 Ref. [66] 2.9 2.9 0.79 61 177 3.83 |e| [%] 0.0 0.0 0.0 0.0 0.0 0.0 SOF. 1.5 1.5 0.79 61 89 3.83 0.30 Ref. [66] 1.5 1.5 0.79 61 89 3.83 |e| [%] 0.0 0.0 0.0 0.0 0.0 0.0 SOF. 1.0 1.0 0.79 44 44 2.78 SOFiSTiK 2014 | VERiFiCATiON MANUAL 231 BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure Table 54.2: (continued) g μ R yp T y S dy S dp S p [g] [−] [−] [s] [mm] [mm] [m/ s 2 ] 0.15 Ref. [66] 1.0 1.0 0.79 44 44 2.76 |e| [%] 0.0 0.0 0.0 0.0 0.0 0.7 μ displacement ductility factor R yp reduction factor due to ductility at performance point T y period associated with yielding point S dy , S dp spectral displacements at yielding and performance point S p pseudo spectral acceleration at performance point The results are in excellent agreement with the reference solution. 54.4 Conclusion Excellent agreement between the reference and the results computed by SOFiSTiK verifies that the pro- cedure for the calculation of the performance point according to Eurocode 8 is adequately implemented. 54.5 Literature [37] SOFiLOAD Manual: Loadgenerator for Finite Elements and Frameworks. Version 2014.1. SOFiSTiK AG. Oberschleißheim, Germany, 2013. [66] P. Fajfar. “A Nonlinear Analysis Method for Performance-Based Seismic Design”. In: Earthquake Spectra 16.3 (2000), pp. 573–592. [67] EN1998-1:2004. Eurocode 8: Design of structures for earthquake resistance, Part 1: General rules, seismic actions and rules for buildings. CEN. 2004. [68] P. Fajfar. “Capacity Spectrum Method Based on Inelastic Demand Spectra”. In: Earthquake engi- neering and structural dynamics 28.9 (1999), pp. 979–993. 232 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE52: Verification of Wave Kinematics 55 BE52: Verification of Wave Kinematics Overview Element Type(s): Analysis Type(s): Procedure(s): Topic(s): WAVE Module(s): SOFiLOAD Input file(s): stokes.dat 55.1 Problem Description This benchmark is concerned with the validation of wave kinematics of regular nonlinear Stokes 5 th order wave theory. In Fig. 55.1 the properties of a wave can be visualised. z x H L d S W L c η h Figure 55.1: Wave 55.2 Reference Solution The reference solution is provided in [69]. This article investigates the solution of the dispersion relation of Stokes fifth order wave theory, which is governed by two coupled nonlinear equations in two variables, through a Newton-Raphson iterative scheme. Different waves are investigated and their wave profile and horizontal velocitiy is computed and plotted. The interest of this benchmark focuses on the provided solution for the corrected coefficient in the original expression for C 2 (the factor +2592 should be replaced by −2592), which is employed also from SOFiSTiK. For more information on this correction please refer to Nishimura & al. (1977), Fenton (1985) [70], Bhattacharyya (1995) [69] and SOFiLOAD manual [37]. 55.3 Model and Results The properties of the considered wave are defined in Table 55.1. SOFiSTiK 2014 | VERiFiCATiON MANUAL 233 BE52: Verification of Wave Kinematics Table 55.1: Model Properties Wave Properties d = 107 ƒ t H = 70 ƒ t T = 16.30 s The wave profile, i.e. the phase angle θ versus the surface elevation η, is computed and shown in Fig 55.2 and the horizontal velocity under the wave crest versus the elevation from the seabed (z − d), in Fig 55.3. Both results are compared to the reference solution, as peresented in Bhattacharyya (1995) [69]. 0 20 40 60 80 100 120 140 160 180 200 − 40 − 30 − 20 − 10 0 10 20 30 40 50 60 θ [ o ] η [ ƒ t ] Reference SOFiSTiK Figure 55.2: Wave profile 234 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE52: Verification of Wave Kinematics 0 5 10 15 20 25 30 35 40 45 50 0 20 40 60 80 100 120 [ƒ t/ s] e l e v a t i o n f r o m s e a b e d ( z − d ) [ ƒ t ] Reference SOFiSTiK Figure 55.3: Horizontal velocity under wave crest 55.4 Conclusion The very good agreement between the reference and the results computed by SOFiSTiK verifies that the Stokes fifth order wave theory is adequately implemented. 55.5 Literature [37] SOFiLOAD Manual: Loadgenerator for Finite Elements and Frameworks. Version 2014.1. SOFiSTiK AG. Oberschleißheim, Germany, 2013. [69] S. K. Bhattacharyya. “On two solutions of fifth order Stokes waves”. In: Applied Ocean Research 17 (1995), pp. 63–68. [70] J. D. Fenton. “A fifth order Stokes theory for steady waves”. In: J. Waterways, Port, Coastal & Ocean Engineering 111(2) (1985), pp. 216–234. SOFiSTiK 2014 | VERiFiCATiON MANUAL 235 BE52: Verification of Wave Kinematics 236 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE53: Verification of Wave Loading 56 BE53: Verification of Wave Loading Overview Element Type(s): Analysis Type(s): Procedure(s): Topic(s): WAVE Module(s): SOFiLOAD Input file(s): wave loading.dat 56.1 Problem Description This benchmark is concerned with the validation of wave loading on a structure. In this example the linear Airy wave theory with Wheeler stretching is applied to one exemplary wave on a monopile, as presented in Fig. 56.1. The surface elevation and the accumulated forces produced by the wave theory are compared with the results calculated by WaveLoads. WaveLoads is a well-known software developed within the research project GIGAWIND at Hannover University for calculating wave induced loading on hydrodynamically transparent structures [71]. z x H L d S W L d c η h L p T Dp Figure 56.1: Wave 56.2 Reference Solution The reference example is calculated with WaveLoads. Further information on the model can be found in the WaveLoads manual [71]. This benchmark aims at verifying three important components: the Airy wave theory, the Wheeler stretching scheme and the Morison equation [37]. 56.3 Model and Results The properties of the considered wave and the structure are defined in Table 56.1. The wave profile, i.e. the surface elevation η over time of one period, is computed and shown in Fig 56.2 and the accumulated SOFiSTiK 2014 | VERiFiCATiON MANUAL 237 BE53: Verification of Wave Loading forces over time of one period, in Fig 56.3. Both results are compared to the calculated reference solution [71]. Table 56.1: Model Properties Wave Properties Structure Properties d = 34 m D p = 6 m H = 17.5 m L p = 54 m T = 15 s C m = 2.0 SWL = 0 m C d = 0.7 The pile is modeled with 500 elements as in the reference example. The Wheeler stretching is applied. The calculated wave length is L = 246.013 mand the calculated depth criterion d/ L = 0.138 indicates that the examined case falls into finite water. -5.000 0 2 4 6 8 10 12 14 16 − 10 − 8 − 6 − 4 − 2 0 2 4 6 8 10 T [s] η [ m ] Reference SOFiSTiK Figure 56.2: Wave profile 238 VERiFiCATiON MANUAL | SOFiSTiK 2014 BE53: Verification of Wave Loading 0 2 4 6 8 10 12 14 16 18 20 −5,000 −4,000 −3,000 −2,000 −1,000 0 1,000 2,000 3,000 4,000 5,000 T [s] F [ k N ] Reference SOFiSTiK Figure 56.3: Accumulated Force for Airy linear wave theory in combination with Wheeler Stretching 56.4 Conclusion The very good agreement between the reference and the results computed by SOFiSTiK verifies that the linear Airy wave theory, the Wheeler stretching scheme and the Morison equation are adequately implemented. 56.5 Literature [37] SOFiLOAD Manual: Loadgenerator for Finite Elements and Frameworks. Version 2014.1. SOFiSTiK AG. Oberschleißheim, Germany, 2013. [71] K. Mittendorf, B. Nguyen, and M. Bl ¨ umel. WaveLoads - A computer program to calculate wave loading on vertical and inclined tubes. ISEB - Fluid Mechanics Institute, University of Hannover. 2005. SOFiSTiK 2014 | VERiFiCATiON MANUAL 239 BE53: Verification of Wave Loading 240 VERiFiCATiON MANUAL | SOFiSTiK 2014 Part III Design Code Benchmark Examples SOFiSTiK 2014 | VERiFiCATiON MANUAL 241 DCE-EN1: Design of Slab for Bending 57 DCE-EN1: Design of Slab for Bending Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): slab bending.dat 57.1 Problem Description The problem consists of a slab section of depth h, as shown in Fig. 57.1. The cross-section is designed for an ultimate moment m Ed and the required reinforcement is determined. m Ed s1 s1 b d h Figure 57.1: Problem Description 57.2 Reference Solution This example is concerned with the design of sections for ULS, subject to pure flexure, such as beams or slabs. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Basic assumptions for section design (Section 6.1) • Reinforcement (Section 9.3.1.1, 9.2.1.1) ε c ε s z F c F s d A s A c Figure 57.2: Stress and Strain Distributions in the Design of Cross-sections SOFiSTiK 2014 | VERiFiCATiON MANUAL 243 DCE-EN1: Design of Slab for Bending In singly reinforced beams and slabs, the conditions in the cross-section at the ultimate limit state, are assumed to be as shown in Fig. 57.2. The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 57.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c = 525N/ mm 2 ε ε d = 25 0 / 00 ƒ yd = ƒ yk / γ s σ Figure 57.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 57.3 Model and Results The rectangular slab section, with properties as defined in Table 57.1, is to be designed, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72] [73], to carry an ultimate moment of 25 kNm. The calculation steps with different design methods [74] [75] [76] are presented below and the results are given in Table 57.2. Here, it has to be mentioned that these standard methods employed in order to calculate the reinforcement are approximate, and therefore deviations often occur. Table 57.1: Model Properties Material Properties Geometric Properties Loading C 25/ 30 h = 20.0 cm m Ed = 25 kNm/ m B 500A d = 17.0 cm b = 1.0 m Table 57.2: Results SOF. General Chart [74] ω−Table [74] k d −Table [74] s1 [cm 2 / m] 3.334 3.328 3.334 3.333 244 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN1: Design of Slab for Bending 57.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 25 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 25/ 1.5 = 14.17 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: M Ed = m Ed · b = 25 kNm; N Ed = 0 M Eds = M Ed − N Ed · z s1 = 25 kNm Design with respect to General Design Chart Bending with axial force for rectangular cross-sections: μ Eds = M Eds b · d 2 · ƒ cd = 25 · 10 −3 1.0 · 0.17 2 · 14.17 = 0.061 Tab. 9.1 [74]: General Chart for up to C50/ 60 - Cross-section without com- pression reinforcement ε = 25 · 10 −3 ; ζ = 0.97 → σ s1d = 456.52 MP s1 = 1 σ s1d · _ M Eds ζ · d + N Ed _ = 3.328 cm 2 / m Design with respect to ω− (or μ s − ) Design Table for rectangular cross-sections: μ Eds = M Eds b · d 2 · ƒ cd = 25 · 10 −3 1.0 · 0.17 2 · 14.17 = 0.061 Tab. 9.2 [74]: ω−Table for up to C50/ 60 - Rectangular section without compression reinforcement ω = 0.0632 (interpolated) and σ sd = 456.52 MP s1 = 1 σ sd · (ω· b · d · ƒ cd + N Ed ) = 3.334 cm 2 / m Design with respect to k d − Design Table for rectangular cross- sections: k d = d _ M Eds / b = 17 _ 25/ 1.0 = 3.40 Tab. 9.3 [74]: k d −Table for up to C50/ 60 - Rectangular section without compression reinforcement k s = 2.381, κ s = 0.952 (interpolated values) s1 = _ k s · M Eds d + N Ed σ s1d _ · κ s = 3.333 cm 2 / m 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 57.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 245 DCE-EN1: Design of Slab for Bending 57.5 Conclusion This example shows the calculation of the required reinforcement for a slab section under bending. Various different reference solutions are employed in order to compare the SOFiSTiK results to. It has been shown that the results are reproduced with excellent accuracy. 57.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. M¨ uller, and F. Lobisch. Bemessungshilfsmittel f ¨ ur Betonbauteile nach Eu- rocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. [76] R. S. Narayanan and A. W. Beeby. Designers’ Guide to EN 1992-1-1 and EN 1992-1-2 - Eurocode 2: Design of Concrete Structures. Thomas Telford, 2005. 246 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN2: Design of a Rectangular CS for Bending 58 DCE-EN2: Design of a Rectangular CS for Bending Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): rectangular bending.dat 58.1 Problem Description The problem consists of a rectangular section, as shown in Fig. 58.1. The cross-section is designed for an ultimate moment M Ed and the required reinforcement is determined. M Ed A s1 A s2 b d d 2 h Figure 58.1: Problem Description 58.2 Reference Solution This example is concerned with the design of doubly reinforced sections for ULS, subject to pure flexure, such as beams. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Basic assumptions for section design (Section 6.1) • Reinforcement (Section 9.3.1.1, 9.2.1.1) ε c ε s1 ε s2 σ s1d σ s2d σ cd z F c F s2d F s1d z s2 d 2 z s1 d 1 d A s1 A s2 Figure 58.2: Stress and Strain Distributions in the Design of Doubly Reinforced Cross-sections SOFiSTiK 2014 | VERiFiCATiON MANUAL 247 DCE-EN2: Design of a Rectangular CS for Bending In doubly reinforced rectangular beams, the conditions in the cross-section at the ultimate limit state, are assumed to be as shown in Fig. 58.2. The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 58.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c = 525N/ mm 2 ε ε d = 25 0 / 00 ƒ yd = ƒ yk / γ s σ Figure 58.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 58.3 Model and Results The rectangular cross- section, with properties as defined in Table 58.1, is to be designed, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73], to carry an ultimate moment of 135 kNm. The calculation steps with different design methods [74] [75] [76] are presented below and the results are given in Table 58.2. Here, it has to be mentioned that these standard methods employed in order to calculate the reinforcement are approximate, and therefore deviations often occur. Table 58.1: Model Properties Material Properties Geometric Properties Loading C 20/ 25 h = 40.0 cm M Ed = 135 kNm B 500A d = 35.0 cm d 2 = 5.0 cm b = 25 cm Table 58.2: Results SOF. General Chart [74] ω−Table [74] k d −Table [74] A s1 [cm 2 / m] 10.73 10.73 10.77 10.79 A s2 [cm 2 / m] 2.47 2.47 2.52 2.43 248 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN2: Design of a Rectangular CS for Bending 58.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 20 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 20/ 1.5 = 11.33 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: N Ed = 0 M Eds = M Ed − N Ed · z s1 = 135 kNm Design with respect to General Design Chart Bending with axial force for rectangular cross-sections: μ Eds = M Eds b · d 2 · ƒ cd = 135 · 10 −3 0.25 · 0.35 2 · 11.33 = 0.389 μ Eds > μ Eds,m = 0.296 5.4: (NA.5): Linear elastic analysis ξ = height of comression zone / d ≤ 0.45 for C12/ 15 − C50/ 60 → compression reinforcement required from design chart for μ Eds,m = 0.296 and d 2 / d = 0.143 : Tab. 9.1 [74]: General Chart for up to C50/ 60 - Section with compression re- inforcement ε s1 = 4.30 · 10 −3 ; ε s2 = −2.35 · 10 −3 ; ζ = z/ d = 0.813 for ε s1 = 4.30 · 10 −3 → σ s1d = 436.8 MP for ε s2 = −2.35 · 10 −3 → σ s1d = −434.9 MP M Eds,m = μ Eds,m · b · d 2 · ƒ cd = 102.7 kNm ΔM Eds = M Eds − M Eds,m = 135 − 102.7 = 32.3 kNm A s1 = 1 σ s1d · _ M Eds,m ζ · d + ΔM Eds d − d 2 + N Ed _ = 10.73 cm 2 A s2 = 1 |σ s2d | · ΔM Eds d − d 2 = 2.47 cm 2 Design with respect to ω− (or μ s − )Table for rectangular cross- sections: μ Eds = M Eds b · d 2 · ƒ cd = 135 · 10 −3 0.25 · 0.35 2 · 11.33 = 0.389 5.4: (NA.5): Linear elastic analysis ξ = height of comression zone / d ≤ 0.45 for C12/ 15 − C50/ 60 Because the internal force determination is done on the basis of a linear 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 58.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 249 DCE-EN2: Design of a Rectangular CS for Bending elastic calculation, then ξ m = 0.45 is chosen. Referring to the design table with compression reinforcement and for d 2 / d = 0.15: Tab. 9.2 [74]: ω−Table for up to C50/ 60 - Rectangular section with compression reinforcement ω 1 = 0.4726; ω 1 = 0.1104 A s1 = 1 ƒ yd · (ω 1 · b · d · ƒ cd + N Ed ) = 10.77 cm 2 A s2 = ƒ cd ƒ yd · (ω 2 · b · d) = 2.52 cm 2 Design with respect to k d − Design Table for rectangular cross- sections: k d = d _ M Eds / b = 35 _ 135/ 0.25 = 1.51 Tab. 9.3 [74]: k d −Table for up to C50/ 60 - Rectangular section with compression reinforcement Not able to read values from k d −table for simply reinforced rectangular cross-sections → compression reinforcement is required 5.4: (NA.5): Linear elastic analysis ξ = height of comression zone / d ≤ 0.45 for C12/ 15 − C50/ 60 Because the internal force determination is done on the basis of a lin- ear elastic calculation, then ξ m = 0.45 is chosen. Referring to the k d −table with compression reinforcement: k s1 = 2.740; k s2 = 0.575 (interpolated values for k d = 1.51) ρ 1 = 1.021; ρ 2 = 1.097 (interpolated values for d 2 / d = 0.143 and k s1 = 2.740) A s1 = ρ 1 · k s1 · M Eds d + N Ed σ s1d = 10.79 cm 2 A s2 = ρ 2 · k s2 · M Eds d = 2.43 cm 2 250 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN2: Design of a Rectangular CS for Bending 58.5 Conclusion This example shows the calculation of the required reinforcement for a rectangular beam cross-section under bending. Various different reference solutions are employed in order to compare the SOFiSTiK results to. It has been shown that the results are reproduced with excellent accuracy. 58.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. M¨ uller, and F. Lobisch. Bemessungshilfsmittel f ¨ ur Betonbauteile nach Eu- rocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. [76] R. S. Narayanan and A. W. Beeby. Designers’ Guide to EN 1992-1-1 and EN 1992-1-2 - Eurocode 2: Design of Concrete Structures. Thomas Telford, 2005. SOFiSTiK 2014 | VERiFiCATiON MANUAL 251 DCE-EN2: Design of a Rectangular CS for Bending 252 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN3: Design of a T-section for Bending 59 DCE-EN3: Design of a T-section for Bending Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): t-beam bending.dat 59.1 Problem Description The problem consists of a T-beam section, as shown in Fig. 59.1. The cross-section is designed for an ultimate moment M Ed and the required reinforcement is determined. M Ed A s1 d 1 z s A s1 h ƒ b h d Figure 59.1: Problem Description 59.2 Reference Solution This example is concerned with the design of sections for ULS, subject to pure flexure. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Basic assumptions for section design (Section 6.1) • Reinforcement (Section 9.3.1.1, 9.2.1.1) SOFiSTiK 2014 | VERiFiCATiON MANUAL 253 DCE-EN3: Design of a T-section for Bending ε c ε s1 z s A s1 Figure 59.2: Stress and Strain Distributions in the Design of T-beams In doubly reinforced rectangular beams, the conditions in the cross-section at the ultimate limit state, are assumed to be as shown in Fig. 59.2. The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 59.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c = 525N/ mm 2 ε ε d = 25 0 / 00 ƒ yd = ƒ yk / γ s σ Figure 59.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 59.3 Model and Results The T-beam, with properties as defined in Table 59.1, is to be designed, with respect to DIN EN 1992-1- 1:2004 (German National Annex) [72], [73], to carry an ultimate moment of 425 kNm. The calculation steps with different design methods [74] [75] [76] are presented below and the results are given in Table 59.2. Here, it has to be mentioned that these standard methods employed in order to calculate the reinforcement are approximate, and therefore deviations often occur. Table 59.1: Model Properties Material Properties Geometric Properties Loading C 20/ 25 h = 65.0 cm M Ed = 425 kNm B 500A d = 60.0 cm d 1 = 5.0 cm b = 30 cm b eƒ ƒ = 258 cm 254 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN3: Design of a T-section for Bending Table 59.1: (continued) Material Properties Geometric Properties Loading h ƒ = 18 cm Table 59.2: Results SOF. ω−Table [74] k d −Table [74] A s1 [cm 2 / m] 15.90 15.74 15.85 SOFiSTiK 2014 | VERiFiCATiON MANUAL 255 DCE-EN3: Design of a T-section for Bending 59.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 20 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 20/ 1.5 = 11.33 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: N Ed = 0 M Eds = M Ed − N Ed · z s1 = 425 kNm Design with respect to ω− (or μ s − )Table for T-beams: μ Eds = M Eds b eƒ ƒ · d 2 · ƒ cd = 425 · 10 −3 2.58 · 0.60 2 · 11.33 = 0.040 Tab. 9.4 [74]: ω−Table for up to C50/ 60 - T-beam Referring to the design table for T-beams for: μ Eds = 0.040 and h ƒ d = 0.18 0.60 = 0.30; b eƒ ƒ b = 2.58 0.30 = 8.6 → ω 1 = 0.039 A s1 = 1 ƒ yd · _ ω 1 · b eƒ ƒ · d · ƒ cd + N Ed _ = 15.74 cm 2 Design with respect to k d − Design Table for T-beams: Alternatively, the k d −Tables can be applied, demonstrated that the neu- tral line lies inside the flange. k d = d _ M Eds / b = 60 _ 425/ 2.58 = 4.67 Tab. 9.3 [74]: k d −Table for up to C50/ 60 - Rectangular section without compression reinforcement Referring to the table for k d = 4.67 and after interpolation → k s = 2.351; ξ = 0.060 ; κ s = 0.952 = ξ · d = 0.060 · 60 = 3.6 cm ¡ h ƒ = 18 cm k d −Table is applicable since the neutral line lies inside the flange A s1 = _ k s · M Eds d + N Ed σ s1d _ · κ s = 15.85 cm 2 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 59.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. 256 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN3: Design of a T-section for Bending SOFiSTiK 2014 | VERiFiCATiON MANUAL 257 DCE-EN3: Design of a T-section for Bending 59.5 Conclusion This example shows the calculation of the required reinforcement for a T-beam under bending. Two different reference solutions are employed in order to compare the SOFiSTiK results to. It has been shown that the results are reproduced with excellent accuracy. 59.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. M¨ uller, and F. Lobisch. Bemessungshilfsmittel f ¨ ur Betonbauteile nach Eu- rocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. [76] R. S. Narayanan and A. W. Beeby. Designers’ Guide to EN 1992-1-1 and EN 1992-1-2 - Eurocode 2: Design of Concrete Structures. Thomas Telford, 2005. 258 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN4: Design of a Rectangular CS for Bending and Axial Force 60 DCE-EN4: Design of a Rectangular CS for Bending and Axial Force Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): rectangular bending axial.dat 60.1 Problem Description The problem consists of a rectangular section, as shown in Fig. 60.1. The cross-section is designed for an ultimate moment M Ed and a compressive force N Ed and the required reinforcement is determined. M Ed N Ed A s,tot / 2 A s,tot / 2 b d d 2 d 1 = d 2 h Figure 60.1: Problem Description 60.2 Reference Solution This example is concerned with the design of sections for ULS, subject to bending with axial force. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Basic assumptions for section design (Section 6.1) • Reinforcement (Section 9.3.1.1, 9.2.1.1) ε c ε s1 ε s2 σ s1d σ s2d σ cd z F c F s2d F s1d z s2 d 2 z s1 d 1 d A s1 A s2 Figure 60.2: Stress and Strain Distributions in the Design of Doubly Reinforced Cross-sections SOFiSTiK 2014 | VERiFiCATiON MANUAL 259 DCE-EN4: Design of a Rectangular CS for Bending and Axial Force In doubly reinforced rectangular beams, the conditions in the cross-section at the ultimate limit state, are assumed to be as shown in Fig. 60.2. The design stress-strain diagram for reinforcing steel considered in this example, consists of an horizontal top branch, as presented in Fig. 60.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c ε ƒ yd = ƒ yk / γ s σ Figure 60.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 60.3 Model and Results The rectangular cross- section, with properties as defined in Table 60.1, is to be designed, with respec- t to DIN EN 1992-1-1:2004 (German National Annex) [72], [73], to carry an ultimate moment of 382 kNm with an axial compressive force of 1785 kN. The calculation steps with a commonly used design method [74] [75] are presented below and the results are given in Table 60.2. Here, it has to be men- tioned that the standard methods employed in order to calculate the reinforcement are approximate, and therefore deviations often occur. Table 60.1: Model Properties Material Properties Geometric Properties Loading C 30/ 37 h = 50.0 cm M Ed = 382 kNm B 500A d = 45.0 cm N Ed = −1785 kN d 1 = d 2 = 5.0 cm b = 30 cm Table 60.2: Results SOF. Interaction Diagram [74] A s,tot [cm 2 / m] 35.03 35.19 260 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN4: Design of a Rectangular CS for Bending and Axial Force 60.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 30 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 30/ 1.5 = 17.0 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: N Ed = −1785 kN M Ed = 382 kNm e d h = _ _ _ _ M Ed N Ed · h _ _ _ _ = _ _ _ _ 382 −1785 · 0.50 _ _ _ _ = 0.428 < 3.5 → Axial force dominant → Design with respect to μ − ν interaction diagram is suggested Design with respect to Interaction diagram for Bending with axial force for rectangular cross-sections: μ Ed = M Ed b · h 2 · ƒ cd = 382 · 10 −3 0.30 · 0.50 2 · 17.0 = 0.30 Tab. 9.6 [74]: μ − ν Interaction diagram for concrete C12/ 15− C50/ 60 - Rect- angular cross-section with double sym- metric reinforcement. ν Ed = N Ed b · h 2 · ƒ cd = −1785 · 10 −3 0.30 · 0.50 · 17.0 = −0.70 from design chart for d 1 / h = 0.05/ 0.5 = 0.10: ω tot = 0.60 A s,tot = ω tot · b · h ƒ yd / ƒ cd = 35.19 cm 2 A s1 = A s2 = A s,tot 2 = 17.6 cm 2 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 60.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 261 DCE-EN4: Design of a Rectangular CS for Bending and Axial Force 60.5 Conclusion This example shows the calculation of the required reinforcement for a rectangular beam cross-section under bending with axial force. It has been shown that the results are reproduced with excellent accura- cy. 60.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. M¨ uller, and F. Lobisch. Bemessungshilfsmittel f ¨ ur Betonbauteile nach Eu- rocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. 262 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force 61 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): rectangular double bending axial.dat 61.1 Problem Description The problem consists of a rectangular section, as shown in Fig. 61.1. The cross-section is designed for double axially bending moments M Edy , M Edz and a compressive force N Ed . M Edy N Ed M Edz each A s,tot / 4 b b 1 = b 2 d 1 = d 2 h Figure 61.1: Problem Description 61.2 Reference Solution This example is concerned with the design of sections for ULS, subject to double bending with axial force. The content of the problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Basic assumptions for section design (Section 6.1) • Reinforcement (Section 9.3.1.1, 9.2.1.1) SOFiSTiK 2014 | VERiFiCATiON MANUAL 263 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force ε c ε s1 ε s2 σ s1d σ s2d σ cd z F c F s2d F s1d z s2 d 2 z s1 d 1 d A s1 A s2 Figure 61.2: Stress and Strain Distributions in the Design of Doubly Reinforced Cross-sections The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 61.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c ε ƒ yd = ƒ yk / γ s σ Figure 61.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 61.3 Model and Results The rectangular cross- section, with properties as defined in Table 61.1, is to be designed, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73], under double axial bending and an axial compressive force of 1600 kN. The calculation steps with a commonly used design method [74] [75] are presented below and the results are given in Table 61.2. Here, it has to be mentioned that the standard methods employed in order to calculate the reinforcement are approximate, and therefore deviations often occur. Table 61.1: Model Properties Material Properties Geometric Properties Loading C 35/ 45 h = 50.0 cm M Edy = 500 kNm B 500A b 1 = b 2 = 5.0 cm M Edz = 450 kNm d 1 = d 2 = 5.0 cm N Ed = −1600 kN b = 40 cm 264 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force Table 61.2: Results SOF. Interaction Diagram [74] A s,tot [cm 2 / m] 115.9 113.1 SOFiSTiK 2014 | VERiFiCATiON MANUAL 265 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force 61.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 30 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 35/ 1.5 = 19.8 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: N Ed = −1600 kN M Edy = 500 kNm M Edz = 450 kNm Design with respect to Interaction diagram for Double Bending with axial force for rectangular cross-sections: μ Edy = M Ed b · h 2 · ƒ cd = 500 · 10 −3 0.40 · 0.50 2 · 19.8 = 0.252 μ Edz = M Ed b · h 2 · ƒ cd = 450 · 10 −3 0.40 · 0.50 2 · 19.8 = 0.284 ν Ed = N Ed b · h 2 · ƒ cd = −1600 · 10 −3 0.40 · 0.50 · 19.8 = −0.403 from design chart →ω tot = 1.24 for: Tab. 9.7 [74]: μ − ν Interaction diagram for concrete C12/ 15− C50/ 60 - Rect- angular cross-section with all-round symmetric reinforcement. • d 1 / h = d 2 / h = 0.05/ 0.5 = 0.10 • b 1 / b = b 2 / b = 0.05/ 0.4 = 0.08 ≈ 0.10 • ν = −0.4 • μ 1 = m _ μ Edy ; μ Edz _ = 0.284 • μ 2 = mn _ μ Edy ; μ Edz _ = 0.252 A s,tot = ω tot · b · h ƒ yd / ƒ cd = 113.1 cm 2 A s,tot/ 4 = 28.28 cm 2 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 61.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. 266 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force 61.5 Conclusion This example shows the calculation of the required reinforcement for a rectangular beam cross-section under double axial bending with compressive axial force. It has been shown that the results are repro- duced with excellent accuracy. 61.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. M¨ uller, and F. Lobisch. Bemessungshilfsmittel f ¨ ur Betonbauteile nach Eu- rocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. SOFiSTiK 2014 | VERiFiCATiON MANUAL 267 DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force 268 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN6: Design of a Rectangular CS for Shear Force 62 DCE-EN6: Design of a Rectangular CS for Shear Force Overview Design Code Family(s): EN, DIN Design Code(s): EN1992 Module(s): AQB Input file(s): rectangular shear.dat 62.1 Problem Description The problem consists of a rectangular section, symmetrically reinforced for bending, as shown in Fig. 62.1. The cross-section is designed for shear force V Ed and the required shear reinforcement is deter- mined. V Ed z d h b A s,tot / 2 A s,tot / 2 Figure 62.1: Problem Description 62.2 Reference Solution This example is concerned with the design of sections for ULS, subject to shear force. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72] [73]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Guidelines for shear design (Section 6.2) • Reinforcement (Section 9.2.2) V N M V(cot θ − cot α) V F td F cd θ α d z s shear reinforcement Figure 62.2: Shear Reinforced Members SOFiSTiK 2014 | VERiFiCATiON MANUAL 269 DCE-EN6: Design of a Rectangular CS for Shear Force The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 62.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c ε ƒ yd = ƒ yk / γ s σ Figure 62.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 62.3 Model and Results The rectangular section, with properties as defined in Table 62.1, is to be designed, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73], under shear force of 343.25 kN. The reference calculation steps are presented below and the results are given in Table 62.2. Then, the same section is designed with repsect to EN 1992-1-1:2004 [77]. The same angle θ(= 1.60) is chosen, as calculated with respect to DIN EN 1992-1-1:2004, in order to compare the results. If no θ value is input, then the calculation starts with the upper limit cot θ = 2.5 and through an optimization process the right angle is selected. In this case, the reinforcement is determined with cot θ = 2.5, giving a shear reinforcement of 7.80 cm 2 / m. Also in order to demonstrate that the correct value of V Rd,m = 734.4 kN (reference value) with repsect to DIN EN 1992-1-1:2004 is calculated in SOFiSTiK, we input a design shear force of 734.3 delivering a shear reinforcement, but when a value of 734.4 is input then AQB gives the warning of ’no shear design possible’ showing that the maximum shear resistance is exceeded. Table 62.1: Model Properties Material Properties Geometric Properties Loading C 30/ 37 h = 50.0 cm V Ed = 343.25 kN B 500A b = 30 cm d = 45.0 cm A s,tot = 38.67 cm 2 Table 62.2: Results A s,tot [cm 2 / m] Design Code SOF. Ref. DIN EN [72] 12.84 12.84 EN [77] 12.18 12.18 270 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN6: Design of a Rectangular CS for Shear Force 62.4 Design Process 1 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 30 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 30/ 1.5 = 17.0 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: V Ed = 343.25 kN Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 z = mx _ d − c V, − 30 mm; d − 2 c V, _ (NDP) 6.2.3 (1): Inner lever arm z z = mx{384; 378} = 384 mm 1.0 ≤ cot θ ≤ 1.2 + 1.4 σ cd / ƒ cd 1 − V Rd,cc / V Ed ≤ 3.0 (NDP) 6.2.3 (2): Eq. 6.7aDE V Rd,cc = c · 0.48 · ƒ 1/ 3 ck · _ 1 − 1.2 σ cd ƒ cd _ · b · z (NDP) Eq. 6.2.3 (2): 6.7bDE c = 0.5 V Rd,cc = 0.5 · 0.48 · 30 1/ 3 · (1 − 0) · 0.3 · 0.384 (NDP) 6.2.3 (2): σ cd = N Ed / A c V Rd,cc = 0.08591 MN = 85.91kN cot θ = 1.2 + 0 1 − 85.91 / 343.25 = 1.60 (NDP) 6.2.3 (2): The angle θ should be limited by Eq. 6.7DE A s,req / s = V Ed / (ƒ yd · z · cot θ) = 12.84 cm 2 / m 6.2.3 (3): Eq. 6.8 ƒ yd = ƒ yk / γ s = 435 MP V Rd,m = b · z · ν 1 · ƒ cd / (cot θ + tnθ) (NDP) 6.2.3 (3): Eq. 6.9 Maximum shear force V Rd,m occurs for θ = 45 ◦ : cot θ = tnθ = 1 ν 1 = 0.75 · ν 2 = 0.75, ν 2 = 1 for ≤ C50/ 60 V Rd,m = 0.3 · 0.384 · 0.75 · 17 / (1 + 1) = 734.4 kN Design with respect to EN 1992-1-1:2004 [77]: 3 z = 0.9 · d 6.2.3 (1): Inner lever arm z z = 0.9 · 450 = 405 mm 1.0 ≤ cot θ ≤ 2.5 → cot θ = 1.60 (choose for comparison) 6.2.3 (2): Eq. 6.7N The angle θ should be limited by Eq. 6.7N A s,req / s = V Ed / (ƒ yd · z · cot θ) = 12.18 cm 2 / m 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 62.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. 3 The sections mentioned in the margins refer to EN 1992-1-1:2004 [77], unless other- wise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 271 DCE-EN6: Design of a Rectangular CS for Shear Force 62.5 Conclusion This example shows the calculation of the required reinforcement for a rectangular cross-section under shear force. It has been shown that the results are reproduced with excellent accuracy. 62.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [77] EN 1992-1-1: Eurocode 2: Design of concrete structures, Part 1-1: General rules and rules for buildings. CEN. 2004. 272 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN7: Design of a T-section for Shear 63 DCE-EN7: Design of a T-section for Shear Overview Design Code Family(s): EN Design Code(s): EN1992 Module(s): AQB Input file(s): t-beam shear.dat 63.1 Problem Description The problem consists of a T-section, as shown in Fig. 63.1. The cross-section is designed for an ultimate shear force V Ed and the required reinforcement is determined. V Ed A s1 d 1 z s A s1 h ƒ b eƒ ƒ b h d Figure 63.1: Problem Description 63.2 Reference Solution This example is concerned with the design of sections for ULS, subject to shear force. The content of this problem is covered by the following parts of EN 1992-1-1:2004 [77]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Guidelines for shear design (Section 6.2) • Reinforcement (Section 9.2.2) V N M V(cot θ − cot α) V F td F cd θ α d z s shear reinforcement Figure 63.2: Shear Reinforced Members SOFiSTiK 2014 | VERiFiCATiON MANUAL 273 DCE-EN7: Design of a T-section for Shear The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 63.3 and as defined in EN 1992-1-1:2004 [77] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c ε ƒ yd = ƒ yk / γ s σ Figure 63.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 63.3 Model and Results The T-section, with properties as defined in Table 63.1, is to be designed, with respect to EN 1992-1- 1:2004 [77] to carry an ultimate shear force of 450 kN. The reference calculation steps are presented below and the results are given in Table 63.2. Table 63.1: Model Properties Material Properties Geometric Properties Loading C 30/ 37 h = 60.0 cm V Ed = 450 kN B 500A d = 53.0 cm d 1 = 7.0 cm b = 30 cm b eƒ ƒ = 180 cm h ƒ = 15 cm A s1 = 15 cm 2 The intermediate steps of calculating the required reinforcement are also validated in this example. First we calculate the design value for the shear resistance V Rd,c for members not requiring shear reinforcement. It gives a value of V Rd,c = 93.76 kN. Checking the results in AQB, we can see that SOFiSTiK outputs also V rd1,c = 93.76. Just to test this result, if we input a shear force of V Ed = 93.75 just below the value for V Rd,c , AQB will not output any value for cot θ and the minimum reinforcement will be printed (M). If we now give a value of V Ed = 93.77 just larger than V Rd,c , then AQB will start increasing cot θ and the minimum reinforcement will be printed. If we continue increasing V Ed , AQB will continue increasing cot θ until it reaches the upper limit of cot θ = 2.5 with using the minimum reinforcement. If now the minimum reinforcement is exceeded, AQB starts calculating a value for the required reinforcement. Another option to test this limit of V Rd,c = 93.76, would be to keep cot θ = 1.0 274 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN7: Design of a T-section for Shear and now with V Ed = 93.77, AQB calculates a value for the required reinforcement larger than the minimum reinforcement. For the maximum value of the angle θ, hence cot θ = 1.0, the maximum value allowed for V Ed can be calculated as 642.23 kN. This can be found in AQB results as the V rd2,c = 642.23 for the case of cot θ = 1.0. Giving as an input a shear force just above this value V Ed = 642.24 triggers a warning ”Shear design not possible”. Next step is the validation of V Rd,m . When the design shear force V Ed exceeds V Rd,m then cot θ must be decreased so that V Ed = V Rd,m . The reference result for V Rd,m is 442.92 kN. Inputing a value just below that, should give a cot θ = 2.5, whereas for a value just above should give cot θ < 2.5. This can be verified easily in AQB output for V Ed = 442.91 and 442.93 kN, respectively. Also the minimum reinforcement is calculated exactly by AQB with a value of 2.63 cm 2 / m. Table 63.2: Results SOF. Ref. A s,req / s [cm 2 / m] 8.90 8.90 A s,mn / s [cm 2 / m] 2.63 2.63 V Rd,c [kN] 93.76 93.76 V Rd,m [kN] 442.92 442.92 V Ed,m [kN] 642.23 642.23 SOFiSTiK 2014 | VERiFiCATiON MANUAL 275 DCE-EN7: Design of a T-section for Shear 63.4 Design Process 1 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 30 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 30/ 1.5 = 17.0 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: V Ed = 450.0 kN Design with respect to EN 1992-1-1:2004 [77]: 2 z = 0.9 · d = 0.9 · 450 = 405 mm 6.2.3 (1): Inner lever arm z V Rd,c = _ C Rd,c · k · (100 · ρ 1 · ƒ ck ) 1/ 3 + k 1 · σ cp _ · b · d 6.2.2 (1): Design value for shear resis- tance V Rd,c for members not requiring design shear reinforcement C Rd,c = 0.18/ γ c = 0.12 k = 1 + _ _ 200 d = 1.6143 < 2.0 ρ 1 = A s b d = 0.0094 < 0.02 V Rd,c = _ 0.12 · 1.6143 · (100 · 0.0094 · 30) 1/ 3 + 0 _ · 0.3 · 0.53 V Rd,c = 93.76 kN V Ed > V Rd,c → shear reinforcement is required 1.0 ≤ cot θ ≤ 2.5 → start with cot θ = 2.50 6.2.3 (2): Eq. 6.7N: The angle θ should be limited by Eq. 6.7N V Rd,m = b · z · ν · ƒ cd / (cot θ + tnθ) 6.2.3 (3): Eq. 6.9 ν = 0.6 · _ 1 − ƒ ck 2500 _ = 0.528 6.2.2 (6): Eq. 6.6N V Rd,m = 0.3 · 0.477 · 0.528 · 17 / (2.5 + 0.4) = 442.92 kN V Ed > V Rd,m : increase θ so that V Ed = V Rd,m ⇒cot θ = 2.44 A s,req / s = V Ed / (ƒ yd · z · cot θ) = 8.90 cm 2 / m ρ ,mn = 0.08 · _ ƒ ck / ƒ yk = 0.08 · 30 / 500 = 0.0008764 9.2.2 (5): Eq. 9.5N A s,mn / s = ρ ,mn · b · sinα = 2.63 cm 2 / m 9.2.2 (5): Eq. 9.4 V Ed ≤ 0.5 · b · d · ν · ƒ cd = 642.23 kN 6.2.2 (6): Eq. 6.5 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [77] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 63.3. 2 The sections mentioned in the margins refer to EN 1992-1-1:2004 [77], unless other- wise specified. 276 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN7: Design of a T-section for Shear 63.5 Conclusion This example shows the calculation of the required reinforcement for a T-beam under shear force. It has been shown that the results are reproduced with excellent accuracy. 63.6 Literature [77] EN 1992-1-1: Eurocode 2: Design of concrete structures, Part 1-1: General rules and rules for buildings. CEN. 2004. SOFiSTiK 2014 | VERiFiCATiON MANUAL 277 DCE-EN7: Design of a T-section for Shear 278 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force 64 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): rectangular shear axial.dat 64.1 Problem Description The problem consists of a rectangular section, symmetrically reinforced for bending, as shown in Fig. 64.1. The cross-section is designed for a shear force V Ed and a compressive force N Ed and and the required reinforcement is determined. V Ed N Ed z d h b A s,tot / 2 A s,tot / 2 Figure 64.1: Problem Description 64.2 Reference Solution This example is concerned with the design of sections for ULS, subject to shear force and axial force. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72] [73]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Guidelines for shear design (Section 6.2) • Reinforcement (Section 9.2.2) V N M V(cot θ − cot α) V F td F cd θ α d z s shear reinforcement Figure 64.2: Shear Reinforced Members SOFiSTiK 2014 | VERiFiCATiON MANUAL 279 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 64.3 and defined in DIN EN 1992-1-1:2004 [72]. A B A B Idealised Design ƒ yk ε ƒ yd = ƒ yk / γ s σ Figure 64.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 64.3 Model and Results The rectangular cross-section, with properties as defined in Table 64.1, is to be designed, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73], under a shear force of 343.25 kN and a compressive axial force of 500.0 kN . The reference calculation steps are presented below and the results are given in Table 64.2. Table 64.1: Model Properties Material Properties Geometric Properties Loading C 30/ 37 h = 50.0 cm V Ed = 343.25 kN B 500A b = 30 cm N Ed = 500.0 kN d = 45.0 cm A s,tot = 38.67 cm 2 c V, = 3.6 cm Table 64.2: Results SOF. Ref. A s / s [cm 2 / m] 11.27 11.27 V Rd,c [kN] 132.71 132.71 cot θ 1.82 1.82 280 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force 64.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 30 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 30/ 1.5 = 17.0 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: V Ed = 343.25 kN N Ed = −500.0 kN z = mx _ d − c V, − 30 mm; d − 2 c V, _ (NDP) 6.2.3 (1): Inner lever arm z z = mx{384; 378} = 384 mm V Rd,c = _ C Rd,c · k · (100 · ρ 1 · ƒ ck ) 1/ 3 + 0.12 · σ cp _ · b · d (NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance V Rd,c for member- s not requiring design shear reinforce- ment with a minimum of _ ν mn + 0.12 · σ cp _ · b · d (NDP) 6.2.2 (1): Eq. 6.2b C Rd,c = 0.15/ γ c = 0.1 (NDP) 6.2.2 (1): C Rd,c = 0.15/ γ c k = 1 + _ _ 200 d = 1 + _ _ 200 450 = 1.6667 < 2.0 ρ 1 = A s,tot / 2 b d = 0.01432 < 0.02 V Rd,c,mn = _ ν mn + 0.12 · σ cp _ · b · d ν mn = (0.0525/ γ c ) · k 3/ 2 · ƒ 1/ 2 ck = 0.41249 (NDP) 6.2.2 (1): Eq. 6.3aDE: ν mn for d ≤ 600 mm V Rd,c,mn = 109.68 kN σ cp = N Ed / A c < 0.2 · ƒ cd (NDP) 6.2.2 (1): Eq. 6.2 σ cp > 0 for compression σ cp = −500 · 10 −3 / 0.15 · 10 6 = −3.3333 N/ mm 2 < 3.4 V Rd,c = _ 0.1 · 1.6667 · (1.432 · 30) 1/ 3 + 0.112 · 3.3333 _ · 0.3 · 0.45 = 132.71 kN V Ed > V Rd,c → shear reinforcement is required 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 64.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 281 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force 1.0 ≤ cot θ ≤ 1.2 + 1.4 σ cd / ƒ cd 1 − V Rd,cc / V Ed ≤ 3.0 (NDP) 6.2.3 (2): Eq. 6.7aDE V Rd,cc = c · 0.48 · ƒ 1/ 3 ck · _ 1 − 1.2 σ cd ƒ cd _ · b · z (NDP) 6.2.3 (2): Eq. 6.7bDE c = 0.5 σ cp = N Ed / A c (NDP) 6.2.3 (2): σ cd = N Ed / A c σ cd = −500 · 10 −3 / 0.15 · 10 6 = −3.3333 N/ mm 2 V Rd,cc = 0.5 · 0.48 · 30 1/ 3 · _ 1 − 1.2 3.3333 17.0 _ · 0.3 · 0.384 6.7DE: σ cd > 0 for compression V Rd,cc = 65.6948 kN cot θ = 1.2 + 1.4 · 3.3333 / 17.0 1 − 65.6948 / 343.25 = 1.823 (NDP) 6.2.3 (2): The angle θ should be limited by Eq. 6.7DE A s,req / s = V Ed / (ƒ yd · z · cot θ) = 11.27 cm 2 / m 6.2.3 (3): Eq. 6.8 ƒ yd = ƒ yk / γ s = 435 MP V Rd,m = b · z · ν 1 · ƒ cd / (cot θ + tnθ) (NDP) 6.2.3 (3): Eq. 6.9 Maximum shear force V Rd,m occurs for θ = 45 ◦ : cot θ = tnθ = 1 (NDP) ν 1 = 0.75 · ν 2 = 0.75, ν 2 = 1 for ≤ C50/ 60 V Rd,m = 0.3 · 0.384 · 0.75 · 17 / (1 + 1) = 734.4 kN 282 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force 64.5 Conclusion This example shows the calculation of the required reinforcement for a rectangular beam cross-section under shear with compressive axial force. It has been shown that the results are reproduced with excellent accuracy. 64.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. SOFiSTiK 2014 | VERiFiCATiON MANUAL 283 DCE-EN8: Design of a Rectangular CS for Shear and Axial Force 284 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN9: Design of a Rectangular CS for Shear and Torsion 65 DCE-EN9: Design of a Rectangular CS for Shear and Torsion Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): rectangular shear torsion.dat 65.1 Problem Description The problem consists of a rectangular section, symmetrically reinforced for bending, as shown in Fig. 65.1. The cross-section is designed for shear force V Ed and torsion T Ed and the required shear and torsion reinforcement is determined. T Ed V Ed d h b A s,tot / 2 A s,tot / 2 Figure 65.1: Problem Description 65.2 Reference Solution This example is concerned with the design of sections for ULS, subject to shear force and torsion. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72] [73]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Guidelines for shear (Section 6.2) and torsion design (Section 6.3) • Reinforcement (Section 9.2.2, 9.2.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 285 DCE-EN9: Design of a Rectangular CS for Shear and Torsion V T Ed t eƒ t eƒ / 2 centre-line A k cover z Figure 65.2: Torsion Reinforced Members The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 65.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c ε ƒ yd = ƒ yk / γ s σ Figure 65.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 65.3 Model and Results The rectangular cross-section, with properties as defined in Table 65.1, is to be designed, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73], under shear force of 175.0 kN and torsional moment 35 kNm. The reference calculation steps [75] are presented below and the results are given in Table 65.2. Table 65.1: Model Properties Material Properties Geometric Properties Loading C 35/ 45 h = 70.0 cm V Ed = 175.0 kN B 500A b = 30 cm T Ed = 35.0 kNm d = 65.0 cm A s,tot = 26.8 cm 2 286 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN9: Design of a Rectangular CS for Shear and Torsion Table 65.2: Results SOF. Ref. A s / s (T) [cm 2 / m] 3.35 3.35 A s (T) [cm 2 ] 5.37 5.37 A s,tot / s [cm 2 / m] 13.60 13.59 V Rd,m [kN] 1303.04 1303.03 T Rd,m [kNm] 124.95 124.95 SOFiSTiK 2014 | VERiFiCATiON MANUAL 287 DCE-EN9: Design of a Rectangular CS for Shear and Torsion 65.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NA) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 30 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 35/ 1.5 = 19.833 MP (NDP) 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects ƒ yk = 500 MP (NDP) 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: V Ed = 175.0 kN, T Ed = 35.0 kN T Ed ≤ V Ed · b 4.5 (NDP) 6.3.2 (5): Eq. (NA.6.31.1): For approximately rectangular solid sections no shear and torsion rein- forcement is required except from the minimum reinforcement, provided that the condition is satisfied and (NDP) 6.3.2 (5): Eq. (NA.6.31.2): When the condition equation is not fulfilled then shear and torsion design has to be reverified 35 > 135 · 0.3 4.5 = 9.0 → Eq. NA.6.31.1 is not fulfilled V Ed · _ 1 + 4.5 · T Ed V Ed · b _ ≤ V Rd,c V Rd,c = _ C Rd,c · k · (100 · ρ 1 · ƒ ck ) 1/ 3 + 0.12 · σ cp _ · b · d (NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance V Rd,c for member- s not requiring design shear reinforce- ment C Rd,c = 0.15/ γ c = 0.1 (NDP) 6.2.2 (1): C Rd,c = 0.15/ γ c k = 1 + _ _ 200 d = 1.5773 < 2.0 ρ 1 = A s b d = 0.01276 < 0.02 V Rd,c = _ 0.1 · 1.5773 · (100 · 0.01276 · 35) 1/ 3 + 0 _ · 0.3 · 0.65 V Rd,c = 109.13 kN 175 · _ 1 + 4.5 · 35 135 · 0.3 _ = 700 > 109.13 → requirement of Eq. NA.6.31.2 is not met Torsional reinforcement t eƒ ƒ ,1 = t eƒ ƒ ,2 = 2 · 50 = 100 mm (s o = s = s s = 50 mm) 6.3.1 (3): Solid sections may be mod- elled by equivalent thin-walled sections (Fig. 65.2) A k = (h − s − s o ) · _ b − t eƒ ƒ ,1 _ = 100 mm A k : area enclosed by the centre-line 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 65.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. 288 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN9: Design of a Rectangular CS for Shear and Torsion A k = (700 − 50 − 50) · (300 − 100) = 120000 mm 2 = 0.12 m 2 k = 2 · [(700 − 50 − 50) + (300 − 100)] = 1600 mm = 1.6 m k : circumference of area A k 6.3.2 (2): The effects of torsion and shear may be superimposed, assuming the same value for θ Simplifying, the reinforcement for torsion may be determined alone un- der the assumption of cot θ = 1.0, θ = 45 ◦ and be added to the independently calculated shear force reinforcement. A s,req / s = T Ed · tnθ / (ƒ yd · 2A k ) (NDP) 6.3.2 (3): Eq. (NA.6.28.1) A s,req / s = 350 · 1.0 / (435 · 2 · 0.12) = 3.35 cm 2 / m (T) A s,req = T Ed · k · cot θ / (ƒ yd · 2A k ) = 5.37 cm 2 (T) 6.3.2 (3): Eq. 6.28 Torsional resistance moment T Rd,m = 2 · ν · ƒ cd · A k · t eƒ ƒ , · sinθ · cos θ 6.3.2 (4): Eq. 6.30 (NDP): ν = 0.525 sinθ · cos θ = 0.5 since θ = 45 ◦ T Rd,m = 124.95 kNm Check of the concrete compressive strut bearing capacity for the load combination of shear force and torsion The maximum resistance of a member subjected to torsion and shear is limited by the capacity of the concrete struts. The following condition should be satisfied: _ T Ed T Rd,m _ 2 + _ V Ed V Rd,m _ 2 ≤ 1 (NDP) 6.3.2 (4): Eq. (NA.6.29.1) for sol- id cross-sections z = mx _ d − c V, − 30 mm; d − 2 c V, _ (NDP) 6.2.3 (1): Inner lever arm z c V, = s o − D o / 2 = 50 − 28/ 2 = 36 mm s o : offset of reinforcement D o : bar diameter z = mx{584; 578} = 584 mm V Rd,m = b · z · ν 1 · ƒ cd / (cot θ + tnθ) (NDP) 6.2.3 (3): Eq. 6.9 Maximum shear force V Rd,m occurs for θ = 45 ◦ : cot θ = tnθ = 1 (NDP) 6.2.3 (3):ν 1 = 0.75· ν 2 = 0.75, ν 2 = 1 for ≤ C50/ 60 V Rd,m = 0.3 · 0.584 · 0.75 · 19.833 / (1 + 1) = 1303.03 kN _ 35 124.95 _ 2 + _ 175 1303.04 _ 2 = 0.0965 < 1 Shear reinforcement A s,req / s = V Ed / (ƒ yd · z · cot θ) = 6.89 cm 2 / m 6.2.3 (3): Eq. 6.8 ƒ yd = ƒ yk / γ s = 435 MP Total required reinforcement Required torsional reinforcement: 2 · A s / s = 2 · 3.35 = 6.7 cm 2 / m (double-shear connection) Total reinforcement: A s,tot / s = 6.7 + 6.89 = 13.59 cm 2 / m SOFiSTiK 2014 | VERiFiCATiON MANUAL 289 DCE-EN9: Design of a Rectangular CS for Shear and Torsion 65.5 Conclusion This example shows the calculation of the required reinforcement for a rectangular beam cross-section under shear and torsion. It has been shown that the results are reproduced with excellent accuracy. 65.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. 290 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN10: Shear between web and flanges of T-sections 66 DCE-EN10: Shear between web and flanges of T- sections Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): t-beam shear web flange.dat 66.1 Problem Description The problem consists of a T-beam section, as shown in Fig. 66.1. The cs is designed for shear, the shear between web and flanges of T-sections is considered and the required reinforcement is determined. V Ed A s1 d 1 A s1 h ƒ b eƒ ƒ h d Figure 66.1: Problem Description 66.2 Reference Solution This example is concerned with the shear design of T-sections, for the ultimate limit state. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Guidelines for shear design (Section 6.2) SOFiSTiK 2014 | VERiFiCATiON MANUAL 291 DCE-EN10: Shear between web and flanges of T-sections F d F d b eƒ ƒ s ƒ h ƒ b F d + ΔF d F d + ΔF d Δ θ ƒ A sƒ longitudinal bar anchored beyond this point compressive struts Figure 66.2: Connection between flange and web in T-sections The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 66.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c = 525N/ mm 2 ε ε d = 25 0 / 00 ƒ yd = ƒ yk / γ s σ Figure 66.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 66.3 Model and Results The T-section, with properties as defined in Table 66.1, is to be designed for shear, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73]. The structure analysed, consists of a single span beam with a distributed load in gravity direction. The cross-section geometry, as well as the shear cut under consideration can be seen in Fig. 66.4. 292 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN10: Shear between web and flanges of T-sections 135 29 106 39.4 95.6 40 250 1 1 1 1 104 104 Figure 66.4: Cross-section Geometry, Properties and Shear Cuts Table 66.1: Model Properties Material Properties Geometric Properties Loading C 20/ 25 h = 135.0 cm P g = 500 kN/ m B 500A h ƒ = 29 cm, h = 106 cm d 1 = 7.0 cm b = 40 cm b eƒ ƒ , = 104 cm, b eƒ ƒ = 250 cm The system with its loading as well as the moment and shear force are shown in Fig. 66.5. The reference calculation steps [75] are presented in the next section and the results are given in Table 66.2. 2 5 0 0 . 0 - 2 5 0 0 . 0 - 2 0 0 0 . 0 2 0 0 0 . 0 - 1 5 0 0 . 0 1 5 0 0 . 0 - 1 0 0 0 . 0 1 0 0 0 . 0 - 5 0 0 . 0 5 0 0 . 0 6 2 5 0 . 0 6 0 0 0 . 0 6 0 0 0 . 0 5 2 5 0 . 0 5 2 5 0 . 0 4 0 0 0 . 0 4 0 0 0 . 0 2 2 5 0 . 0 2 2 5 0 . 0 500.0 Figure 66.5: Loaded Structure, Resulting Moment and Shear Force SOFiSTiK 2014 | VERiFiCATiON MANUAL 293 DCE-EN10: Shear between web and flanges of T-sections Table 66.2: Results At beam 1001 SOF. Ref. A s1 [cm 2 ] 39.44 39.44 A sƒ / s ƒ [cm 2 / m] 10.46 10.45 V Rd,c [kN] at = 0.0 m 73.41 73.41 V Rd,c [kN] at = 1.0 m 154.37 154.51 V Rd,c,mn [kN] 73.41 73.41 V Rd,m / s [kN] 1366.65 1367.16 cot θ 1.65 1.65 z [m] at = 1.0 m 1.25 1.25 V Ed [kN] 748.99 748.80 294 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN10: Shear between web and flanges of T-sections 66.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 25 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 25/ 1.5 = 14.17 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 σ sd = 456.52 MP Design Load for beam 1001: M Eds = 2250 kNm μ Eds = M Eds b eƒ ƒ · d 2 · ƒ cd = 2250 · 10 −3 2.5 · 1.28 2 · 14.17 = 0.03876 ω = 0.03971 and ξ = 0.9766 (interpolated) Tab. 9.2 [74]: ω−Table for up to C50/ 60 without compression rein- forcement A s1 = 1 σ sd · (ω· b · d · ƒ cd + N Ed ) = 39.44 cm 2 N Ed = 0 z = ξ · d = 1.25 m The shear force, is determined by the change of the longitudinal force, at the junction between one side of a flange and the web, in the separated flange: 6.2.4 (3): Eq. 6.20 V Ed = ΔF d = ΔM Eds z · b eƒ ƒ , b eƒ ƒ = 2250 1.25 · 1.04 2.5 = 748.8 kN V Rd,c = _ C Rd,c · k · (100 · ρ 1 · ƒ ck ) 1/ 3 + 0.12 · σ cp _ · b · d (NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance V Rd,c for member- s not requiring design shear reinforce- ment with a minimum of _ ν mn + 0.12 · σ cp _ · b · d (NDP) 6.2.2 (1): Eq. 6.2b C Rd,c = 0.15/ γ c = 0.1 (NDP) 6.2.2 (1): C Rd,c = 0.15/ γ c for beam 1001 at = 0.0 m: k = 1 + _ _ 200 d = 1 + _ _ 200 1215 = 1.4057 < 2.0 ρ 1 = A s b d = 0.0 < 0.02 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 66.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 295 DCE-EN10: Shear between web and flanges of T-sections V Rd,c = 0 V Rd,c,mn = _ ν mn + 0.12 · σ cp _ · b · d ν mn = (0.0375/ γ c ) · k 3/ 2 · ƒ 1/ 2 ck = 0.20833 (NDP) 6.2.2 (1): Eq. 6.3bDE V Rd,c,mn = 73.41 kN →V Rd,c = 73.41 kN V Ed > V Rd,c → shear reinforcement is required for beam 1001 at = 1.0 m: k = 1 + _ _ 200 d = 1 + _ _ 200 1280 = 1.3953 < 2.0 ρ 1 = A s b d = 39.44 29 · 128 = 0.01062 < 0.02 V Rd,c = _ 0.1 · 1.3953 · (100 · 0.01062 · 25) 1/ 3 + 0 _ · 0.29 · 1280 V Rd,c = 154.51 kN V Ed > V Rd,c → shear reinforcement is required 1.0 ≤ cot θ ≤ 1.2 + 1.4 σ cd / ƒ cd 1 − V Rd,cc / V Ed ≤ 3.0 (NDP) 6.2.3 (2): Eq. 6.7aDE V Rd,cc = c · 0.48 · ƒ 1/ 3 ck · _ 1 − 1.2 σ cd ƒ cd _ · b · z (NDP) Eq. 6.2.3 (2): 6.7bDE c = 0.5 b = h ƒ = 0.29 m and z = Δ = 1 m (NDP) 6.2.4 (4) V Rd,cc = 0.5 · 0.48 · 25 1/ 3 · (1 − 0) · 0.29 · 1.0 (NDP) 6.2.3 (2): σ cd = N Ed / A c N Ed = 0 V Rd,cc = 203.5kN cot θ = 1.2 + 0 1 − 203.5 / 748.8 = 1.647 cot θ ƒ = 1.647 (NDP) 6.2.3 (2): The angle θ should be limited by Eq. 6.7DE A sƒ ,req / s ƒ = V Ed / (ƒ yd · z · cot θ) 6.2.3 (3): Eq. 6.8 ƒ yd = ƒ yk / γ s = 435 MP A sƒ ,req / s ƒ = 748.8 · 10 −3 / (435 · 1.0 · 1.647) A sƒ ,req / s ƒ = 10.45 cm 2 / m V Ed ≤ ν · ƒ cd · sinθ ƒ · cos θ ƒ · h ƒ · Δ (NDP) 6.2.4 (4): Eq. 6.22 Maximum shear force V Rd,m (NDP) 6.2.3 (3): ν = ν 1 = 0.75 for ≤ C50/ 60 sinθ ƒ · cos θ ƒ = 0.4436 V Rd,m = 0.75 · 14.17 · 0.4436 · 0.29 · 1.0 V Rd,m = 1367.16 kN 296 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN10: Shear between web and flanges of T-sections 66.5 Conclusion This example is concerned with the calculation of the shear between web and flanges of T-sections. It has been shown that the results are reproduced with excellent accuracy. 66.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. M¨ uller, and F. Lobisch. Bemessungshilfsmittel f ¨ ur Betonbauteile nach Eu- rocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. SOFiSTiK 2014 | VERiFiCATiON MANUAL 297 DCE-EN10: Shear between web and flanges of T-sections 298 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN11: Shear at the interface between concrete cast 67 DCE-EN11: Shear at the interface between concrete cast Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): shear interface.dat 67.1 Problem Description The problem consists of a T-beam section, as shown in Fig. 67.1. The cs is designed for shear, the shear at the interface between concrete cast at different times is considered and the required reinforcement is determined. V Ed A s1 Shear section d 1 z s A s1 h ƒ b eƒ ƒ b h d Figure 67.1: Problem Description 67.2 Reference Solution This example is concerned with the shear design of T-sections, for the ultimate limit state. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.3) • Guidelines for shear design (Section 6.2) SOFiSTiK 2014 | VERiFiCATiON MANUAL 299 DCE-EN11: Shear at the interface between concrete cast ≤ 30 ◦ h 2 ≤ 10d h 1 ≤ 10d anchorage anchorage V Ed N Ed α 45 ◦ ≤ α ≤ 90 ◦ new concrete old concrete b b Figure 67.2: Indented Construction Joint - Examples of Interfaces The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as presented in Fig. 67.3 and as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). A B A B Idealised Design ƒ yk ƒ tk,c ε ƒ yd = ƒ yk / γ s σ Figure 67.3: Idealised and Design Stress-Strain Diagram for Reinforcing Steel 67.3 Model and Results The T-section, with properties as defined in Table 67.1, is to be designed for shear, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73]. The reference calculation steps [75] are presented in the next section and the results are given in Table 67.2. Table 67.1: Model Properties Material Properties Geometric Properties Loading C 20/ 25 h = 135.0 cm V z = 800 kN B 500A h ƒ = 29 cm M y = 2250 kNm d 1 = 7.0 cm b = 40 cm , b eƒ ƒ = 250 cm A s1 = 1.0 cm 2 z s = 95.56 cm 300 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN11: Shear at the interface between concrete cast Table 67.2: Results s [cm 2 / m] SOF. Ref. state 7.06 7.07 state only V 4.91 4.90 state V + M 4.99 4.99 SOFiSTiK 2014 | VERiFiCATiON MANUAL 301 DCE-EN11: Shear at the interface between concrete cast 67.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 25 MP Tab. 3.1: Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 25/ 1.5 = 14.17 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 σ sd = 456.52 MP τ = T b = V · S y · b where S is the static moment of the separated area S = h · b · (z s − h / 2) = 0.18058 m 3 τ = 0.8 · 0.18058 0.16667 · 0.4 = 2.1669 MP T = 0.8 · 0.18058 0.16667 = 0.86676 MN/ m = 866.76 kN/ m The shear section with a length of 0.4 m is split into two equal parts with b = 0.2 m T = 866.76 / 2 = 433.38 kN/ m State : Design Load: The associated design shear flow V Ed is: V Ed = T = 433.38 kN/ m Ed = τ = 2.1669 MP V Rd,c = _ C Rd,c · k · (100 · ρ 1 · ƒ ck ) 1/ 3 + 0.12 · σ cp _ · b · d (NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance V Rd,c for member- s not requiring design shear reinforce- ment Rd,c = C Rd,c · k · (100 · ρ 1 · ƒ ck ) 1/ 3 + 0.12 · σ cp ρ 1 = A s b d = 0.0 → Rd,c = 0.0 with a minimum of V Rd,c,mn = _ ν mn + 0.12 · σ cp _ · b · d (NDP) 6.2.2 (1): Eq. 6.2b Rd,c,mn = ν mn + 0.12 · σ cp ν mn = (0.0375/ γ c ) · k 3/ 2 · ƒ 1/ 2 ck = 0.20833 MP (NDP) 6.2.2 (1): Eq. 6.3bDE 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8, which can be seen in Fig. 67.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. 302 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN11: Shear at the interface between concrete cast Rd,c,mn = 0.20833 → Rd,c = 0.20833 MP Ed > Rd,c → shear reinforcement is required Ed ≤ Rd 6.2.5 (1): Eq. 6.23: The design shear stress at the interface should satisfy this Rd = c · ƒ ctd + μ · σ n + ρ · ƒ yd · (1.2 · μ · sinα + cos α) (NDP) 6.2.5 (1): Eq. 6.25 and Rd ≤ 0.5 · ν · ƒ cd Maximum shear stress Rd,m (NDP) 6.2.5 (1): ν = 0.70 for indented surface Rd,m = 0.5 · ν · ƒ cd = 4.9585 MP c = 0.50 and μ = 0.9 for indented surface 6.2.5 (2): c, μ: factors depending on the roughness of the interface ƒ ctd = α ct · ƒ ctk;0.05 / γ c (NDP) 3.1.6 (2)P: Eq. 3.16 α ct = 0.85 3.1.2 (3): Tab. 3.1 - Strength for con- crete: ƒ ctk;0.05 = 1.8 MP: ƒ ctd = 0.85 · 1.80 / 1.5 = 1.02 Rd = 0.5 · 1.02 + 0 + s 0.2 · 1.0 · 435 · (1.2 · 0.9 · 1 + 0) ρ = s b · : area of reinforcement cross- ing interface / area of joint Rd = 0.51 + s 0.2 · 469.56 = 2.1669 s = 7.07 cm 2 / m State only shear force V: Design Load: From the calculated inner lever arms for the two states we get a ratio: z z = 0.7664 The associated design shear flow V Ed is: V Ed = 0.7664 · 433.38 = 332.15 kN/ m and Ed = 332.15/ 0.2 = 1.66 MP Following the same calculation steps as for state we have: Rd,c = 0.20833 MP (as above) Ed > Rd,c → shear reinforcement is required Ed ≤ Rd Rd = c · ƒ ctd + μ · σ n + ρ · ƒ yd · (1.2 · μ · sinα + cos α) Rd = 0.5 · 1.02 + 0 + s 0.2 · 1.0 · 435 · (1.2 · 0.9 · 1 + 0) Rd = 0.51 + s 0.2 · 469.56 = 1.66 s = 4.90 cm 2 / m SOFiSTiK 2014 | VERiFiCATiON MANUAL 303 DCE-EN11: Shear at the interface between concrete cast State shear force V and moment M: M Eds = 2250 kNm μ Eds = M Eds b eƒ ƒ · d 2 · ƒ cd = 2250 · 10 −3 2.5 · 1.28 2 · 14.17 = 0.03876 ω = 0.03971 and ξ = 0.9766 (interpolated) Tab. 9.2 [74]: ω−Table for up to C50/ 60 without compression rein- forcement A s1 = 1 σ sd · (ω· b · d · ƒ cd + N Ed ) = 39.44 cm 2 N Ed = 0 z = mx _ d − c V, − 30 mm; d − 2 c V, _ (NDP) 6.2.3 (1): Inner lever arm z z = mx{1160; 1190} = 1190 mm Design Load: T = V / z = 800 / 1.19 = 672.268 kN/ m The shear section with a length of 0.4 m is split into two equal parts with b = 0.2 m T = 672.268 / 2 = 336.134 kN/ m V Ed = 336.134 kN/ m and Ed = 336.134 / 0.2 = 1.68 MP Rd,c = C Rd,c · k · (100 · ρ 1 · ƒ ck ) 1/ 3 + 0.12 · σ cp C Rd,c = 0.15/ γ c = 0.1 (NDP) 6.2.2 (1): C Rd,c = 0.15/ γ c k = 1 + _ _ 200 d = 1 + _ _ 200 1280 = 1.3953 < 2.0 ρ 1 = A s b d = 39.44 40 · 128 = 0.007703 < 0.02 Rd,c = 0.1 · 1.3953 · (100 · 0.007703 · 25) 1/ 3 + 0 Rd,c = 0.373229 MP Ed > Rd,c → shear reinforcement is required Rd = c · ƒ ctd + μ · σ n + ρ · ƒ yd · (1.2 · μ · sinα + cos α) Rd = 0.5 · 1.02 + 0 + s 0.2 · 1.0 · 435 · (1.2 · 0.9 · 1 + 0) Rd = 0.51 + s 0.2 · 469.56 = 1.68 s = 4.99 cm 2 / m 304 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN11: Shear at the interface between concrete cast 67.5 Conclusion This example shows the calculation of the required reinforcement for a T-section under shear at the interface between concrete cast at different times. It has been shown that the results are reproduced with excellent accuracy. 67.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. M¨ uller, and F. Lobisch. Bemessungshilfsmittel f ¨ ur Betonbauteile nach Eu- rocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. SOFiSTiK 2014 | VERiFiCATiON MANUAL 305 DCE-EN11: Shear at the interface between concrete cast 306 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN12: Calculation of crack widths 68 DCE-EN12: Calculation of crack widths Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB Input file(s): crack widths.dat 68.1 Problem Description The problem consists of a rectangular section, asymmetrically reinforced, as shown in Fig. 68.1. Differ- ent loading conditions are examined, always consisting of a bending moment M Ed , and in addition with or without a compressive or tensile axial force N Ed . The crack width is determined. N Ed M Ed z d h b A s A s Figure 68.1: Problem Description 68.2 Reference Solution This example is concerned with the control of crack widths. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Design stress-strain curves for concrete and reinforcement (Section 3.1.7, 3.2.7) • Basic assumptions for calculation of crack widths (Section 7.3.2, 7.3.3, 7.3.4) A B 5(c + ϕ/ 2) h − D E C ϕ c h c,eƒ d h A ε 1 ε 2 = 0 B Figure 68.2: Stress and Strain Distributions in the Design of Cross-sections SOFiSTiK 2014 | VERiFiCATiON MANUAL 307 DCE-EN12: Calculation of crack widths The design stress-strain diagram for reinforcing steel considered in this example, consists of an inclined top branch, as defined in DIN EN 1992-1-1:2004 [72] (Section 3.2.7). 68.3 Model and Results The rectangular cross-section, with properties as defined in Table 68.1, is to be designed for crack width, with respect to DIN EN 1992-1-1:2004 (German National Annex) [72] [73]. The calculation steps with different loading conditions and calculated with different sections of DIN EN 1992-1-1:2004 are presented below and the results are given in Table 68.2. Table 68.1: Model Properties Material Properties Geometric Properties Loading C 25/ 30 h = 100.0 cm N Ed = 0 or ±300 kN B 500A d = 96.0 cm M Ed = 562.5 kNm b = 30.0 cm ϕ s = 25.0 mm, A s = 24.50 cm 2 ϕ s = 12.0 mm, A s = 2.26 cm 2 k = 0.3 mm Table 68.2: Results Case Load A s given [cm 2 ] Result SOF. Ref. M Ed , N Ed = 0 24.50 A s,req [cm 2 ] 6.93 6.93 σ s [MP] 207.85 207.85 M Ed , N Ed = 300 24.50 A s,req [cm 2 ] 10.39 10.39 M Ed , N Ed = −300 24.50 A s,req [cm 2 ] 4.04 4.04 V M Ed , N Ed = 0 24.50 A s [cm 2 ] passed with given reinforcement σ s [MP] 440.57 440.53 V M Ed , N Ed = 0 12.0 A s [cm 2 ] not passed with given reinforcement → new: 14.54 σ s [MP] 436.30 436.28 V M Ed , N Ed = 0 24.50 k [mm] 0.13 0.13 308 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN12: Calculation of crack widths 68.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: γ c = 1.50 (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials Steel: γ s = 1.15 ƒ ck = 25 MP Tab. 3.1 : Strength for concrete ƒ cd = cc · ƒ ck / γ c = 0.85 · 25/ 1.5 = 14.17 MP 3.1.6: (1)P, Eq. (3.15): cc = 0.85 con- sidering long term effects ƒ yk = 500 MP 3.2.2: (3)P: yield strength ƒ yk = 500 MP ƒ yd = ƒ yk / γ s = 500/ 1.15 = 434.78 MP 3.2.7: (2), Fig. 3.8 Design Load: M Ed = 562.5 kNm N Ed = 0.0 or ±300 kN Minimum reinforcement areas 7.3.2: If crack control is required, a min- imum amount of bonded reinforcement is required to control cracking in areas where tension is expected • Case : M Ed , N Ed = 0.0 A s,mn · σ s = k c · k · ƒ ct,eƒ ƒ · A ct 7.3.2 (2): Eq. 7.1: A s,mn minimum area of reinforcing steel within tensile zone ƒ ct,eƒ ƒ = ƒ ctm ≥ 3.0 MP (NDP) 7.3.2 (2): ƒ ct,eƒ ƒ mean value of concrete tensile strength When the cracking time can’t be placed with certainty in the first 28 days then ƒ ct,eƒ ƒ ≥ 3.0 MP for normal concrete ƒ ct,eƒ ƒ = 2.6 < 3.0 MP →ƒ ct,eƒ ƒ = 3.0 MP Tab. 3.1: ƒ ctm = 2.6 MP for C 25 7.3.2 (2): A ct is the area of concrete within tensile zone, for pure bending of a rectangular section its height is h/ 2 A ct = b · h eƒ ƒ = 0.3 · 0.5 = 0.15 m 2 f ct,eff h eƒ ƒ = h/ 2 7.3.2 (2): Eq. 7.2: k c for bending of rectangular sections k c = 0.4 · _ 1 − σ c k 1 · (h/ h ∗ ) · ƒ ct,eƒ ƒ _ ≤ 1 7.3.2 (2): Eq. 7.4: σ c mean stress of the concrete σ c = N Ed / (b · h) = 0.0 → k c = 0.4 7.3.2 (2): h is the lesser value of b, h h = min{300, 1000} = 300 mm k = 0.8 for h ≤ 300 7.3.2 (NA.5): Eq. NA.7.5.2: For deter- mination of the reinforcement stress the maximum bar diameter has to be modi- fied ϕ s = ϕ ∗ s · ƒ ct,eƒ ƒ / 2.9 N/ mm 2 ϕ ∗ s = 25 · 2.9 / 3.0 = 24.16667 σ s = _ k · 3.48 · 10 6 / ϕ ∗ s = 207.846 MP (NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width A s,req = 0.4 · 0.8 · 3.0 · 0.15 · 10 4 / 207.846 = 6.928cm 2 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.2.7:(2), Fig. 3.8. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 309 DCE-EN12: Calculation of crack widths • Case : N Ed = 300 kN A s,mn · σ s = k c · k · ƒ ct,eƒ ƒ · A ct 7.3.2 (2): Eq. 7.1: A s,mn minimum area of reinforcing steel within tensile zone ƒ ct,eƒ ƒ = ƒ ctm ≥ 3.0 MP ƒ ct,eƒ ƒ = 2.6 < 3.0 MP →ƒ ct,eƒ ƒ = 3.0 MP Tab. 3.1: ƒ ctm = 2.6 MP for C 25 A ct = b · h eƒ ƒ = 0.15 m 2 7.3.2 (2): A ct is the area of concrete within tensile zone σ c = N Ed b · h = 300 · 10 3 300 · 1000 = 1.0 MP 7.3.2 (2): Eq. 7.4: σ c mean stress of the concrete, tensile stress σ c < 0 k = 0.8 for h ≤ 300 7.3.2 (2): h is the lesser value of b, h h = min{300, 1000} = 300 mm (NDP) 7.3.2 (NA.5): Eq. NA.7.5.2: For determination of the reinforcemen- t stress the maximum bar diameter has to be modified ϕ s = ϕ ∗ s · ƒ ct,eƒ ƒ / 2.9 N/ mm 2 ϕ ∗ s = 25 · 2.9 / 3.0 = 24.16667 σ s = _ k · 3.48 · 10 6 / ϕ ∗ s = 207.846 MP (NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width k c = 0.4 · _ 1 − σ c k 1 · (h/ h ∗ ) · ƒ ct,eƒ ƒ _ ≤ 1 7.3.2 (2): Eq. 7.2: k c for bending with axial force of rectangular sections k 1 = 2 · h ∗ / (3 · h) = 2/ 3 h ∗ = 1.0 m for h ≥ 1.0 m k 1 = 2 · h ∗ / (3 · h) if N Ed tensile force k c = 0.4 · _ 1 + 1.0 (2/ 3) · 1 · 3.0 _ = 0.6 ≤ 1 7.3.2 (2): Eq. 7.1: Tensile stress σ c < 0 A s,req = 0.6 · 0.8 · 3.0 · 0.15 · 10 4 / 207.846 = 10.39cm 2 • Case : N Ed = −300 kN A s,mn · σ s = k c · k · ƒ ct,eƒ ƒ · A ct 7.3.2 (2): Eq. 7.1: A s,mn minimum area of reinforcing steel within tensile zone ƒ ct,eƒ ƒ = ƒ ctm ≥ 3.0 MP ƒ ct,eƒ ƒ = 2.6 < 3.0 MP →ƒ ct,eƒ ƒ = 3.0 MP Tab. 3.1: ƒ ctm = 2.6 MP for C 25 A ct = b · h eƒ ƒ 7.3.2 (2): A ct is the area of concrete within tensile zone The height of the tensile zone is determined through the stresses: σ c = N Ed b · h = 300 · 10 3 300 · 1000 = 1.0 MP 7.3.2 (2): Eq. 7.4: σ c mean stress of the concrete, compressive stress σ c > 0 σ = ƒ ct,eƒ ƒ = 3.0 MP ⇒ h eƒ ƒ = 3.0 · 50 3.0 + 1.0 = 37.5cm A ct = 0.3 · 0.375 = 0.1125 m 2 k = 0.8 for h ≤ 300 7.3.2 (2): h is the lesser value of b, h h = min{300, 1000} = 300 mm (NDP) 7.3.2 (NA.5): Eq. NA.7.5.2: For determination of the reinforcemen- t stress the maximum bar diameter has to be modified ϕ s = ϕ ∗ s · ƒ ct,eƒ ƒ / 2.9 N/ mm 2 ϕ ∗ s = 25 · 2.9 / 3.0 = 24.16667 310 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN12: Calculation of crack widths σ s = _ k · 3.48 · 10 6 / ϕ ∗ s = 207.846 MP (NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width k c = 0.4 · _ 1 − σ c k 1 · (h/ h ∗ ) · ƒ ct,eƒ ƒ _ ≤ 1 7.3.2 (2): Eq. 7.2: k c for bending with axial force of rectangular sections k 1 = 1.5 h ∗ = 1.0 m for h ≥ 1.0 m k 1 = 1.5 if N Ed compressive force k c = 0.4 · _ 1 − 1.0 1.5 · 1 · 3.0 _ = 0.3111 ≤ 1 A s,req = 0.3111 · 0.8 · 3.0 · 0.1125 · 10 4 / 207.846 = 4.04 cm 2 Control of cracking without direct calculation 7.3.3: Control of cracking without direct calculation Examples calculated in this section are w.r.t. Table 7.2DE, here Table 7.3N is not relevant • Case V: N Ed = 0.0 ƒ ct,eƒ ƒ = ƒ ctm ≥ 3.0 MP (NDP) 7.3.2 (2): ƒ ct,eƒ ƒ mean value of concrete tensile strength ƒ ct,eƒ ƒ = 2.6 < 3.0 MP →ƒ ct,eƒ ƒ = 3.0 MP Tab. 3.1: ƒ ctm = 2.6 MP for C 25 (NDP) 7.3.3: Eq. 7.7.1DE: The maxi- mum bar diameters should be modified for load action ϕ s = ϕ ∗ s · σ s · A s 4(h − d) · b · 2.9 ≥ ϕ ∗ s · ƒ ct,eƒ ƒ 2.9 A s = 24.5 cm 2 prescribed reinforce- ment ϕ s = 25 mm = ϕ ∗ s · 264.06 · 24.50 4(100 − 96) · 30 · 2.9 = ϕ ∗ s · 4.6476 → ϕ ∗ s = 5.3791 mm σ s = _ k · 3.48 · 10 6 / ϕ ∗ s = 440.53 MP (NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width σ s = 264.06 < 440.53 MP which corresponds to the value calculated in SOFiSTiK [ssr] → crack width control is passed with given reinforcement. In case the usage factor becomes 1.0 then the stresses σ s are equal, as it can be seen in Case V below. and ϕ s = 25 mm > ϕ ∗ s · ƒ ct,eƒ ƒ 2.9 = 5.3791 · 3.0 2.9 = 5.5646 also control the steel stress with respect to the calculated strains ε s = 0.440 + 1.913 · (0.50 − 0.04) = 1.31998 → σ s = 0.00131998 · 200000 = 264.0 MP or from Tab. 7.2DE and for σ s = 264.04 ≈ 260.0 MP (NDP) 7.3.3 Tab. 7.2DE (a) → ϕ ∗ s = 15 mm → ϕ s = ϕ ∗ s · σ s · A s 4(h − d) · b · 2.9 ≥ ϕ ∗ s · ƒ ct,eƒ ƒ 2.9 (NDP) 7.3.3: Eq. 7.7.1DE: The maxi- mum bar diameters should be modified for load action ϕ s = 15 · 264.06 · 24.50 4(100 − 96) · 30 · 2.9 = 69.69 mm > 25 mm → crack width control is passed with given reinforcement. SOFiSTiK 2014 | VERiFiCATiON MANUAL 311 DCE-EN12: Calculation of crack widths • Case V: N Ed = 0.0, A s = 12.0 cm 2 In this case, the prescribed reinforcement is decreased from A s = 24.5 cm 2 to A s = 12.0 cm 2 in order to examine a case where the the crack width control is not passed with the given reinforcement. ƒ ct,eƒ ƒ = ƒ ctm ≥ 3.0 MP (NDP) 7.3.2 (2): ƒ ct,eƒ ƒ mean value of concrete tensile strength ƒ ct,eƒ ƒ = 2.6 < 3.0 MP →ƒ ct,eƒ ƒ = 3.0 MP Tab. 3.1: ƒ ctm = 2.6 MP for C 25 from Tab. 7.2DE and for σ s = 509.15 ≈ 510.0 MP (NDP) 7.3.3 Tab. 7.2DE (a) → ϕ ∗ s ≈ 3.9 mm → ϕ s = ϕ ∗ s · σ s · A s 4(h − d) · b · 2.9 (NDP) 7.3.3: Eq. 7.7.1DE: The maxi- mum bar diameters should be modified for load action ϕ s = 3.9 · 509.15 · 12.0 4(100 − 96) · 30 · 2.9 = 17.12 mm < 25 mm → crack width control is not passed with given reinforcement. →start increasing reinforcement in order to be in the limits of admissible steel stresses from Tab. 7.2DE and for σ s = 436.43 MP (NDP) 7.3.3 Tab. 7.2DE (a) → ϕ ∗ s ≈ 5.6 mm → ϕ s = ϕ ∗ s · σ s · A s 4(h − d) · b · 2.9 7.3.3: Eq. 7.7.1DE: The maximum bar diameters should be modified for load action ϕ s = 5.6 · 436.43 · 14.54 4(100 − 96) · 30 · 2.9 = 25.45 mm≥ 25 mm → crack width control passed with additional reinforcement. If we now input as prescribed reinforcement the reinforcement that is calculated in order to pass crack control, i.e. A s = 14.54 cm 2 we get a steel stress of σ s = 436.36 MP which gives ϕ s = 25 mm = ϕ ∗ s · 436.36 · 14.54 4(100 − 96) · 30 · 2.9 → ϕ ∗ s = 5.4849 mm σ s = _ k · 3.48 · 10 6 / ϕ ∗ s (NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width σ s = _ 0.3 · 3.48 · 10 6 / 5.4849 = 436.28 MP Here we can notice that the stresses are equal leading to a usage factor of 1.0 Control of cracking with direct calculation 7.3.4: Control of cracking with direct cal- culation • Case V: N Ed = 0.0 α e = E s / E cm = 200000 / 31476 = 6.354 7.3.4 (1): α e is the ratio E s / E cm ρ p,eƒ ƒ = (A s + ξ 2 1 · A p ) / A c,eƒ ƒ 7.3.4 (1): Eq. 7.10: where A p and A c,eƒ ƒ are defined in 7.3.2 (3) 312 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN12: Calculation of crack widths A c,eƒ ƒ = h c,eƒ · b 7.3.2 (3): h c,eƒ see Fig. 7.1DE (d) h / d 1 = 100/ 4 = 25.00 → h c,eƒ / d 1 = 3.25 A c,eƒ ƒ = (3.25 · 4) · 30 = 13 · 30 = 390 cm 2 A p = 0.0 cm 2 7.3.2 (3): where A p is the area of pre or post-tensioned tendons within A c,eƒ ƒ ρ p,eƒ ƒ = 24.50 / 390 = 0.06282 ε sm − ε cm = σ s − k t · ƒ ct,eƒ ƒ ρ p,eƒ ƒ · (1 + α e · ρ p,eƒ ƒ ) E s ≥ 0.6 · σ s E s 7.3.4 (1): Eq. 7.9: the difference of the mean strain in the reinforcement and in the concrete ƒ ct,eƒ ƒ = ƒ ctm = 0.30 · ƒ 2/ 3 ck = 2.565 ≈ 2.6 MP (NDP) 7.3.2 (2): ƒ ct,eƒ ƒ mean value of concrete tensile strength, here no mini- mum value of ƒ ct,eƒ ƒ ≥ 3.0 MP is set ƒ ct,eƒ ƒ = 2.6 < 3.0 MP →ƒ ct,eƒ ƒ = 3.0 MP Tab. 3.1: ƒ ctm = 2.6 MP for C 25 or ƒ ctm = 0.30 · ƒ 2/ 3 ck = 2.565 MP ε sm − ε cm = 264.06 − 0.4 · 2.565 0.06282 · (1 + 6.354 · 0.06282) 200000 k t : factor dependent on the duration of the load, k t = 0.4 for long term loading ε sm − ε cm = 1.2060 · 10 −3 > 0.6 · 264.06 200000 = 0.79218 · 10 −3 s r,m = ϕ 3.6 · ρ p,eƒ ƒ ≤ σ s · ϕ 3.6 · ρ p,eƒ ƒ (NDP) 7.3.4 (3): Eq. 7.11: s r,m is the maximum crack spacing s r,m = 25 3.6 · 0.06282 = 110.545 mm s r,m ≤ σ s · ϕ 3.6 · ρ p,eƒ ƒ = 611.25 mm k = s r,m · (ε sm − ε cm ) = 110.545 · 1.2060 · 10 −3 7.3.4 (1): Eq. 7.8: k the crack width k = 0.133 < 0.3 mm → Check for crack width passed with given reinforcements SOFiSTiK 2014 | VERiFiCATiON MANUAL 313 DCE-EN12: Calculation of crack widths 68.5 Conclusion This example shows the calculation of crack widths. Various ways of reference calculations are demon- strated, in order to compare the SOFiSTiK results to. It has been shown that the results are reproduced with excellent accuracy. 68.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. 314 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN13: Design of a Steel I-section for Bending and Shear 69 DCE-EN13: Design of a Steel I-section for Bending and Shear Overview Design Code Family(s): EN Design Code(s): EN1993 Module(s): AQB Input file(s): usage steel.dat 69.1 Problem Description The problem consists of a steel I- section, as shown in Fig. 69.1. The cross-section is designed for bending and shear. y z b h V y M z t r t ƒ Figure 69.1: Problem Description 69.2 Reference Solution This example is concerned with the resistance of steel cross-sections for bending and shear. The content of this problem is covered by the following parts of EN 1993-1-1:2005 [78]: • Structural steel (Section 3.2 ) • Resistance of cross-sections (Section 6.2) 69.3 Model and Results The I-section, a HEA 200, with properties as defined in Table 69.1, is to be designed for an ultimate moment M z and a shear force V y , with respect to EN 1993-1-1:2005 [78]. The calculation steps are SOFiSTiK 2014 | VERiFiCATiON MANUAL 315 DCE-EN13: Design of a Steel I-section for Bending and Shear presented below and the results are given in Table 69.2. The utilisation level of allowable plastic forces are calculated and compared to SOFiSTiK results. Table 69.1: Model Properties Material Properties Geometric Properties Loading S 235 HEA 200 V y = 200 or 300 kN b = 20.0 cm M z = 20 kNm h = 19.0 cm t ƒ = 1.0 cm, t = 0.65 cm r = 1.8 cm Table 69.2: Results Case Result SOF. Ref. V p,Rd,y [kN] 542.71 542.71 V p,Rd,z [kN] 245.32 245.32 Util. level V y 0.369 0.369 Util. level M z 0.418 0.418 Util. level V y 0.553 0.553 Util. level M z,red 0.422 0.422 316 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN13: Design of a Steel I-section for Bending and Shear 69.4 Design Process 1 Material: Structural Steel S 235 ƒ y = 235 N/ mm 2 Tab. 3.1 : Nominal values of yield strength ƒ y and ultimate tensile strength ƒ for hot rolled structural steel. Cross-sectional properties: W p,y = 429.5 cm 3 Plastic section modulus of HEA 200 w.r.t. y− and z−axis W p,z = 203.8 cm 3 Where the shear force is present allowance should be made for its effect on the moment resistance. This effect may be neglected, where the 6.2.8: Bending and shear shear force is less than half the plastic shear resistance. V Ed ≤ 0.5 V p,Rd 6.2.8 (2) V p,Rd = A V · _ ƒ y / 3 _ γ M0 6.2.6 (2): Eq. 6.18: V p,Rd the design plastic shear resistance A V the shear area w.r.t. y− and z−axis, respectively A V y = 2 · A ƒ nge (in the y−direction only contribution the two flanges ) A V y = 2 · t ƒ · b = 40 cm 2 A V z = A − 2 · b · t ƒ + (t + 2 · r) · t ƒ but not less than η · h · t 6.2.6 (3): The shear area A V may be taken as follows for rolled I-sections with load parallel to the web h = h − 2 · t ƒ = 17 cm A V z = 18.08 cm 2 > 1 · 17 · 0.65 = 11.05 6.2.6 (3): η may be conservatively taken equal to 1.0 6.1 (1): Partial factor γ M0 = 1.00 is recommended V p,Rd,y = 40 · _ 235 / 3 _ 1.00 = 542.71 kN V p,Rd,z = 18.08 · _ 235 / 3 _ 1.00 = 245.32 kN • Case : V Ed = V y = 200 kN, M z = 20 kNm V Ed V p,Rd,y = 0.369 < 0.5 → no reduction of moment resistance due to shear M z,V,Rd = W p,z · ƒ y γ M0 = 47.89 kNm 6.2.8 (5): Eq. 6.30: The reduced design plastic resistance moment (for ρ = 0) M z M z,V,Rd = 0.418 1 The sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless other- wise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 317 DCE-EN13: Design of a Steel I-section for Bending and Shear • Case : V Ed = V y = 300 kN, M z = 20 kNm V Ed V p,Rd,y = 0.553 > 0.5 → reduction of moment resistance due to shear 6.2.8 (3): The reduced moment resis- tance should be taken as the design re- sistance of the cross-section, calculated using a reduced yield strength (1 − ρ) · ƒ y for the shear area ρ = _ 2 V Ed V p,Rd,y − 1 _ 2 = 0.01114 M z,V,Rd = (1 − ρ) · W p,z · ƒ y γ M0 M z,V,Rd = (1 − 0.01114) · 47.89 = 47.36 kNm M z M z,V,Rd = 0.422 318 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN13: Design of a Steel I-section for Bending and Shear 69.5 Conclusion This example shows the calculation of the resistance of steel cross-sections for bending and shear. It has been shown that the results are reproduced with excellent accuracy. 69.6 Literature [78] EN 1993-1-1: Eurocode 3: Design of concrete structures, Part 1-1: General rules and rules for buildings. CEN. 2005. SOFiSTiK 2014 | VERiFiCATiON MANUAL 319 DCE-EN13: Design of a Steel I-section for Bending and Shear 320 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN14: Classification of Steel Cross-sections 70 DCE-EN14: Classification of Steel Cross-sections Overview Design Code Family(s): EN Design Code(s): EN1993 Module(s): AQB Input file(s): class steel.dat 70.1 Problem Description The problem consists of a steel I- section, as shown in Fig. 70.1. The cross-section is classified for bending and compression. y z b h t r t ƒ Figure 70.1: Problem Description 70.2 Reference Solution This example is concerned with the classification of steel cross-sections. Section classification is a vital step in checking the suitability of a section to sustain any given design actions. It is concerned with the local buckling susceptibility and is invovled on the resistance checks of the section. The content of this problem is covered by the following parts of EN 1993-1-1:2004 [78]: • Structural steel (Section 3.2 ) • Classification of cross-sections (Section 5.5) • Cross-section requirements for plastic global analysis (Section 5.6) • Resistance of cross-sections (Section 6.2) • Buckling resistance of members (Section 6.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 321 DCE-EN14: Classification of Steel Cross-sections A diagrammatic representation of the four classes of section is given in Fig. 70.2, where a cross-section is subjected to an increasing major axis bending moment until failure [79]. Class 1: Fullly plastic moment is reached Class 2: Fullly plastic moment reached but unloading occurs Class 3: Partial yielding occurs, parts of section still elastic Class 4: Local buckling causes unloading in the elastic range Curvature Bending Moment Fully Plastic Moment Figure 70.2: Idealized Moment Curvature Behaviour for Four Classes of Cross-sections 70.3 Model and Results The I-section, a UB 457x152x74, with properties as defined in Table 70.1, is to be classified for bending and compression, with respect to EN 1993-1-1:2005 [78]. For the compression case, an axial load of N = 3000 kN is applied and for the bending case a moment of M y = 500 kNm. In AQB the classification of the cross-sections is done taking into account the stress levels and the respective design request. Thus, a Class 1 cross-section can be reached, only if a nonlinear design (plastic-plastic) is requested and if the loading is such as to cause the yield stress to be exceeded. Therefore, in order to derive the higher Class possible for this cross-section, we consider these loads, which will cause higher stresses than the yield stress. The calculation steps are presented below and the results are given in Table 70.2. Table 70.1: Model Properties Material Properties Geometric Properties Loading S 275 UB 457x152x74 N = −3000 kN b = 154.4 mm M y = 500 kNm h = 462.0 mm t ƒ = 17.0 mm t = 9.6 mm r = 10.2 mm 322 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN14: Classification of Steel Cross-sections Table 70.2: Results Case Part Result SOF. Ref. Flange c/ t 3.66 3.66 Web 42.46 42.46 Bending Flange Class 1 1 Web 1 1 Compression Flange Class 1 1 Web 4 4 SOFiSTiK 2014 | VERiFiCATiON MANUAL 323 DCE-EN14: Classification of Steel Cross-sections 70.4 Design Process 1 Material: Structural Steel S 275 Tab. 3.1 : Nominal values of yield strength ƒ y and ultimate tensile strength ƒ for hot rolled structural steel. ƒ y = 275 N/ mm 2 for maximum thickness ≤ 40 mm ε = _ 235/ ƒ y = 0.924 Tab. 5.2: Maximum width-to-thickness ratios for compression parts The role of cross-section classification is to identify the extent to which the resistance and rotation capacity of cross-sections is limited by its local buckling resistance. Tab. 5.5.1(1): Classification of cross- section basis • Bending: For the flange: c = b/ 2 − t / 2 − r Tab. 5.2 (sheet 2): Outstand flanges 5.5.2 (3): The classification depends on the width to thickness ratio of the parts subject to compression c / t = (154.4/ 2 − 9.6/ 2 − 10.2) / 17 = 3.66 c / t ≤ 9 ε ≤ 8.32 Tab. 5.2 (sheet 2): Outstand flanges - part subject to compression → Flange classification: Class 1 For the web: c = h − 2t ƒ − 2r Tab. 5.2 (sheet 1): Internal compres- sion parts c / t = (462 − 2 · 17 − 2 · 10.2) / 9.6 = 42.46 c / t ≤ 72 ε ≤ 66.53 Tab. 5.2 (sheet 1): Internal compres- sion parts - part subject to bending → Web classification: Class 1 Overall classification for bending: Class 1 Class 1 cross-sections are those which can form a plastic hinge with the rotation capacity required from plastic analysis without reduction of the resistance. 5.5.2 (1): Classification • Compression: For the flange as above → Flange classification: Class 1 For the web as above : c / t = 42.46 Class 3: c / t ≤ 42 ε ≤ 38.8 Tab. 5.2 (sheet 1): Internal compres- sion parts - part subject to compression Tab. 5.5.2 (8): A part which fails to sat- isfy the limits of Class 3 should be taken as Class 4 c / t = 42.46 > 38.8 → Web classification: Class 4 Overall classification for bending: Class 4 Tab. 5.5.2 (6): A cross-section is clas- sified according to the highest least favourable class Class 4 cross-sections are those in which local buckling will occur be- fore the attainment of yield stress in one or more parts of the cross- section. 5.5.2 (1): Classification 1 The sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless other- wise specified. 324 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN14: Classification of Steel Cross-sections 70.5 Conclusion This example shows the classification of steel cross-sections for bending and compression. It has been shown that the results are reproduced with excellent accuracy. 70.6 Literature [78] EN 1993-1-1: Eurocode 3: Design of concrete structures, Part 1-1: General rules and rules for buildings. CEN. 2005. [79] Structural Eurocodes - Extracts from the structural Eurocodes for students of structural design. BSI - British Standards Institution. 2007. SOFiSTiK 2014 | VERiFiCATiON MANUAL 325 DCE-EN14: Classification of Steel Cross-sections 326 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN15: Buckling Resistance of Steel Members 71 DCE-EN15: Buckling Resistance of Steel Members Overview Design Code Family(s): EN Design Code(s): EN1993 Module(s): BDK Input file(s): buckling steel.dat 71.1 Problem Description The problem consists of a simply supported beam with a steel I-section subject to uniform end moments, as shown in Fig. 71.1. The cross-section is checked against buckling. b h M y M y y L t r t ƒ Figure 71.1: Problem Description 71.2 Reference Solution This example is concerned with the buckling resistance of steel members. Lateral torsional buckling occurs in unrestrained, or inadequately restrained beams bent about the major axis and this causes the compression flange to buckle and deflect sideways, thus inducing twisting of the section. The content of this problem is covered by the following parts of EN 1993-1-1:2005 [78]: • Structural steel (Section 3.2 ) • Classification of cross-sections (Section 5.5) • Buckling resistance of members (Section 6.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 327 DCE-EN15: Buckling Resistance of Steel Members 71.3 Model and Results The I-section, a UB 457x152x74, with properties as defined in Table 71.1, is to be ckecked for buckling, with respect to EN 1993-1-1:2005 [78]. The calculation steps [79] are presented below and the results are given in Table 71.2. Table 71.1: Model Properties Material Properties Geometric Properties Loading S 275 L = 8 m M y = 150 kNm E = 210000 N/ mm 2 UB 457x152x74 h = 462.0 mm b = 154.4 mm t ƒ = 17.0 mm t = 9.6 mm r = 10.2 mm A = 94.48 cm 2 z = 1046.5 cm 4 T = 66.23 cm 4 = 516297.12 cm 6 Table 71.2: Results SOF. Ref. M cr [kNm] 154.26 154.26 λ LT 1.703 1.703 LT 1.907 1.907 χ LT 0.321 0.321 M Ed / M b,Rd (BDK) 1.045 1.045 328 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN15: Buckling Resistance of Steel Members 71.4 Design Process 1 Material: Structural Steel S 275 Tab. 3.1 : Nominal values of yield strength ƒ y and ultimate tensile strength ƒ for hot rolled structural steel. ƒ y = 275 N/ mm 2 for maximum thickness ≤ 40 mm ε = _ 235/ ƒ y = 0.924 Tab. 5.2: Maximum width-to-thickness ratios for compression parts Design Load: M Ed = 150kNm The cross-section is classified as Class 1, as demonstrated in Bench- mark DCE-EN14. Tab. 5.5: Classification of cross-section M c,Rd = M p,Rd,y = W p,y · ƒ y γ M0 = 447.31 kNm 6.2.5 (2):Eq. 6.13: Bending resistance M c,Rd for Class 1 cross-section 6.1 (1): γ M0 = 1.00 M cr = π _ E z G T L _ _ _ _ 1 + π 2 E G T L 2 _ 6.3.2.2 (2): M cr , the elastic critical mo- ment for ltb is based on gross cross sec- tional properties Various empirical or approximate formu- lae exist for the determination of M cr M cr = 3.14 · 2197.74 · 53.496 8 _ _ _ _ 1 + 3.14 2 · 108.42 53.496 · 8 2 _ M cr = 154.26 kNm λ LT = _ _ _ W y ƒ y M cr = _ _ 447.31 154.26 = 1.703 6.3.2.2 (1): λ LT non dimensional slen- derness for lateral torsional buckling LT = 0.5 _ 1 + α LT _ λ LT − λ LT,0 + βλ 2 LT __ 6.3.2.3 (1): LT for rolled sections in bending. Recommended values: λ LT,0 = 0.4 (maximum value) β = 0.75 (minimum value) 6.3.2.3 (1): Table 6.5: Recommenda- tion for the selection of ltb curve for cross-sections using Eq. 6.57 h / b = 462/ 154.4 = 2.99 > 2 for rolled I-sections and h / b > 2 → buckling curve c for buckling curve c → α LT = 0.49 6.3.2.2 (2): Table 6.5: Recommenda- tion values for imperfection factors α LT for ltb curves LT = 0.5 _ 1 + 0.49 _ 1.703 − 0.4 + 0.75 · 1.703 2 __ = 1.907 χ LT = 1 LT + _ 2 LT − β λ 2 LT = 0.321 6.3.2.3 (1): Eq. 6.57: χ LT reduction fac- tor for ltb but χ LT = 0.321 ≤ 1.0 and χ LT = 0.321 ≤ 1 λ 2 LT = 0.345 M b,Rd = χ LT W y ƒ y γ M1 = 143.587 kNm 6.3.2.1 (3): Eq. 6.55: M b,Rd The design buckling resistance moment of laterally unrestrained beam For Class 1 section W y = W p,y 6.1 (1): γ M1 = 1.00 recommended val- ue M Ed M b,Rd = 1.045 → Beam fails in LTB 1 The sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless other- wise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 329 DCE-EN15: Buckling Resistance of Steel Members 71.5 Conclusion This example shows the ckeck for lateral torsional buckling of steel members. It has been shown that the results are reproduced with excellent accuracy. 71.6 Literature [78] EN 1993-1-1: Eurocode 3: Design of concrete structures, Part 1-1: General rules and rules for buildings. CEN. 2005. [79] Structural Eurocodes - Extracts from the structural Eurocodes for students of structural design. BSI - British Standards Institution. 2007. 330 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force 72 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force Overview Design Code Family(s): EN Design Code(s): EN1993 Module(s): AQB Input file(s): usage interact.dat 72.1 Problem Description The problem consists of a steel I- section, as shown in Fig. 72.1. The cross-section is designed for bending, shear and axial force. y z b h V y M z t r t ƒ Figure 72.1: Problem Description 72.2 Reference Solution This example is concerned with the resistance of steel cross-sections for the combination of bending, shear and axial force. Where the shear and axial force are present allowance should be made for the effect of both shear force and axial force on the resistance moment.The content of this problem is covered by the following parts of EN 1993-1-1:2005 [78]: • Structural steel (Section 3.2 ) • Resistance of cross-sections (Section 6.2) SOFiSTiK 2014 | VERiFiCATiON MANUAL 331 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force 72.3 Model and Results The I-section, a HEB 300, with properties as defined in Table 72.1, is to be designed for an ultimate moment M y , a shear force V z and an axial force N = −2000 kN, with respect to EN 1993-1-1:2005 [78]. The calculation steps [43] are presented below and the results are given in Table 72.2. The utilisation level of allowable plastic forces are calculated and compared to SOFiSTiK results. The insignificant deviation at the utilisation level of the plastic moment resistance ( Util. level M y,red = 0.879) arises from the fact that in EN 1993-1-1:2005 approximate formulas are given for the contribution of the area parts of the I-section to the allowance of the shear and the axial force over the moment resistance. Table 72.1: Model Properties Material Properties Geometric Properties Loading S 235 HEB 300 V z = 500 kN b = 30.0 cm M y = 150 kNm h = 30.0 cm N = −2000 kN t ƒ = 1.9 cm, t = 1.1 cm r = 2.7 cm A = 149.078 cm 2 Table 72.2: Results Result SOF. Ref. Util. level N 0.571 0.571 Util. level N red 0.633 0.633 Util. level V z 0.777 0.777 Util. level M y 0.342 0.342 Util. level M y,red 0.879 0.882 332 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force 72.4 Design Process 1 Material: Structural Steel S 235 ƒ y = 235 N/ mm 2 Tab. 3.1 : Nominal values of yield strength ƒ y and ultimate tensile strength ƒ for hot rolled structural steel. Cross-sectional properties: W p,y = 1868.6 cm 3 Plastic section modulus of HEB 300 w.r.t. y− A V z = A − 2 · b · t ƒ + (t + 2 · r) · t ƒ but not less than η · h · t 6.2.6 (3): The shear area A V may be taken as follows for rolled I-sections with load parallel to the web h = h − 2 · t ƒ = 26.2 cm A V z = 47.428 cm 2 > 1 · 26.2 · 1.1 = 28.82 6.2.6 (3): η may be conservatively taken equal to 1.0 6.1 (1): Partial factor γ M0 = 1.00 is recommended V p,Rd,z = 47.428 · _ 235 / 3 _ 1.00 = 643.49 kN M c,Rd = W p,y · ƒ y γ M0 = 439.12 kNm 6.2.5 (2): Eq. 6.13: The design re- sistance for bending for Class 1 cross- section Where the shear and axial force are present allowance should be made for the effect of both shear force and axial force on resistance moment. 6.2.10 (1): Bending, shear and axial force Provided that the design value of the shear force V Ed does not exceed the 50% of the design plastic shear resistance V p,Rd no reduction of the resistances for bending and axial force need to be made. 6.2.10 (2) V Ed ≤ 0.5 V p,Rd V Ed V p,Rd,z = 0.777 > 0.5 → shear resistance exceeded Where V Ed exceeds 50% of V p,Rd , the design resistance of the cross- section to combinations of moment and axial force should be calculated using a reduced yield strength (1 − ρ) · ƒ y for the shear area 6.2.10 (3) ρ = _ 2 V Ed V p,Rd,z − 1 _ 2 = 0.3069 6.2.8 (5): Eq. 6.30: The reduced design plastic resistance moment al- lowing for shear force may alternitavely be obtained for I-sections with equal flanges and bending about major axis where A = h · t M y,V,Rd = _ W p,y − ρ · A 2 4t _ · ƒ y γ M0 = 425.503 kNm M y M y,V,Rd = 0.353 M y M c,Rd = 0.342 (with no reduction) 1 The sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless other- wise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 333 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force N p,Rd = A · ƒ y γ M0 = 3503.3 kN 6.2.3 (2): The design plastic resistance of gross cross-section N p R d N V,Rd = N p,Rd · (1 − α · ρ) 6.2.10 (3): The reduced plastic resis- tance ue to shear α = A Vz / A N V,Rd = 3161.249 kN N Ed N V,Rd = 0.633 N Ed N p,Rd = 0.571 (with no reduction) Where an axial force is present, allowance should be made for its effect on the plastic moment resistance. 6.2.9.1 (1): Bending and axial force - Class 1 and 2 cross-sections 6.2.9.1 (4): For doubly symmetri- cal I-sections, allowance need not to be made for the effect of the axial force on the plastic resistance moment about the y − y axis when the criteria are satisfied N Ed ≤ _ _ _ _ _ _ _ _ _ _ _ 0.25 N V,Rd 0.5 h t ƒ y (1 − ρ) γ M0 N Ed = 2000 > _ _ _ 790.31 kN 234.71 kN → consideration of axial force in interaction M N,y,Rd = M p,y,Rd · 1 − n 1 − 0.5 α 6.2.9.1 (5): Eq 6.36: The following ap- proximations may be used for standard rolled I-sections with equal flanges and M N,y,Rd ≤ M p,y,Rd n = N Ed / N V,Rd = 0.6327 α = (1 − ρ) · (A − 2 · b · t ƒ ) / A = 0.1631 M N,y,Rd = 425.503 · 1 − 0.6327 1 − 0.5 0.1631 = 170.164 kNm M y,Ed M NV,y,Rd = 0.882 334 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force 72.5 Conclusion This example shows the calculation of the resistance of steel cross-sections for bending, shear and axial force. It has been shown that the results are reproduced with very good accuracy. 72.6 Literature [43] Schneider. Bautabellen f ¨ ur Ingenieure. 19th. Werner Verlag, 2010. [78] EN 1993-1-1: Eurocode 3: Design of concrete structures, Part 1-1: General rules and rules for buildings. CEN. 2005. SOFiSTiK 2014 | VERiFiCATiON MANUAL 335 DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force 336 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS 73 DCE-EN17: Stress Calculation at a Rectangular Pre- stressed Concrete CS Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB, TENDON Input file(s): stress prestress.dat 73.1 Problem Description The problem consists of a rectangular cross-section of prestressed concrete, as shown in Fig. 73.1. The stresses developed at the section due to prestress and bending are verified. M y N p A p h d z p Figure 73.1: Problem Description 73.2 Reference Solution This example is concerned with the design of prestressed concrete cs, subject to bending and prestress force. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Stress-strain curves for concrete and prestressing steel (Section 3.1.7, 3.3.6) • Verification by the partial factor method - Design values (Section 2.4.2) • Prestressing force (Section 5.10.2, 5.10.3) SOFiSTiK 2014 | VERiFiCATiON MANUAL 337 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS M/ W 2 N P / A N P · z p / W 2 + + = A p z p M/ W 1 N P / A N P · z p / W 1 Figure 73.2: Stress Distribution in Prestress Concrete Cross-section In rectangular cs, which are prestressed and loaded, stress conditions are developed, as shown in Fig. 73.2, where the different contributions of the loadings can be seen. The design stress-strain diagrams for prestressing steel is presented in Fig. 73.3, as defined in DIN EN 1992-1-1:2004 [72] (Section 3.3.6). A B A B Idealised Design ƒ p0,1k ƒ pk ε p ƒ pd = ƒ p0,1k / γ s σ p Figure 73.3: Idealised and Design Stress-Strain Diagram for Prestressing Steel 73.3 Model and Results The simply supported beam of Fig. 73.4, consists of a rectangular cross-section with properties as defined in Table 73.1 and is prestressed and loaded with its own weight. A verification of the stresses is performed in the middle of the span with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73]. The geometry of the tendon can be visualised in Fig. 73.5. The calculation steps [75] are presented below and the results are given in Table 73.2. Table 73.1: Model Properties Material Properties Geometric Properties Loading (at = 10 m) C 35/ 45 h = 100.0 cm M g = 1250 kNm Y 1770 b = 100.0 cm N p = −3651.1 kN d = 95.0 cm L = 20.0 m 338 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS Table 73.1: (continued) Material Properties Geometric Properties Loading (at = 10 m) A p = 28.5 cm 2 L Figure 73.4: Simply Supported Beam 1 10001 0 .0 0 1001 10002 1 .0 0 1002 10003 2 .0 0 1003 10004 3 .0 0 1004 10005 4 .0 0 1005 10006 5 .0 0 1006 10007 6 .0 0 1007 10008 7 .0 0 1008 10009 8 .0 0 1009 10010 9 .0 0 2 10011 1 0 .0 0 1010 10012 1 1 .0 0 1011 10013 1 2 .0 0 1012 10014 1 3 .0 0 1013 10015 1 4 .0 0 1014 10016 1 5 .0 0 1015 10017 1 6 .0 0 1016 10018 1 7 .0 0 1017 10019 1 8 .0 0 1018 10020 1 9 .0 0 3 2 0 .0 0 0 .0 0 0 0 .0 5 7 0 .0 5 7 0 .0 5 7 0 .1 1 4 0 .0 5 7 0 .1 7 1 0 .0 5 5 0 .2 2 3 0 .0 5 0 0 .2 7 1 0 .0 4 5 0 .3 1 3 0 .0 3 8 0 .3 4 7 0 .0 3 1 0 .3 7 4 0 .0 2 2 0 .3 9 0 0 .0 1 1 0 .3 9 6 0 .0 0 0 0 .3 9 0 - 0 .0 1 1 0 .3 7 4 - 0 .0 2 2 0 .3 4 7 - 0 .0 3 1 0 .3 1 3 - 0 .0 3 8 0 .2 7 1 - 0 .0 4 5 0 .2 2 3 - 0 .0 5 0 0 .1 7 1 - 0 .0 5 5 0 .1 1 5 - 0 .0 5 7 0 .0 5 7 - 0 .0 5 7 0 .0 0 0 - 0 .0 5 7 Figure 73.5: Tendon Geometry 0.916 0.932 0 .9 1 6 3 5 7 2 .3 0 .9 1 7 3 5 7 4 .2 0 .9 1 7 3 5 7 6 .1 0 .9 1 8 3 5 7 8 .1 0 .9 1 8 3 5 8 0 .2 0 .9 1 9 3 5 8 2 .6 0 .9 2 0 3 5 8 5 .6 0 .9 2 0 3 5 8 8 .9 0 .9 2 1 3 5 9 2 .4 0 .9 2 2 3 5 9 6 .1 0 .9 2 3 3 6 0 0 .0 0 .9 2 4 3 6 0 4 .2 0 .9 2 6 3 6 0 8 .5 0 .9 2 7 3 6 1 3 .1 0 .9 2 8 3 6 1 8 .0 0 .9 2 9 3 6 2 3 .0 0 .9 3 1 3 6 2 8 .3 0 .9 3 2 3 6 3 3 .8 0 .9 3 3 3 6 3 9 .5 0 .9 3 5 3 6 4 5 .4 0 .9 3 7 3 6 5 1 .6 0 .9 3 8 3 6 5 7 .8 0 .9 4 0 3 6 6 3 .8 0 .9 4 1 3 6 6 9 .6 0 .9 4 3 3 6 7 5 .1 0 .9 4 4 3 6 8 0 .5 0 .9 4 4 3 6 7 8 .8 0 .9 4 2 3 6 7 3 .9 0 .9 4 1 3 6 6 9 .2 0 .9 4 0 3 6 6 4 .8 0 .9 3 9 3 6 6 0 .6 0 .9 3 8 3 6 5 6 .6 0 .9 3 7 3 6 5 2 .8 0 .9 3 6 3 6 4 9 .2 0 .9 3 5 3 6 4 5 .9 0 .9 3 4 3 6 4 2 .9 0 .9 3 4 3 6 4 0 .4 0 .9 3 3 3 6 3 8 .2 0 .9 3 3 3 6 3 6 .3 0 .9 3 2 3 6 3 4 .3 0 . 9 4 4 3 6 8 2 . 2 0.973 1.000 = 1368 N/mm2 0.944 Figure 73.6: Prestress Forces and Stresses SOFiSTiK 2014 | VERiFiCATiON MANUAL 339 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS Table 73.2: Results Case CS Result SOF. Ref. 0 σ c,b [MP] −12.455 −12.454 M y [kNm] −1451.02 −1451.02 0 −4.882 −4.882 M y [kNm] −201.02 −201.02 1 −11.866 −11.866 M y [kNm] −1426.65 −1426.65 V 1 −4.626 −4.626 M y [kNm] −176.65 −176.65 340 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS 73.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: C 35/ 45 3.1: Concrete E cm = 34000 N/ mm 2 3.1.2: Tab. 3.1: E cm for C 35/ 45 Prestressing Steel: Y 1170 3.3: Prestressing Steel E p = 195000 N/ mm 2 3.3.6 (3): E p for wires ƒ pk = 1770 N/ mm 2 3.3.2, 3.3.3: ƒ pk Characteristic tensile strength of prestressing steel ƒ p0,1k = 1520 N/ mm 2 3.3.2, 3.3.3: ƒ p0,1k 0.1% proof-stress of prestressing steel, yield strength Prestressing system: BBV L19 150 mm 2 19 wires with area of 150 mm 2 each, giving a total of A p = 28.5 cm 2 Cross-section: A c = 1.0 · 1.0 = 1 m 2 Diameter of duct ϕ dct = 67 mm Ratio α E,p = E p / E cm = 195000 / 34000 = 5.74 A c,netto = A c − π · (ϕ dct / 2) 2 = 0.996 m 2 A de = A c + A p · α E,p = 1.013 m 2 The force applied to a tendon, i.e. the force at the active end during tensioning, should not exceed the following value 5.10.2.1 (1)P: Prestressing force during tensioning - Maximum stressing force P m = A p · σ p,m 5.10.2.1 (1)P: Eq. 5.41: P m maximum stressing force where σ p,m = min _ 0.80ƒ pk ; 0.90ƒ p0,1k _ (NDP) 5.10.2.1 (1)P: σ p,m maximum stress applied to the tendon P m = A p · 0.80 · ƒ pk = 28.5 · 10 −4 · 0.80 · 1770 = 4035.6 kN P m = A p · 0.90 · ƒ p0,1k = 28.5 · 10 −4 · 0.90 · 1520 = 3898.8 kN → P m = 3898.8 kN and σ p,m = 1368 N/ mm 2 The value of the initial prestress force at time t = t 0 applied to the concrete immediately after tensioning and anchoring should not exceed the following value 5.10.3 (2): Prestress force P m0 () = A p · σ p,m0 () 5.10.3 (2): Eq. 5.43: P m0 initial pre- stress force at time t = t 0 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.3.6: Fig. 3.10, which can be seen in Fig 73.3. 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 341 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS where σ p,m0 () = min _ 0.75ƒ pk ; 0.85ƒ p0,1k _ (NDP) 5.10.3 (2): σ p,m0 () stress in the tendon immediately after tensioning or transfer P m0 = A p · 0.75 · ƒ pk = 28.5 · 10 −4 · 0.75 · 1770 = 3783.4 kN P m0 = A p · 0.85 · ƒ p0,1k = 28.5 · 10 −4 · 0.85 · 1520 = 3682.2 kN → P m0 = 3682.2 kN and σ p,m0 = 1292 N/ mm 2 Further calculations for the distribution of prestress forces and stresses along the beam are not in the scope of this Benchmark and will not be described here. The complete diagram can be seen in Fig. 73.5, after the consideration of losses at anchorage and due to friction, as calculated by SOFiSTiK. There the values of σ p,m = 1368 N/ mm 2 and P m0 = 3682.2 kN can be visualised. Load Actions: Self weight per length: γ = 25 kN/ m → g 1 = γ · A = 25 · 1 = 25 kNm Safety factors at ultimate limit state DIN EN 1990/NA [80]: (NDP) A.1.3.1 (4): Tab. NA.A.1.2 (B): Partial factors for actions (NDP) 2.4.2.2 (1): Partial factors for pre- stress Actions (unfavourable) Safety factor at final state • permanent γ G = 1.35 • prestress γ P = 1.00 Combination coefficients at serviceability limit state g 1 = 25 kNm: for rare, frequent and quasi-permanent combination (for stresses) At = 10.0 m middle of the span: M g = g 1 · L 2 / 8 = 1250 kNm N p = P m0 ( = 10.0 m) = −3651.1 kN (from SOFiSTiK) Calculation of stresses σ c,b at = 10.0 m middle of the span: The concrete stresses may be deter- mined for each construction stage un- der the total quasi-permanent combina- tion σ c {G+ P m0 + ψ 2 · Q} In this Benchmark no variable load Q is examined Position of the tendon: z = 0, 396 m • Case : Prestress at construction stage section 0 (P cs0) M p z p N p P m0,=10 −σ c +σ c N p = −3651.1 kN 342 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS M p 1 = N P · z = −3651.1 · 0.396 = −1445.86 kNm z s the new position of the center of the cross-section for cs0 z p = z + z s M p 2 = N P · z s = −3651.1 · 0.001415 = −5.16 kNm M p = −1445.86 − 5.16 = −1451.02 kNm = M y M p bending moment caused by pre- stressing σ c,b = N p A c,netto + M y W 1,cs0 W 1,cs0 cross-section moduli for con- truction stage 0 at the bottom left and right point σ c,b stress at the concrete at the bottom of the cross section σ c,b = −3651.1 0.996 + −1451.02 0.165 = −12.454 MP • Case : Prestress and self-weight at con. stage sect. 0 (P+G cs0) M g + M p z p N p P m0,=10 −σ c +σ c N p = −3651.1 kN and M g = 1250 kNm As computed above: M p = −1451.02 kNm M y = 1250 − 1451.02 = −201.02 kNm σ c,b = −3651.1 0.996 + −201.02 0.165 = −4.882 MP • Case : Prestress at con. stage sect. 1 (P cs1) N p = −3651.1 kN and M p 1 = −1445.86 kNm (as above) M p 2 = N P · z s = −3651.1 · (−0.005259) = 19.2 kNm M p = −1445.86 + 19.2 = −1426.65 kNm = M y σ c,b = N p A de + M y W 1,cs1 W 1,cs1 cross-section moduli for con- truction stage 1 σ c,b = −3651.1 1.013 + −1426.65 0.173 = −11.866 MP • Case V: Prestress and self-weight at con. stage sect. 1 (P+G cs1) N p = −3651.1 kN and M g = 1250 kNm As computed above: M p = −1426.65 kNm M y = 1250 − 1426.65 = −176.65 kNm σ c,b = −3651.1 1.013 + −176.65 0.173 = −4.626 MP SOFiSTiK 2014 | VERiFiCATiON MANUAL 343 DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS 73.5 Conclusion This example shows the calculation of the stresses, developed in the concrete cross-section due to prestress and bending. It has been shown that the results are reproduced with excellent accuracy. 73.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. [80] DIN EN 1990/NA: Eurocode: Basis of structural design, Nationaler Anhang Deutschland DIN EN 1990/NA:2010-12. CEN. 2010. 344 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS 74 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS Overview Design Code Family(s): DIN Design Code(s): EN1992 Module(s): AQB, CSM Input file(s): creep shrinkage.dat 74.1 Problem Description The problem consists of a simply supported beam with a rectangular cross-section of prestressed con- crete, as shown in Fig. 74.1. The time dependent losses are calculated, considering the reduction of stress caused by the deformation of concrete due to creep and shrinkage, under the permanent loads. M y N p A p h z p Figure 74.1: Problem Description 74.2 Reference Solution This example is concerned with the calculation of creep and shrinkage on a prestressed concrete cs, subject to bending and prestress force. The content of this problem is covered by the following parts of DIN EN 1992-1-1:2004 [72]: • Creep and Shrinkage (Section 3.1.4) • Annex B: Creep and Shrinkage (Section B.1, B.2) • Time dependent losses of prestress for pre- and post-tensioning (Section 5.10.6) The time dependant losses may be calculated by considering the following two reductions of stress [72]: • due to the reduction of strain, caused by the deformation of concrete due to creep and shrinkage, under the permanent loads • the reduction of stress in the steel due to the relaxation under tension. In this Benchmark the stress loss due to creep and shrinkage will be examined. SOFiSTiK 2014 | VERiFiCATiON MANUAL 345 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS 74.3 Model and Results Benchmark 17 is here extended for the case of creep and shrinkage developing on a prestressed con- crete simply supported beam. The analysed system can be seen in Fig. 74.2, with properties as defined in Table 74.1. Further information about the tendon geometry and prestressing can be found in Bench- mark 17. The beam consists of a rectangular cs and is prestressed and loaded with its own weight. A calculation of the creep and shrinkage is performed in the middle of the span with respect to DIN EN 1992-1-1:2004 (German National Annex) [72], [73]. The calculation steps [75] are presented below and the results are given in Table 74.2 for the calculation with AQB. For CSM only the results of the creep coefficients and the final losses are given, since the calculation is performed in steps. Table 74.1: Model Properties Material Properties Geometric Properties Loading (at = 10 m) Time C 35/ 45 h = 100.0 cm M g = 1250 kNm t 0 = 28 dys Y 1770 b = 100.0 cm N p = −3651.1 kN t s = 0 dys RH = 80 L = 20.0 m t eƒ ƒ = 1000000 dys A p = 28.5 cm 2 L Figure 74.2: Simply Supported Beam Table 74.2: Results Result AQB CSM+AQB Ref. ε cs −18.85 · 10 −5 - −18.85 · 10 −5 ε −31.58 · 10 −5 - −31.58 · 10 −5 ϕ 0 1.463 1.463 1.463 ϕ(t, t 0 ) 1.393 1.393 1.393 Δσ p,c+s [MP] −66.94 −68.20 −68.79 ΔP c+s [kN] 190.8 194.4 196.05 346 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS 74.4 Design Process 1 Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]: 2 Material: Concrete: C 35/ 45 3.1: Concrete E cm = 34077 N/ mm 2 3.1.2: Tab. 3.1: E cm , ƒ ck and ƒ cm for C 35/ 45 ƒ ck = 35 N/ mm 2 ƒ cm = 43 N/ mm 2 Prestressing Steel: Y 1170 3.3: Prestressing Steel E p = 195000 N/ mm 2 3.3.6 (3): E p for wires ƒ pk = 1770 N/ mm 2 3.3.2, 3.3.3: ƒ pk Characteristic tensile strength of prestressing steel Prestressing system: BBV L19 150 mm 2 19 wires with area of 150 mm 2 each, giving a total of A p = 28.5 cm 2 Cross-section: A c = 1.0 · 1.0 = 1 m 2 Diameter of duct ϕ dct = 67 mm Ratio α E,p = E p / E cm = 195000 / 34077 = 5.7223 A c,netto = A c − π · (ϕ dct / 2) 2 = 0.996 m 2 A de = A c + A p · α E,p = 1.013 m 2 Load Actions: Self weight per length: γ = 25 kN/ m At = 10.0 m middle of the span: M g = g 1 · L 2 / 8 = 1250 kNm N p = P m0 ( = 10.0 m) = −3651.1 kN (from SOFiSTiK) Calculation of stresses at = 10.0 m midspan: Position of the tendon: z cp = 0, 396 m Prestress and self-weight at con. stage sect. 0 (P+G cs0) N p = −3651.1 kN and M g = 1250 kNm 1 The tools used in the design process are based on steel stress-strain diagrams, as defined in [72] 3.3.6: Fig. 3.10 2 The sections mentioned in the margins refer to DIN EN 1992-1-1:2004 (German Na- tional Annex) [72], [73], unless otherwise specified. SOFiSTiK 2014 | VERiFiCATiON MANUAL 347 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS M g + M p z p N p P m0,=10 −σ c +σ c M p 1 = N P · z cp = −3651.1 · 0.396 = −1445.86 kNm z s the new position of the center of the cross-section for cs0 z p = z cp + z s M p 2 = N P · z s = −3651.1 · 0.001415 = −5.16 kNm M p = −1445.86 − 5.16 = −1451.02 kNm M p bending moment caused by pre- stressing M y = 1250 − 1451.02 = −201.02 kNm σ c,QP = −3651.1 0.996 + −201.02 0.165 = −4.882 MP σ c,QP stress in concrete Calculation of creep and shrinkage at = 10.0 m midspan: t 0 = 28 days t 0 minimun age of concrete for loading t s age of concrete at start of drying shrinkage t age of concrete at the moment consid- ered t s = 0 days t = t eƒ ƒ + t 0 = 1000000 + 28 = 1000028 days ε cs = ε cd + ε c 3.1.4 (6): Eq. 3.8: ε cs total shrinkage strain ε cd (t) = β ds (t, ts) · k h · ε cd,0 3.1.4 (6): Eq. 3.9: ε cd drying shrinkage strain The development of the drying shrinkage strain in time is strongly de- pends on β ds (t, ts) factor. SOFiSTiK accounts not only for the age at start of drying t s but also for the influence of the age of the prestressing t 0 . Therefore, the calculation of factor β ds reads: β ds = β ds (t, t s ) − β ds (t 0 , t s ) 3.1.4 (6): Eq. 3.10: β ds β ds = (t − t s ) (t − t s ) + 0.04 · _ h 3 0 − (t 0 − t s ) (t 0 − t s ) + 0.04 · _ h 3 0 3.1.4 (6): h 0 the notional size (mm) of the cs h 0 = 2A c / = 500 mm β ds = (1000028 − 0) (1000028 − 0) + 0.04 · _ 500 3 − (28 − 0) (28 − 0) + 0.04 · _ 500 3 β ds = 0.99955 − 0.05892 = 0.94063 k h = 0.70 for h 0 ≥ 500 mm 3.1.4 (6): Tab. 3.3: k h coefficient de- pending on h 0 ε cd,0 = 0.85 _ (220 + 110 · α ds1 ) · exp _ −α ds2 · ƒ cm ƒ cmo __ · 10 −6 · β RH Annex B.2 (1): Eq. B.11: ε cd,0 basic drying shrinkage strain β RH = 1.55 _ 1 − _ RH RH 0 _ 3 _ = 1.55 _ 1 − _ 80 100 _ 3 _ = 0.7564 Annex B.2 (1): Eq. B.12: β RH RH the ambient relative humidity (%) ε cd,0 = 0.85 _ (220 + 110 · 4) · exp _ −0.12 · 43 10 __ · 10 −6 · 0.7564 Annex B.2 (1): α ds1 , α ds1 coefficients depending on type of cement. For class N α ds1 = 4, α ds2 = 0.12 348 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS ε cd,0 = 2.533 · 10 −4 ε cd = 0.94063 · 0.70 · 2.533 · 10 −4 = 0.0001668 ε cd = 1.668 · 10 −4 = 0.1668 ◦ / ◦◦ ε c (t) = β s (t) · ε c (∞) 3.1.4 (6): Eq. 3.11: ε c autogenous shrinkage strain ε c (∞) = 2.5(ƒ ck − 10) · 10 −6 = 2.5(35 − 10) · 10 −6 3.1.4 (6): Eq. 3.12: ε c (∞) ε c (∞) = 6.25 · 10 −5 = 0.0625 ◦ / ◦◦ Proportionally to β ds (t, ts), SOFiSTiK calculates factor β s as follows: β s = β s (t) − β s (t 0 ) 3.1.4 (6): Eq. 3.13: β s β s = 1 − e −0.2· t − _ 1 − e −0.2· t 0 _ = e −0.2· t 0 − e −0.2· t β s = 0.347 ε = ε cd,0 + ε c (∞) = 2.533 · 10 −4 + 6.25 · 10 −5 ε = −31.58 · 10 −5 ε absolute shrinkage strain negative sign to declare losses ε c = 0.347 · 6.25 · 10 −5 = 2.169 · 10 −5 = 0.02169 ◦ / ◦◦ ε cs = 1.668 · 10 −4 + 2.169 · 10 −5 = −18.85 · 10 −5 negative sign to declare losses ϕ(t, t 0 ) = ϕ 0 · β c (t, t 0 ) Annex B.1 (1): Eq. B.1: ϕ(t, t 0 ) creep coefficient ϕ 0 = ϕ RH · β(ƒ cm ) · β(t 0 ) Annex B.1 (1): Eq. B.2: ϕ 0 notional creep coefficient ϕ RH = _ 1 + 1 − RH/ 100 0.1 · 3 _ h 0 · α 1 _ · α 2 Annex B.1 (1): Eq. B.3: ϕ RH factor for effect of relative humidity on creep β(ƒ cm ) = 16.8 _ ƒ cm = 16.8/ 43 = 2.562 Annex B.1 (1): Eq. B.4: β(ƒ cm ) factor for effect of concrete strength on creep α 1 = _ 35 ƒ cm _ 0.7 ≤ 1 = 0.8658 Annex B.1 (1): Eq. B.8c: α 1 , α 2 , α 3 coefficients to consider influence of con- crete strength α 2 = _ 35 ƒ cm _ 0.2 ≤ 1 = 0.9597 α 3 = _ 35 ƒ cm _ 0.5 ≤ 1 = 0.9022 ϕ RH = _ 1 + 1 − 80/ 100 0.1 · 3 500 · 0.8658 _ · 0.9597 = 1.1691 β(t 0 ) = 1 _ 0.1 + t 0.20 0 _ Annex B.1 (1): Eq. B.5: β(t 0 ) factor for effect of concrete age at loading on creep t 0 = t 0,T · _ 9 2 + t 1.2 0,T + 1 _ α ≥ 0.5 Annex B.1 (2): Eq. B.9: t 0,T tempera- ture adjusted age of concrete at loading adjusted according to expression B.10 SOFiSTiK 2014 | VERiFiCATiON MANUAL 349 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS t T = n =1 e −(4000/ [273+T(Δt )]−13.65) · Δt Annex B.1 (3): Eq. B.10: t T temper- ature adjusted concrete age which re- places t in the corresponding equations t 0,T = 28 · e −(4000/ [273+20]−13.65) = 28 · 1.0 = 28.0 ⇒t 0 = 28.0 · _ 9 2 + 28.0 1.2 + 1 _ 0 = 28.0 Annex B.1 (2): Eq. B.9: α a power which depends on type of cement For class N α = 0 β(t 0 ) = 1 _ 0.1 + 28.0 0.20 _ = 0.48844 β c (t, t 0 ) = _ (t − t 0 ) (β H + t − t 0 ) _ 0.3 Annex B.1 (1): Eq. B.7: β c (t, t 0 ) co- efficient to describe the development of creep with time after loading β H = 1.5 · _ 1 + (0.012 · RH) 18 _ · h 0 + 250 · α 3 ≤ 1500 · α 3 Annex B.1 (1): Eq. B.8: β H coefficient depending on relative humidity and no- tional member size β H = 1.5 · _ 1 + (0.012 · 80) 18 _ · 500 + 250 · 0.9022 β H = 1335.25 ≤ 1500 · 0.9022 = 1353.30 ⇒β c (t, t 0 ) = 0.9996 ϕ 0 = 1.1691 · 2.562 · 0.48844 = 1.463 ϕ(t, t 0 ) = 1.463 · 0.9996/ 1.05 = 1.393 Annex B.1 (3): The values of ϕ(t, t 0 ) given above should be associated with the tangent modulus E c 3.1.4 (2): The values of the creep coef- ficient, ϕ(t, t 0 ) is related to E c , the tan- gent modulus, which may be taken as 1.05 · E cm According to EN, the creep value is related to the tangent Young’s modulus E c , where E c being defined as 1.05 · E cm . To account for this, SOFiSTiK adopts this scaling for the computed creep coefficient (in SOFiSTiK, all computations are consistently based on E cm ). ΔP c+s+r = A p ·Δσ p,c+s+r = A p ε cs · E p + 0.8Δσ pr + E p E cm ϕ(t, t 0 ) · σ c,QP 1 + E p E cm A p A c _ 1 + A c c z 2 cp _ [1 + 0.8ϕ(t, t 0 )] 5.10.6 (2): Eq. 5.46: ΔP c+s+r time de- pendent losses of prestress In this example only the losses due to creep and shrinkage are taken into account, the reduction of stress due to relaxation (Δσ pr ) is ignored. Δσ p,c+s = −0.1885 · 10 −3 · 195000 + 5.7223 · 1.393 · (−4.882) 1 + 5.7223 28.5 · 10 −4 0.996 _ 1 + 0.996 0.08277 0.396 2 _ [1 + 0.8 · 1.393] Δσ p,c+s variation of stress in tendons due to creep and shrinkage at location , at time t Δσ p,c+s = −68.79 MP ΔP c+s = A p · Δσ p,c+s = 28.5 · 10 −4 · 68.79 · 10 3 = 196.05 kN 350 VERiFiCATiON MANUAL | SOFiSTiK 2014 DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS 74.5 Conclusion This example shows the calculation of the time dependent losses due to creep and shrinkage. It has been shown that the results are in very good agreement with the reference solution. 74.6 Literature [72] DIN EN 1992-1-1/NA: Eurocode 2: Design of concrete structures, Part 1-1/NA: General rules and rules for buildings - German version EN 1992-1-1:2005 (D), Nationaler Anhang Deutschland - Stand Februar 2010. CEN. 2010. [73] F. Fingerloos, J. Hegger, and K. Zilch. DIN EN 1992-1-1 Bemessung und Konstruktion von Stahlbeton- und Spannbetontragwerken - Teil 1-1: Allgemeine Bemessungsregeln und Regeln f ¨ ur den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Beton- und Bautechnik-Verein E.V. 2011. SOFiSTiK 2014 | VERiFiCATiON MANUAL 351