VERiFiCATiON MANUAL SOFiSTiK 2014
VERiFiCATiON MANUAL
VERiFiCATiON MANUAL, Version 2014.4 Software Version: SOFiSTiK 2014 c 2013 by SOFiSTiK AG, Oberschleissheim, Germany. Copyright
SOFiSTiK AG HQ Oberschleissheim Bruckmannring 38 85764 Oberschleissheim Germany
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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 ¨ Moermann und partner | Architect: Asymptote Architecture | Photo: Bjorn
Contents
Contents
Introduction
3
1
About this Manual 1.1 Layout and Organization of a Benchmark .............................................. 1.2 Finding the Benchmark of interest ....................................................... 1.3 Symbols ....................................................................................
3 3 3 5
2
Index by Categories 2.1 Mechanical Benchmarks ................................................................. 2.2 Design Code Benchmarks................................................................
7 7 9
I 3
II
SOFiSTiK Software Quality Assurance (SQA) SOFiSTiK SQA Policy 3.1 Objectives .................................................................................. 3.1.1 About SOFiSTiK ........................................................................ 3.1.2 Innovation and Reliability .............................................................. 3.2 Organisation ............................................................................... 3.2.1 Software Release Schedule ........................................................... 3.2.2 SQA Modules - Classification ......................................................... 3.2.3 Responsibilities ......................................................................... 3.2.4 Software Release Procedure .......................................................... 3.3 Instruments................................................................................. 3.3.1 CRM System ............................................................................ 3.3.2 Tracking System (internal) ............................................................ 3.3.3 Continuous Integration – Continuous Testing........................................ 3.4 Additional Provisions ...................................................................... 3.4.1 Training................................................................................... 3.4.2 Academia Network ..................................................................... 3.5 Disclaimer ..................................................................................
Benchmark Examples
11 13 13 13 13 13 13 14 14 15 16 16 16 16 17 17 18 18
19
4
BE1: Joint Deflection of Plane Truss 4.1 Problem Description....................................................................... 4.2 Reference Solution ........................................................................ 4.3 Model and Results ........................................................................ 4.4 Conclusion ................................................................................. 4.5 Literature ...................................................................................
21 21 21 21 22 22
5
BE2: Kinematic Coupling Conditions 5.1 Problem Description....................................................................... 5.2 Reference Solution ........................................................................ 5.3 Model and Results ........................................................................
23 23 23 24
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5.4 5.5
Conclusion ................................................................................. Literature ...................................................................................
27 27
6
BE3: Beam Stresses and Deflections 6.1 Problem Description....................................................................... 6.2 Reference Solution ........................................................................ 6.3 Model and Results ........................................................................ 6.4 Conclusion ................................................................................. 6.5 Literature ...................................................................................
29 29 29 30 31 31
7
BE4: Tie Rod with Lateral Loading 7.1 Problem Description....................................................................... 7.2 Reference Solution ........................................................................ 7.3 Model and Results ........................................................................ 7.4 Conclusion ................................................................................. 7.5 Literature ...................................................................................
33 33 33 34 35 35
8
BE5: Bending of a T-beam 8.1 Problem Description....................................................................... 8.2 Reference Solution ........................................................................ 8.3 Model and Results ........................................................................ 8.4 Conclusion ................................................................................. 8.5 Literature ...................................................................................
37 37 37 37 38 38
9
BE6: Warping Torsion Bar 9.1 Problem Description....................................................................... 9.2 Reference Solution ........................................................................ 9.3 Model and Results ........................................................................ 9.4 Conclusion ................................................................................. 9.5 Literature ...................................................................................
39 39 39 41 42 42
10
BE7: Large Deflection of Cantilever Beams I 10.1 Problem Description....................................................................... 10.2 Reference Solution ........................................................................ 10.3 Model and Results ........................................................................ 10.4 Conclusion ................................................................................. 10.5 Literature ...................................................................................
43 43 43 44 46 46
11
BE8: Large Deflection of Cantilever Beams II 11.1 Problem Description....................................................................... 11.2 Reference Solution ........................................................................ 11.3 Model and Results ........................................................................ 11.4 Conclusion ................................................................................. 11.5 Literature ...................................................................................
47 47 47 48 49 49
12
BE9: Verification of Beam and Section Types I 12.1 Problem Description....................................................................... 12.2 Reference Solution ........................................................................ 12.3 Model and Results ........................................................................ 12.4 Conclusion ................................................................................. 12.5 Literature ...................................................................................
51 51 51 53 57 57
13
BE10: Verification of Beam and Section Types II 13.1 Problem Description....................................................................... 13.2 Reference Solution ........................................................................
59 59 59
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13.3 13.4 13.5
Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
60 62 62
14
BE11: 14.1 14.2 14.3 14.4 14.5
Plastification of a Rectangular Beam Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
63 63 63 64 65 65
15
BE12: 15.1 15.2 15.3 15.4 15.5
Cantilever in Torsion Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
67 67 67 68 69 69
16
BE13: 16.1 16.2 16.3 16.4 16.5
Buckling of a Bar with Hinged Ends I Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
71 71 71 72 73 73
17
BE14: 17.1 17.2 17.3 17.4 17.5
Buckling of a Bar with Hinged Ends II Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
75 75 75 76 76 76
18
BE15: 18.1 18.2 18.3 18.4 18.5
Flexural and Torsional Buckling Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
77 77 77 78 79 79
19
BE16: 19.1 19.2 19.3 19.4 19.5
Torsion due to Biaxial Bending Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
81 81 81 81 82 82
20
BE17: 20.1 20.2 20.3 20.4 20.5
Lateral Torsional Buckling Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
83 83 83 84 84 84
21
BE18: Three-storey Column under Large Compressive Force and Torsional Moment
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21.1 21.2 21.3 21.4 21.5
Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
85 85 86 87 87
22
BE19: 22.1 22.2 22.3 22.4 22.5
Two-span Beam with Warping Torsion and Compressive Force Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
89 89 89 90 92 92
23
BE20: 23.1 23.2 23.3 23.4 23.5
Passive Earth Pressure I Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
93 93 93 94 96 96
24
BE21: 24.1 24.2 24.3 24.4 24.5
Passive Earth Pressure II Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
97 97 97 99 100 101
25
BE22: 25.1 25.2 25.3 25.4 25.5
Tunneling - Ground Reaction Line Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
103 103 103 105 106 106
26
BE23: 26.1 26.2 26.3 26.4 26.5
Undamped Free Vibration of a SDOF System Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
107 107 107 108 110 110
27
BE24: 27.1 27.2 27.3 27.4 27.5
Free Vibration of a Under-critically Damped SDOF System Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
111 111 111 113 114 114
28
BE25: 28.1 28.2 28.3 28.4 28.5
Eigenvalue Analysis of a Beam Under Various End Constraints Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
115 115 115 116 116 116
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29
BE26: 29.1 29.2 29.3 29.4 29.5
Response of a SDOF System to Harmonic Excitation Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
117 117 117 119 120 120
30
BE27: 30.1 30.2 30.3 30.4 30.5
Response of a SDOF System to Impulsive Loading Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
121 121 121 123 124 124
31
BE28: 31.1 31.2 31.3 31.4 31.5
Cylindrical Hole in an Infinite Elastic Medium Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
125 125 125 126 128 128
32
BE29: 32.1 32.2 32.3 32.4 32.5
Cylindrical Hole in an Infinite Mohr-Coulomb Medium Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
129 129 129 130 132 132
33
BE30: 33.1 33.2 33.3 33.4 33.5
Strip Loading on an Elastic Semi-Infinite Mass Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
133 133 133 134 136 136
34
BE31: 34.1 34.2 34.3 34.4 34.5
Snap-Through Behaviour of a Truss Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
137 137 137 138 139 139
35
BE32: 35.1 35.2 35.3 35.4 35.5
Thermal Extension of Structural Steel in case of Fire Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
141 141 141 141 143 143
36
BE33: 36.1 36.2 36.3 36.4 36.5
Work Laws in case of Fire for Concrete and Structural Steel Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
145 145 145 145 149 149
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37
BE34: 37.1 37.2 37.3 37.4 37.5
Ultimate Bearing Capacity of Concrete and Steel under Fire Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
151 151 151 151 153 153
38
BE35: 38.1 38.2 38.3 38.4 38.5
Calculation of Restraining Forces in Steel Members in case of Fire Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
155 155 155 155 157 157
39
BE36: 39.1 39.2 39.3 39.4 39.5
Pushover Analysis: Performance Point Calculation by ATC-40 Procedure Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
159 159 159 160 162 162
40
BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory 40.1 Problem Description....................................................................... 40.2 Reference Solution ........................................................................ 40.3 Model and Results ........................................................................ 40.4 Conclusion ................................................................................. 40.5 Literature ...................................................................................
163 163 163 164 165 166
41
BE38: 41.1 41.2 41.3 41.4 41.5
Calculation of Slope Stability by Phi-C Reduction Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
167 167 167 168 169 169
42
BE39: 42.1 42.2 42.3 42.4 42.5
Natural Frequencies of a Rectangular Plate Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
171 171 171 172 173 173
43
BE40: 43.1 43.2 43.3 43.4 43.5
Portal Frame Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
175 175 175 176 177 177
44
BE41: 44.1 44.2 44.3 44.4
Linear Pinched Cylinder Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion .................................................................................
179 179 179 180 181
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44.5
Literature ...................................................................................
182
45
BE42: 45.1 45.2 45.3 45.4 45.5
Thick Circular Plate Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
183 183 183 184 185 185
46
BE43: 46.1 46.2 46.3 46.4 46.5
Panel with Circular Hole Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
187 187 187 188 190 190
47
BE44: 47.1 47.2 47.3 47.4 47.5
Undrained Elastic Soil Layer Subjected to Strip Loading Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
191 191 191 191 193 193
48
BE45: 48.1 48.2 48.3 48.4 48.5
One-Dimensional Soil Consolidation Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
195 195 195 196 197 198
49
BE46: 49.1 49.2 49.3 49.4 49.5
Material Nonlinear Analysis of Reinforced Concrete Beam Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
199 199 199 199 201 201
50
BE47: 50.1 50.2 50.3 50.4 50.5
Pushover Analysis: SAC LA9 Building Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
203 203 203 204 207 207
51
BE48: Triaxial Consolidated Undrained (CU) Test 51.1 Problem Description....................................................................... 51.2 Reference Solution ........................................................................ 51.3 Model and Results ........................................................................ 51.3.1 Hostun-RF Sand, σc = 200 kN/ m2 ................................................. 51.3.2 Hostun-RF Sand, σc = 300 kN/ m2 ................................................. 51.4 Conclusion ................................................................................. 51.5 Literature ...................................................................................
209 209 209 211 213 214 216 216
52
BE49: Triaxial Drained Test 52.1 Problem Description.......................................................................
217 217
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52.2 Reference Solution ........................................................................ 52.3 Model and Results ........................................................................ 52.3.1 Hostun-RF Sand, σc = 100 kN/ m2 ................................................. 52.3.2 Hostun-RF Sand, σc = 300 kN/ m2 ................................................. 52.4 Conclusion ................................................................................. 52.5 Literature ...................................................................................
217 217 219 220 222 222
53
BE50: 53.1 53.2 53.3 53.4 53.5
A Circular Cavity Embedded in a Full-Plane Under Impulse Pressure Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
223 223 223 224 225 226
54
BE51: 54.1 54.2 54.3 54.4 54.5
Pushover Analysis: Performance Point Calculation by EC8 Procedure Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
227 227 227 229 232 232
55
BE52: 55.1 55.2 55.3 55.4 55.5
Verification of Wave Kinematics Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
233 233 233 233 235 235
56
BE53: 56.1 56.2 56.3 56.4 56.5
Verification of Wave Loading Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Conclusion ................................................................................. Literature ...................................................................................
237 237 237 237 239 239
III
Design Code Benchmark Examples
57
DCE-EN1: Design of Slab for Bending 57.1 Problem Description....................................................................... 57.2 Reference Solution ........................................................................ 57.3 Model and Results ........................................................................ 57.4 Design Process ............................................................................ 57.5 Conclusion ................................................................................. 57.6 Literature ...................................................................................
243 243 243 244 245 246 246
58
DCE-EN2: Design of a Rectangular CS for Bending 58.1 Problem Description....................................................................... 58.2 Reference Solution ........................................................................ 58.3 Model and Results ........................................................................ 58.4 Design Process ............................................................................ 58.5 Conclusion ................................................................................. 58.6 Literature ...................................................................................
247 247 247 248 249 251 251
59
DCE-EN3: Design of a T-section for Bending
253
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59.1 59.2 59.3 59.4 59.5 59.6
Problem Description....................................................................... Reference Solution ........................................................................ Model and Results ........................................................................ Design Process ............................................................................ Conclusion ................................................................................. Literature ...................................................................................
253 253 254 256 258 258
60
DCE-EN4: Design of a Rectangular CS for Bending and Axial Force 60.1 Problem Description....................................................................... 60.2 Reference Solution ........................................................................ 60.3 Model and Results ........................................................................ 60.4 Design Process ............................................................................ 60.5 Conclusion ................................................................................. 60.6 Literature ...................................................................................
259 259 259 260 261 262 262
61
DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force 61.1 Problem Description....................................................................... 61.2 Reference Solution ........................................................................ 61.3 Model and Results ........................................................................ 61.4 Design Process ............................................................................ 61.5 Conclusion ................................................................................. 61.6 Literature ...................................................................................
263 263 263 264 266 267 267
62
DCE-EN6: Design of a Rectangular CS for Shear Force 62.1 Problem Description....................................................................... 62.2 Reference Solution ........................................................................ 62.3 Model and Results ........................................................................ 62.4 Design Process ............................................................................ 62.5 Conclusion ................................................................................. 62.6 Literature ...................................................................................
269 269 269 270 271 272 272
63
DCE-EN7: Design of a T-section for Shear 63.1 Problem Description....................................................................... 63.2 Reference Solution ........................................................................ 63.3 Model and Results ........................................................................ 63.4 Design Process ............................................................................ 63.5 Conclusion ................................................................................. 63.6 Literature ...................................................................................
273 273 273 274 276 277 277
64
DCE-EN8: Design of a Rectangular CS for Shear and Axial Force 64.1 Problem Description....................................................................... 64.2 Reference Solution ........................................................................ 64.3 Model and Results ........................................................................ 64.4 Design Process ............................................................................ 64.5 Conclusion ................................................................................. 64.6 Literature ...................................................................................
279 279 279 280 281 283 283
65
DCE-EN9: Design of a Rectangular CS for Shear and Torsion 65.1 Problem Description....................................................................... 65.2 Reference Solution ........................................................................ 65.3 Model and Results ........................................................................ 65.4 Design Process ............................................................................ 65.5 Conclusion ................................................................................. 65.6 Literature ...................................................................................
285 285 285 286 288 290 290
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66
DCE-EN10: Shear between web and flanges of T-sections 66.1 Problem Description....................................................................... 66.2 Reference Solution ........................................................................ 66.3 Model and Results ........................................................................ 66.4 Design Process ............................................................................ 66.5 Conclusion ................................................................................. 66.6 Literature ...................................................................................
291 291 291 292 295 297 297
67
DCE-EN11: Shear at the interface between concrete cast 67.1 Problem Description....................................................................... 67.2 Reference Solution ........................................................................ 67.3 Model and Results ........................................................................ 67.4 Design Process ............................................................................ 67.5 Conclusion ................................................................................. 67.6 Literature ...................................................................................
299 299 299 300 302 305 305
68
DCE-EN12: Calculation of crack widths 68.1 Problem Description....................................................................... 68.2 Reference Solution ........................................................................ 68.3 Model and Results ........................................................................ 68.4 Design Process ............................................................................ 68.5 Conclusion ................................................................................. 68.6 Literature ...................................................................................
307 307 307 308 309 314 314
69
DCE-EN13: Design of a Steel I-section for Bending and Shear 69.1 Problem Description....................................................................... 69.2 Reference Solution ........................................................................ 69.3 Model and Results ........................................................................ 69.4 Design Process ............................................................................ 69.5 Conclusion ................................................................................. 69.6 Literature ...................................................................................
315 315 315 315 317 319 319
70
DCE-EN14: Classification of Steel Cross-sections 70.1 Problem Description....................................................................... 70.2 Reference Solution ........................................................................ 70.3 Model and Results ........................................................................ 70.4 Design Process ............................................................................ 70.5 Conclusion ................................................................................. 70.6 Literature ...................................................................................
321 321 321 322 324 325 325
71
DCE-EN15: Buckling Resistance of Steel Members 71.1 Problem Description....................................................................... 71.2 Reference Solution ........................................................................ 71.3 Model and Results ........................................................................ 71.4 Design Process ............................................................................ 71.5 Conclusion ................................................................................. 71.6 Literature ...................................................................................
327 327 327 328 329 330 330
72
DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force 72.1 Problem Description....................................................................... 72.2 Reference Solution ........................................................................ 72.3 Model and Results ........................................................................ 72.4 Design Process ............................................................................ 72.5 Conclusion .................................................................................
331 331 331 332 333 335
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72.6
Literature ...................................................................................
335
73
DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS 73.1 Problem Description....................................................................... 73.2 Reference Solution ........................................................................ 73.3 Model and Results ........................................................................ 73.4 Design Process ............................................................................ 73.5 Conclusion ................................................................................. 73.6 Literature ...................................................................................
337 337 337 338 341 344 344
74
DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS 74.1 Problem Description....................................................................... 74.2 Reference Solution ........................................................................ 74.3 Model and Results ........................................................................ 74.4 Design Process ............................................................................ 74.5 Conclusion ................................................................................. 74.6 Literature ...................................................................................
345 345 345 346 347 351 351
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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 solutions1 . 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 respect 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.
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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 Continuum 3D Continuum 2D (plane strain) Continuum axisymmetric Shell FE beam 3D Nonlinear FE beam 3D (AQB)
Element Type
Fiber beam 2D Fiber beam 3D Truss element Cable element Spring element Damping element Couplings Geometrically nonlinear Physically nonlinear
Analysis Type
Dynamic Static Potential problem Buckling analysis Eigenvalue/ Modal analysis
Procedure Time stepping
4
VERiFiCATiON MANUAL | SOFiSTiK 2014
About this Manual
Table 1.1: (continued) Categories
Options Load stepping Phi-C reduction Soil related
Topic
Seismic Fire design AQB AQUA ASE
Module
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
er
relative error of the approximate number
|er |
absolute relative error of the approximate number
e
error of the approximate number
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About this Manual
6
|e|
absolute error of the approximate number
ep ()
same as e()
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.
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Index by Categories
8
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
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-EN15, DCE-EN16
German Standards
DIN
DCE-EN1, DCE-EN2, DCE-EN3, DCE-EN4, DCEEN5, 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, DCEEN5, 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, DCEEN5, 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
DCE-EN14,
9
Index by Categories
10
VERiFiCATiON MANUAL | SOFiSTiK 2014
Part I
SOFiSTiK Software Quality Assurance (SQA)
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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 product is an issue of outstanding relevance for SOFiSTiK. To some degree, however, innovation and reliability 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 assurance 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 FCS
BETA
2011
2012
Start of Transition Phase
2013
2014
2015
2016
2017
Discontinuation of Maintenance
SOFiSTiK 2010 · 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
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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 Functional Testing Automated, modular and continuous regression testing Instruments (analytic)
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 managing 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.
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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.
Development Continuous Integration Continuous Testing Implementing feedback
Product Management Functional testing Integrating customer 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 minimum requirements 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.
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SOFiSTiK SQA Policy
Release Requirements
Hotfix
Service Pack
Major Release
Continuous Testing Passed
Continuous Testing Passed
Continuous Testing Passed
Functional 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 regarding 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.
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SOFiSTiK SQA Policy
Automated Continuous Testing procedure
Automated Continuous Integration 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 modifications of the program’s code basis are directly integrated via rebuild into the internal testing environment. 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 promptly. 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 qualified 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.
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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.
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Part II
Benchmark Examples
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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.
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BE1: Joint Deflection of Plane Truss
Table 4.1: Model Properties Material Properties
Geometric Properties
Loading
E = 30 103 ks
tot = 60 ƒ t = 18.288 m
P = 20 kp = 89.0 kN
= 206842.773 MP ν = 0.3
htot = 15 ƒ t = 4.572 m 2 = 11 = 7.5 ƒ t = 2.286 m 1 = 4 = 6 = 10 = 12 = 15 ƒ t = 4.572 m A1 = A4 = 2 n2 = 12.90 cm2 A2 = A11 = A10 = A12 = 1 n2 = 6.45 cm2 A5 = A9 = 1.5 n2 = 9.68 cm2 A3 = A6 = 3 n2 = 19.35 cm2 A7 = A8 = 4 n2 = 25.81 cm2
Table 4.2: Results
δ8 [mm]
SOF.
Ref. [2]
|er | [%]
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.
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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 A00
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 A00 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 + ϕyo (z − zo ) − ϕzo (y − yo )
SOFiSTiK 2014 | VERiFiCATiON MANUAL
(5.1)
23
BE2: Kinematic Coupling Conditions
y = yo + ϕzo ( − o ) − ϕo (z − zo )
(5.2)
z = zo + ϕo (y − yo ) − ϕyo ( − 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)
KF = KP + KMX + KMY + KMZ = KPPX + KPPY + KPPZ + KMX + KMY + KMZ
(5.5)
whereas
describes mechanically a rigid connection with clamped support at the reference node. Further information 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 A00 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 Py , a longitudinal P , a vertical Pz and a rotational M . Table 5.1: Model Properties Material Properties
Geometric Properties
Loading
C 30/ 45
bem = 4 m
Py = 50.0 kN
hA = 0.4 m, bA = 0.2 m
P = −50.0 kN
hA00 = 0.3 m, bA00 = 0.15 m
Pz = 50.0 kN
(A00 − A ) = 2 m
M = 10.0 kN
(yA00 − yA ) = 0 m (zA00 − zA ) = 0 m
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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 ϕz = 3y / (2bem ) will also result as the effect of a prescribed displacement of value y at the beam tip A00 . Table 5.2: Results for KPPX Coupling Condition [mm]
Load Case P
SOF.
Ref.
−0.107
−0.107
Table 5.3: Results for LPX Coupling Condition y [mm]
Load Case Py
ϕz [mrd]
SOF.
Ref.
SOF.
Ref.
165.008
165.008
−61.878
−61.878
Table 5.4: Results for KP and KF Coupling Condition Coupling
KP Py
P
Pz
Py
P
Pz
M
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
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
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
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
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
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
DOF / LC [mm]
y [mm]
z [mm]
ϕ [mrd]
ϕy [mrd]
ϕz [mrd]
KF
The results are presented in Tables 5.2 - 5.4, where they are compared to the reference results calculated 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.
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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
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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, Germany, 2012. [4] SOFiMSHA Manual: Import and Export of Finite Elements and Beam Structures. Version 16.01. SOFiSTiK AG. Oberschleißheim, Germany, 2012.
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27
BE2: Kinematic Coupling Conditions
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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]:
σm =
Mh
σmn = −
and
2
Mh , 2
(6.1)
in which , is the moment of inertia of the cross-section with respect to the neutral axis and M the 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=
q2 2 16E
=
M2 , 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.
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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-beam cross-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
E = 30000 MP
Geometric Properties
Geometric Properties
Rectangular
I-beam
= 200 mm
= 200 mm
= 100 mm
b = 16 mm
h = 30 mm
teb = 2.174 mm
b = 7 mm
tƒ nge = 2 mm
y = 1.575 cm4
y = 1.575 cm4
Loading
q = 10 kN/ m
Table 6.2: Results Rectangular
I-beam
Ref.
47.619
47.620
47.619
0.529
0.529
0.529
σm [MPa] δ [mm]
47.619
47.619
47.620 47.619
41.271
-41.271 -47.619
-47.619
47.620
41.27141.271
41.271
-41.271-41.271
-47.620 -47.619
-41.271 -47.620
1.32
0.970
0.292
0.623
0.231
0.397
0.529
0.496
0.496
0.397
0.231
0.292
0.623
0.970
1.32
Figure 6.2: Distribution of stresses
Figure 6.3: Deformed Structure
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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 [5]
Literature S. Timoshenko. Strength of Materials, Part I, Elementary Theory and Problems. 2nd. D. Van Nostrand Co., Inc., 1940.
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BE3: Beam Stresses and Deflections
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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 δm , the slope θ at the left-hand end and the maximum bending moment Mm . In addition, compare these three quantities for the case of the unstiffned tie rod (N = 0). P
N
N | ←−
−→ |
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:
δm =
P3 , 48E
(7.1)
where is the lenght of the beam and E its flexural rigidity. The slope θ at both ends is:
θ=±
SOFiSTiK 2014 | VERiFiCATiON MANUAL
P2 . 16E
(7.2)
33
BE4: Tie Rod with Lateral Loading
The maximum value of the bending moment at the middle is:
Mm =
P .
(7.3)
4
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]:
δm =
P3 48E
·
− tanh 1 3 3
,
(7.4)
where 2 = N2 / 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
=2m
P = 0.1 kN
h = 30 mm
N = 0.1 kN
b = 30 mm = 6.75 × 10−8 m4
Figure 7.2: Deformed Structure [mm]: N 6= 0
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VERiFiCATiON MANUAL | SOFiSTiK 2014
BE4: Tie Rod with Lateral Loading
Table 7.2: Results
7.4
N=0
Ref.
N 6= 0
Ref.
δm [m]
0.00823
0.00823
0.00807
0.00807
Mm [kNm]
0.05000
0.05000
0.04919
0.04919
θ [rd]
0.01235
0.01235
0.01210
0.01210
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 [6]
Literature S. Timoshenko. Strength of Materials, Part II, Advanced Theory and Problems. 2nd. D. Van Nostrand 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 Mz . Determine the maximum tensile and compressive bending stresses. y
Mz z h2 Mz | ←− h1
−→ |
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 h1 and h2 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 Mz , we obtain the maximum tensile and compressive stresses [5]:
σm =
8.3
Mz h1 z
and
σmn = −
Mz h2 z
.
(8.1)
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.
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BE5: Bending of a T-beam
Table 8.1: Model Properties Material Properties
Geometric Properties
Loading
E = 30000 MP
=1m
Mz = 100 kNm
h = 20 cm h1 = 6 cm, h2 = 14 cm b = 9 cm teb = 1.5 cm tƒ nge = 4 cm z = 2000 cm4
Table 8.2: Results SOF.
Ref.
σm [MPa]
300
300
σmn [MPa]
−700
−700
100.00 300.00 -700.00 -700.00 0.00
100.00 100.00
100.00 300.00
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 [5]
38
Literature S. Timoshenko. Strength of Materials, Part I, Elementary Theory and Problems. 2nd. D. Van Nostrand Co., Inc., 1940.
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 mT [7]. Determine the angle of twist ϕ at the midspan.
mT | ←−
−→ |
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 crosssection 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].
T
z
ϕ y 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
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39
BE6: Warping Torsion Bar
will twist, with its free end rotating through an angle ϕ called the angle of twist [8]:
ϕ=
TL ,
(9.1)
GT
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 = GT
dϕ ,
(9.2)
d
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 = ECM
d3 ϕ (9.3)
d3
where ECM , 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 Mω is defined as:
Mω = ECM
d2 ϕ (9.4)
d2
It should be noted, that the bimoment itself is not measurable, however it serves as a convenient parameter to quantify this prevention of warping.
mT ()
y
m1 MT + MT´d
d mT
mT
(b)
mT = m0 + m1
d
MT (a)
m0
(c)
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
d2 ϕ
d4 ϕ ECM
− GT
d4
d2
= mT ,
(9.5)
where mT the distributed torque along the bar. The natural boundary conditions are:
Mω = ECM
d2 ϕ d2
,
z = 0 or l
(9.6)
and d3 ϕ −ECM
d3
+ GT
dϕ d
= MT ,
z = 0 or l
(9.7)
where MT is the concentrated end torque and Mω 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
0000
00
− λ2 ϕ =
mT
.
(9.8)
ECM
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=
C1
C2
1
λ
λ
2GT
0
snh λ + 2
ϕ =
C1 λ
cosh λ + C3 + C4 − 2
cosh λ +
00
C2 λ
snh λ + C3 −
ϕ = C1 snh λ + C2 cosh λ −
ϕ
000
(m0 +
m1 )2 3
1
(2m0 + m1 ) 2GT 1
(m0 + m1 ) GT 1
= C1 λcosh λ + C2 λsnh λ −
m1 GT
(9.9)
(9.10)
(9.11)
(9.12)
The values of the constants C1 to C4 can be derived with respect to the kinematic boundary conditions of the problem. For the case of warping-free sections, where CM = 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.
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BE6: Warping Torsion Bar
Table 9.1: Model Properties Material Properties
Cross-sectional Properties
Loading
E = 217396.3331684 N/ mm2
=1m
mT = 1 Nmm/ mm
G = 81386.6878 N/ mm2
h = 80 mm
ν = 0.33557673
t = 2 mm b = 40 mm CM = 0.323 × 108 mm6 T = 431.979 mm4
Table 9.2: Results
ϕ [mrd]
Twist in x-direction
Ref. [7]
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 crosssection 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 [6] [7] [8] [9] [10]
42
Literature S. Timoshenko. Strength of Materials, Part II, Advanced Theory and Problems. 2nd. D. Van Nostrand Co., Inc., 1940. 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. F.P. Beer, E.R. Johnston, and J.T. DeWolf. Mechanics of Materials. 4th. McGraw-Hill, 2006. P. Seaburg and C.J. Carter. Steel Design Guide Series 9: Torsional Analysis of Structural Steel Members. AISC. 2003. C. Petersen. Stahlbau. 2nd. Vieweg, 1990.
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].
L Δ
δ
P Figure 10.2: Problem Definition
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BE7: Large Deflection of Cantilever Beams I
The mathematical treatment of the equilibrium of cantilever beams does not involve great difficulty. Nevertheless, 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 0.3 m and a thickness of 0.10261 m, 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
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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
load factor 1.0 0.9
L−Δ
δ
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
9.0
10.0
diplacement [m] load factor 1.0 0.9
L−Δ
δ
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
diplacement [m] Figure 10.4: Load - Deflection: (a). Beam elements (b). Quad elements
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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. Belendez, C. Neipp, and A. Belendez. “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.
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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
E = 100 MP
Geometric Properties
Geometric Properties
Beam elements
Quad elements
= 10 m
= 10 m
D = 0.2 m
B = 0.3 m
t = 0.01 m
t = 0.10261 m
Loading
M = 3.38478 kNm
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.
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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
SC C
SC C
SC C
0.10
0.10
SC C
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
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.
C SC 0.10
0.10
0.10
SC C
C
0.20
SC 0.10 0.10
0.10
0.10
C
C
SC
SC 0.20
SC C
0.10
0.01
0.10
SC C
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 =
Z
z 2 dA,
(12.1)
A
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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 crosssections is:
T =
Z
r 2 dA,
(12.2)
A
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]: 4A2m , Pn s
T =
=1
(12.3)
t
and for open thin-walled non-circular cross-sections is:
T =
n 1X
3
s t3 ,
(12.4)
=1
where Am 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
bt
3
3
t 1 − 0.630
b
+
1 3
(h − 2t) s3 + 2 α D4 ,
(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) s3 + 0.29
s t
s 2 2
+ t2
4 (12.6)
t
b
z
dA r
h y s
Am
t D Figure 12.2: Cross-Sectional Properties
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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 [cm4 ]
|er |
T [cm4 ]
|er |
SOF.
Ref.
[%]
SOF.
Ref.
[%]
833.33
833.33
0.00
1400.00
1400.00
0.00
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
Type square -srec rectangular -srec
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
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BE9: Verification of Beam and Section Types I
Table 12.2: (continued) y [cm4 ]
|er |
T [cm4 ]
|er |
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
207.0 [16]
0.07
1.99 [16]
5.24
206.9 [17]
0.03
2.01 [17]
4.19
23130 [16]
0.01
51.40 [16]
0.76
23128 [17]
0.00
50.41 [17]
1.18
Type
UPE 100 (tabulated)
IPE 400 (tabulated)
206.85
23126.97
2.09
51.01
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. SC SC C
C
SC C
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 cm4 and the reference is 729.00 cm4 . The difference between them is 741.00 - 729.00 = 12 cm4 , which corresponds to the reference value
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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: Tgenersed thn−ed theory = Tcosed,SOFSTK = Tcosed,thn−ed theory + Topen,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 Tgenersed thn−ed theory from the Tcosed,thn−ed theory . 0.2 0.18
T,reƒ − T,cc [cm4 ]
0.16 0.14 0.12 0.1 8 · 10−2 6 · 10−2 4 · 10−2 2 · 10−2 0 10
9.5
9
8.5
7.5 7 8 Thickness [mm]
6.5
6
5.5
5
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.
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BE9: Verification of Beam and Section Types I
Open
Closed
Figure 12.5: Distribution of Stresses
6 -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
5
Deviation [%]
4
3
2
1
0 10
9
8
7
5 6 Thickness [mm]
4
3
2
1
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.
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BE9: Verification of Beam and Section Types I
SC C
SC C
-POLY
-PLAT
SC C
-PLAT and -WELD
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 irrelevantly of their definition with small deviations from the exact solution.
12.5 [5] [10] [15] [16] [17]
Literature
S. Timoshenko. Strength of Materials, Part I, Elementary Theory and Problems. 2nd. D. Van Nostrand Co., Inc., 1940. C. Petersen. Stahlbau. 2nd. Vieweg, 1990. K. Holschemacher. Entwurfs- und Berechnungstafeln fur ¨ Bauingenieure. 3rd. Bauwerk, 2007. M. Schneider-Burger. Stahlbau-Profile. 24th. Verlag Stahleisen, 2004. ¨ R. Kindmann, M. Kraus, and H. J. Niebuhr. Stahlbau Kompakt, Bemessungshilfen, Profiltabellen. Verlag Stahleisen, 2006.
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BE9: Verification of Beam and Section Types I
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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 transverse 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 δm of the cantilever, under the action of a transverse load P, occurs at the tip and is [15]:
δm =
PL3 , 3E
(13.1)
and the rotation ϕz
ϕz =
PL2 . 2E
(13.2)
For the case of the moment M, applied at the -axis the angle of twist ϕ is [10]:
ϕ =
ML , GT
(13.3)
where G is the shear modulus, E the flexural rigidity and T the torsional moment.
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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 crosssection the shear deformation areas Ay and Az 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=1m
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]
|er |
ϕz [mrd]
|er |
SOF.
Ref.
[%]
SOF.
Ref.
[%]
1.333
1.333
0.00
2.000
2.000
0.00
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
Type square -srec rectangular -srec
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BE10: Verification of Beam and Section Types II
Table 13.2: (continued) y [m]
|er |
ϕz [mrd]
|er |
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
Type
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]
|er |
SOF.
Ref.
[%]
6.190
6.190
0.00
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
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
Type square -srec rectangular -srec
square box open -plat
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BE10: Verification of Beam and Section Types II
Table 13.3: (continued) ϕ [mrd]
|er |
SOF.
Ref.
[%]
L-beam -weld
1368.421
1368.421
0.00
L-beam -plat
1368.421
1368.421
0.00
Type
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 yREF , the calculated value is used yCALC 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 [10] [15]
62
Literature
C. Petersen. Stahlbau. 2nd. Vieweg, 1990. K. Holschemacher. Entwurfs- und Berechnungstafeln fur ¨ Bauingenieure. 3rd. Bauwerk, 2007.
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 crosssection fully plastifies. The beam is modelled and analysed with different elements and modules. Myed | ←−
−→ |
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 corresponding limit load can be calculated as:
Myed =
σyed bh2
,
(14.1)
6
where σyed is the yield stress, b and h the dimensions of the beam. The cross-section fully plastifies when the load reaches M = Mt = 1.5 × Myed , where all fibers of the beam are in condition of yielding [6]. σ σyed
ε
−σyed
Figure 14.2: Stress-Strain Curve
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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=1m
Myed = 280 Nm
ν = 0.3
h = 20 mm
σyed = 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/Myed
0.99
1.00
1.48
Fiber Beam
Standard Beam
σ [MP]
σ [MP]
σ
σeƒ ƒ
415.80
415.80
415.80
415.80
Ref.
σ < 420.00 Fully Elastic
≤ 420
≤ 420
≤ 420
≤ 420
σ ≤ 420.00 First Yield
≤ 420
≤ 420
≤ 430.9
≤ 420
σ ≤ 420.00 Elastic-Plastic
Fully-Plastic 1.50
1.51
Quad
Fully-Plastic No Convergence
Fully-Plastic
Fully-Plastic
No Convergence
No Convergence
Fully-Plastic
σ = 420.00
No Convergence
Fully-Plastic
Fully-Plastic
Fully-Plastic
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/Myed = 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/Myed = 1.0 and 1.48. A closer look at the list of results
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VERiFiCATiON MANUAL | SOFiSTiK 2014
BE11: Plastification of a Rectangular Beam
though, reveals that the eƒ ƒ ecte stresses do not exceed the σyed .
415.80
420.00
420.00 415.80
420.00
420.00 415.80
420.00
420.00
4.20
420.00 -420.00
4.24 -4.20
420.00 -420.00
-4.24
-415.80
-420.00
-420.00
-420.00
-420.00
-415.80 -415.80
-420.00
-420.00
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 Nostrand Co., Inc., 1940. [18] Verification Manual for the Mechanical APLD Application, Release 12.0. Ansys, Inc. 2009.
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BE11: Plastification of a Rectangular Beam
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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 Δy = / 200 = 2.5 cm, creating a geometrical imperfection. The beam is loaded with a transverse force Pz 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.
Pz
Δy
z y P
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 equilibrium of moments at the deformed system, with respect to the global -axis, the torsional moment at the support Mgob is: Mgob = Pz (y + Δy ) − Py z ,
(15.1)
whereas by the local -axis the torsional moment Moc is:
Moc = Pz y + P (
Δy
) z ,
(15.2)
where is the length of the beam, Δy the initial geometrical imperfection and P is negative for compression.
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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 γM = 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 CM are neglected. At the support the warping is not constrained. Table 15.1: Model Properties Material Properties
Geometric Properties
Loading
S 355
=5m
Pz = 10 kN
γM = 1.1
RRo/ SH 200 × 100 × 10 [15]
P = 100 kN
CM = 0
Δy = 2.5 cm
Table 15.2: Results
SOF. Ref.[21]
y
z
Mgob
Moc
Pbck
[cm]
[cm]
[kNcm]
[kNcm]
[kN]
3.208
10.204
57.07
26.97
163.7
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]
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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] [19]
K. Holschemacher. Entwurfs- und Berechnungstafeln fur ¨ Bauingenieure. 3rd. Bauwerk, 2007. ¨ V. Gensichen and G. Lumpe. Zur Leistungsfahigkeit, korrekten Anwendung und Kontrolle ¨ raumlicher Stabwerksprogramme. Stahlbau Seminar 07. ¨ ¨ [20] V. Gensichen. Zur Leistungsfahigkeit raumlicher Stabwerksprogramme, Feldstudie in Zusammenarbeit mit maßgebenden Programmherstellern. Stahlbau Seminar 07/08. ¨ [21] V. Gensichen and G. Lumpe. “Zur Leistungsfahigkeit, korrekten Anwendung und Kontrolle von ¨ EDV-Programmen fur Stabwerke im Stahlbau”. In: Stahlbau 77 (Teil 2) ¨ die Berechnung raumlicher (2008).
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BE12: Cantilever in Torsion
70
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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
/ 2
P
y y Figure 16.1: Problem Description
16.2
Reference Solution
The problem of 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]:
Pcr =
SOFiSTiK 2014 | VERiFiCATiON MANUAL
π 2 E (β)2
=
π 2 E 2
,
(16.1)
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
Py = 1 kN
h = 0.5 m
P 300 ◦ C
700
1.01184
1.01184
0.000 %
±1%
900
1.18000
1.18000
0.000 %
Tol.
for Θ ≤ 300
◦
C
± 0.05 mm
Table 35.3: Results for Thermal Elongation of Steel - QUAD Θ [◦ C]
142
Ref. [35]
SOF.
|er | [%]
Δ [mm]
Δ’ [mm]
or e [mm]
100
0.09984
0.09984
0.000 mm
300
0.37184
0.37184
0.000 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 %
Tol.
for Θ ≤ 300
◦
C
± 0.05 mm
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 exposed to fire. CEN. 2010.
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BE32: Thermal Extension of Structural Steel in case of Fire
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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
b
h
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 design 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.
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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:
ƒyk = 355 MP
ƒck = 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: σs(Θ) / ƒyk(Θ) or σc(Θ) / ƒck(Θ) = 0.2, 0.6, 0.9
Table 36.2: Results for Structural Steel - FIBER Θ [◦ C]
20
200
400
600
800
146
er [%]
Ref. [35]
SOF.
σs(Θ) / ƒyk(Θ)
Δ [mm]
Δ’ [mm]
0.2
0.034
0.034
0.560
0.6
0.101
0.101
−0.424
0.9
0.152
0.152
−0.094
0.2
−0.194
−0.194
−0.141
0.6
−0.119
−0.119
−0.119
0.9
0.159
0.156
1.794
0.2
−0.472
−0.472
0.097
0.6
−0.293
−0.294
−0.305
0.9
0.451
0.449
0.524
0.2
−0.789
−0.789
0.053
0.6
−0.581
−0.581
−0.054
0.9
0.162
0.160
1.243
0.2
−1.059
−1.059
0.030
Tol. [%]
±3%
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BE33: Work Laws in case of Fire for Concrete and Structural Steel
Table 36.2: (continued) Θ [◦ C]
er [%]
Ref. [35]
SOF.
σs(Θ) / ƒyk(Θ)
Δ [mm]
Δ’ [mm]
0.6
−0.914
−0.914
−0.028
0.9
−0.170
−0.172
−1.163
Tol. [%]
Table 36.3: Results for Concrete - FIBER Θ [◦ C]
er [%]
Ref. [35]
SOF.
σs(Θ) / ƒyk(Θ)
Δ [mm]
Δ’ [mm]
0.2
0.0334
0.0334
0.074
0.6
0.104
0.1036
0.428
0.9
0.176
0.1763
−0.173
0.2
−0.107
−0.1070
0.024
0.6
0.0474
0.0474
−0.035
0.9
0.2075
0.2075
0.014
0.2
−0.356
−0.3557
0.085
0.6
−0.075
−0.0750
0.016
0.9
0.216
0.2160
−0.008
0.2
−0.685
−0.6850
−0.007
0.6
0.0167
0.0167
−0.182
0.9
0.744
0.7442
−0.033
0.2
−1.066
−1.0662
−0.023
0.6
−0.365
−0.3645
0.145
0.9
0.363
0.363
−0.010
20
200
400
600
800
Tolerance [%]
±3%
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]
er [%]
Ref. [35]
SOF.
σs(Θ) / ƒyk(Θ)
Δ [mm]
Δ’ [mm]
0.2
0.034
0.034
0.560
0.6
0.101
0.101
−0.424
0.9
0.152
0.152
−0.094
20
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BE33: Work Laws in case of Fire for Concrete and Structural Steel
Table 36.4: (continued) Θ [◦ C]
200
400
600
800
er [%]
Ref. [35]
SOF.
σs(Θ) / ƒyk(Θ)
Δ [mm]
Δ’ [mm]
0.2
−0.194
−0.194
−0.173
0.6
−0.119
−0.119
−0.276
0.9
0.159
0.155
2.630
0.2
−0.472
−0.472
0.046
0.6
−0.293
−0.295
−0.669
0.9
0.451
0.437
3.148
0.2
−0.789
−0.789
−0.022
0.6
−0.581
−0.585
−0.640
0.9
0.162
0.146
9.856
0.2
−1.059
−1.059
−0.029
0.6
−0.914
−0.917
−0.358
0.9
−0.170
−0.185
−8.765
Tolerance [%]
±3%
Table 36.5: Results for Concrete - QUAD Θ [◦ C]
20
200
400
600
800
148
er [%]
Ref. [35]
SOF.
σs(Θ) / ƒyk(Θ)
Δ [mm]
Δ’ [mm]
0.2
0.0334
0.0334
0.081
0.6
0.1040
0.1036
0.429
0.9
0.1760
0.1763
−0.173
0.2
−0.1070
−0.1047
2.137
0.6
0.0474
0.0475
−0.108
0.9
0.2075
0.2086
−0.527
0.2
−0.3560
−0.3557
0.082
0.6
−0.0750
−0.0750
0.014
0.9
0.2160
0.2160
−0.008
0.2
−0.6850
−0.6851
−0.010
0.6
0.0167
0.0167
−0.197
0.9
0.7440
0.7442
−0.033
0.2
−1.0660
−1.0663
−0.026
0.6
−0.3650
−0.3645
0.145
0.9
0.3630
0.3630
−0.010
Tolerance [%]
±3%
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BE33: Work Laws in case of Fire for Concrete and Structural Steel
350
800 ◦ C 600 ◦ C 400 ◦ C 200 ◦ C 20 ◦ C
300
σs(Θ) [MP]
250 200 150 100 50 0 0.0
2.0
4.0
6.0
8.0
10.0 12.0 Strain [◦ /◦◦ ]
14.0
16.0
18.0
20.0
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 exposed to fire. CEN. 2010.
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BE34: Ultimate Bearing Capacity of Concrete and Steel under Fire
37 BE34: Ultimate Bearing Capacity of Concrete and Steel 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
b
h
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-
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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:
ƒyk = 355 MP
ƒck = 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]
e [kN]
er [%]
−35.5
0.000
0.000
−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
Ref. [35]
SOF.
NR,ƒ ,k [kN]
NR,ƒ ,k ’ [kN]
20
−35.5
200
Tol.
Table 37.3: Results for Concrete - FIBER beam Θ [◦ C]
e [kN]
er [%]
−20.0
−0.029
0.144
−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
Ref. [35]
SOF.
NR,ƒ ,k [kN]
NR,ƒ ,k ’ [kN]
20
−20.0
200
Tol.
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.
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Table 37.4: Results for Structural Steel - QUAD Θ [◦ C]
e [kN]
er [%]
−35.5
0.000
0.000
−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
Ref. [35]
SOF.
NR,ƒ ,k [kN]
NR,ƒ ,k ’ [kN]
20
−35.5
200
Tol.
Table 37.5: Results for Concrete - QUAD Θ [◦ C]
37.4
e [kN]
er [%]
−20.0
−0.029
0.144
−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
Ref. [35]
SOF.
NR,ƒ ,k [kN]
NR,ƒ ,k ’ [kN]
20
−20.0
200
Tol.
Conclusion
This example verifies the influence of compressive loading on the ultimate bearing capacity under different 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 exposed to fire. CEN. 2010.
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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 restraining forces developed in an immovable steel member due to temperature exposure are investigated for the model of Fig. 38.1.
Θo h 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 ƒyk(20◦ C) = 650 N/ mm2 and thermo-mechanical properties according to EN 1993-1-2. The model is exposed to different temperatures. In the first case the same temperature is assigned across the cross-section height, whereas
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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
ƒyk(20◦ C) = 650 N/ mm2
= 1000 mm
Case 1
E(20◦ C) = 210000 N/ mm2
h = 100 mm
Θo = 120◦ , C Θ = 120◦ C
Stress-strain curve
b = 100 mm
Case 2 Θo = 20◦ , C Θ = 220◦ C
according to DIN EN 1993-1-2
Table 38.2: Results for Structural Steel - FIBER beam Temperature Load
SOF.
X
X’
NZ [kN]
−2585.0
−2584.8
0.006
±1
MZ [kNm]
0.0
0.0
0.000
±1
σZ [N/mm2 ]
−258.5
−258.5
0.006
±5
NZ [kN]
−2511.0
−2503.9
0.282
±1
MZ [kNm]
−40.3
−40.2
0.249
±1
σZ [N/mm2 ]
−479.0
−479.0
0.000
±5
Θ [◦ C]
120/ 120
20/ 220
|er | [%]
Ref. [35]
Tol. [%]
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
SOF.
X
X’
NZ [kN]
−2585.0
−2589.8
0.186
±1
MZ [kNm]
0.0
0.0
0.000
±1
σZ [N/mm2 ]
−258.5
−258.98
0.186
±5
NZ [kN]
−2511.0
−2523.8
0.510
±1
MZ [kNm]
−40.3
−40.73
1.067
±1
σZ [N/mm2 ]
−479.0
−480.23
0.257
±5
Θ [◦ C]
120/ 120
20/ 220
156
|er | [%]
Ref. [35]
Tol. [%]
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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 6= 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 exposed to fire. CEN. 2010.
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BE36: Pushover Analysis: Performance Point Calculation by ATC-40 Procedure
39 BE36: Pushover Analysis: Performance Point Calculation 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 El. Demand Diagram Performance Point Capacity Diagram Sp Demand Diagram
Sdp
Sd
Figure 39.1: Determination of the performance point PP (Sdp , Sp )
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 format1 ) and the capacity diagram are known, it is possible to determine the performance point PP (Sdp , Sp ) (Fig. 39.1). The procedure comprises of a series of trial calculations (trial performance points PPt (Sdp,t , Sp,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 performance 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 PPt . 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 Sd format
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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. ϕcnod = 1
;
=1
;
m=1,
(39.1)
where ϕcnod 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] Sd =
cnod
= cnod , ϕcnod · Vb S = 2 = Vb , ·m
(39.2a) (39.2b)
where Vb is the base shear and cnod is the control node displacement. Since the original system is a SDOF system, Vb 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 Sd , 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 forcedisplacement work law P − of the non-linear spring element (Fig.39.2). Table 39.1: Model Properties [36] Capacity Diagram Point
160
Sd [mm], S [m/ s2 ]
Elastic Demand
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 SB :
CA = 0.40, CV = 0.40
- Type SD :
CA = 0.44, CV = 0.64
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BE36: Pushover Analysis: Performance Point Calculation by ATC-40 Procedure
D
P [kN]
C B 3.0
A
2.0
1.0
200.0
150.0
100.0
0.0
50.0
O
0.0
u [mm]
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 summarized in Table 39.1. Two soil profile types are considered - soil profile type SB and SD . The outcome of the analysis is shown in Figures 39.3 and 39.4. Sa0[m/sec2]
Tb0=00.1
Tc0=00.4
T0=00.5 Ty
SPL01
SPL02
SPL03
SPL4
10.00
T0=01.0 Tp
5.00
PP
Capacity
T0=01.5
PY T0=02.0 UBC,Soil-B Demand, ξ-eff0=0 9.36L T0=04.0 200.000
150.000
100.000
50.000
0.000
0.00
Sd0[mm]
Figure 39.3: Capacity-Demand-Diagram (Soil Profile Type SB )
Sab[m/sec2]
Tbb=b0.1
Tb=b0.5 SPLb1
SPLb2
Tcb=b0.6 SPLb3
SPL4
Ty
10.00
Tb=b1.0
Tp 5.00
PP
PY
Capacity
UBC,Soil-D Tb=b1.5 Demand, ξ-effb=b14.59L Tb=b2.0 Tdb=b3.0 Tb=b4.0
200.000
150.000
100.000
50.000
0.000
0.00
Sdb[mm]
Figure 39.4: Capacity-Demand-Diagram (Soil Profile Type SD )
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The results of the SOFiSTiK calculation and the comparison with the reference solution are summarized in Table 39.2. Table 39.2: Results ξeƒ ƒ
SR
SR
Sdy
Sy
Sdp
Sp
[%]
[−]
[−]
[ mm]
[m/ s2 ]
[mm]
[m/ s2 ]
SOF.
9.4
0.80
0.84
51.30
2.62
85.04
3.23
Ref. [36]
9.2
0.80
0.85
53.34
2.65
83.36
3.24
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
Ref. [36]
14.2
0.66
0.74
58.42
3.04
149.86
3.63
2.8
1.5
1.4
2.5
0.7
0.3
1.7
Soil type
SB
|e|
SD
|e|
[%]
[%]
ξeƒ ƒ effective viscous damping of the equivalent linear SDOF system SR , SR spectral reduction factors in constant acceleration and constant velocity range of spectrum Sdy , Sy spectral displacement and spectral acceleration at yielding point Sdp , Sp 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. Redwood City, CA, 1996. [37] SOFiLOAD Manual: Loadgenerator for Finite Elements and Frameworks. Version 2014.1. SOFiSTiK AG. Oberschleißheim, Germany, 2013.
162
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BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory
40 BE37: Beam Calculation of Varying Cross-Section according 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 compressive 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. V H
qE E ∗
∗ ψ0
H∗ ∗
b
0 t
s
h
ΔT A qA 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
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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/ cm2 , γg = 1.35
V = 500 kN
ψ0 = 1/ 200, 0 = −48 mm
qE = 6 kN/ m, qA = 10 kN/ m
αT = 1.2 × 10−5 1/ ◦ K
ΔT = Trght − Teƒ t = −25 ◦ C
= 12 m, ∗ = 4 m
H = 20 kN, H∗ = 10 kN
Table 40.2: Cross-sectional Properties
Web [mm]
Flange [mm]
Area [cm2 ]
y [cm4 ]
Position 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 [kN m]
N [kN]
[mm]
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 ϕE
E
E
MA
NA
NE
NK
[mrd]
[mm]
[mm]
[MN m]
[MN]
[MN]
[MN]
Ref. [38]
67.70
423.50
−1.9050
−1.10
−0.5203
−0.50
−1.882
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
-
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
-
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
-
Number of Elements 3
6
24
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 increases. 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 ϕE
E
MA
[mrd]
[mm]
[MN m]
Ref. [38]
67.70
423.50
−1.100
ASE
67.657
423.27
−1.0995
STAR2
67.918
424.73
−1.1002
Number of Elements 24
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 satisfying 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.
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BE37: Beam Calculation of Varying Cross-Section according to Second Order Theory
Table 40.5: Results with DYNA ϕE
E
E
MA
NA
NE
[mrd]
[mm]
[mm]
[MN m]
[MN]
[MN]
Ref. [38]
67.70
423.50
−1.9050
−1.10
−0.5203
−0.50
DYNA
67.6569
423.2766
−1.9045
−1.0994
−0.5203
−0.50
Number of Elements 24
Table 40.6: Results with DYNA for Combination of Constant Normal Force and Single Loadings
40.5 [38]
166
Load
ϕE
E
MA
Case
[mrd]
[mm]
[MN m]
H∗ , H
23.6779
148.3151
−0.3972
q
22.5896
163.3797
−0.6130
ΔT
14.8218
77.1333
−0.0401
ψ0
2.6481
15.9532
−0.0395
0 P
3.9194
18.4953
−0.0096
67.6569
423.2766
−1.0994
Literature
¨ H. Rubin. “Ein einheitliches, geschlossenes Konzept zur Berechnung von Staben mit stetig ¨ verandlichem Querschnitt nach Theorie I. und II. Ordnung”. In: Bauingenieur 66 (1991), pp. 465– 477.
VERiFiCATiON MANUAL | SOFiSTiK 2014
BE38: Calculation of Slope Stability by Phi-C Reduction
41 tion
BE38: Calculation of Slope Stability by Phi-C Reduc-
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.
h1
h2 1
sope
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 instability. 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 =
SOFiSTiK 2014 | VERiFiCATiON MANUAL
be sher strength eqbrm sher stress
(41.1)
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]:
ηϕ =
tn ϕnp (41.2)
tn ϕt
ηc =
cnp (41.3)
ct
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 = ηph = ηc . 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
168
Material Properties
Geometric Properties
E = 2600 kN/ m2 , ν = 0.3
h1 = 6.5 m
γ = 16 kN/ m3
h2 = 2.0 m
ϕ = 20◦ , ψ = 20◦
1 = 2 = 3 = 3.0 m
c = 5 kN/ m2
sope = 9.0 m
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.
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169
BE38: Calculation of Slope Stability by Phi-C Reduction
170
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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]
ƒm,n =
v u λm,n 2 t gD 2π
(42.1)
γh
where 2
λm,n = π
SOFiSTiK 2014 | VERiFiCATiON MANUAL
2
m2 2
+
n2 b2
(42.2)
171
BE39: Natural Frequencies of a Rectangular Plate
and D is the flexural rigidity of the plate
D=
Eh3 (42.3)
12(1 − ν 2 )
Combining the above equations gives
ƒm,n =
π
22
m2 + n
2 2 b2
v u t g
Eh3 (42.4)
γh 12(1 − ν 2 )
where , b the dimensions of the plate, h the thickness and γh/ g the mass of the plate per unit area. The values of λm,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 λm,n 2
42.3
m
n
λm,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
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/ m3
b = 3.0 m
ν = 0.3
h = 0.02 m
Table 42.3: Results SOF. [Hz]
Ref. [Hz]
|er | [%]
1
2.941
2.955
0.476
2
5.623
5.682
1.047
3
9.200
9.091
1.197
Eigenfrequency Number
172
VERiFiCATiON MANUAL | SOFiSTiK 2014
BE39: Natural Frequencies of a Rectangular Plate
Table 42.3: (continued) SOF. [Hz]
Ref. [Hz]
|er | [%]
4
10.206
10.228
0.214
5
11.706
11.819
0.954
Eigenfrequency Number
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 [29] [43]
Literature
S. Timoshenko. Vibration Problems in Engineering. 2nd. D. Van Nostrand Co., Inc., 1937. Schneider. Bautabellen fur ¨ 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. δ F
F
q
H b
2 c
c
ψ0
ψ0 h
1 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
H1 = H2 =
q2 (43.1)
4h(k + 2)
M1 = M2 =
Hh (43.2)
3
M3 = M4 = M1 − H1 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. p The moments at nodes 1 − 4 are determined in dependency of the column characteristic ratio ε = c N/ Ec , giving the influence of the normal force N = F + q/ 2 with respect to the column properties, length c and bending stiffness Ec . 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 ψ0 = 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
Ec = 6000 kNm2
=6m
q = 10 kN/ m
Eb = 4000 kNm2
h=5m
H = 20 kN
ψ0 = 1/ 200
F = 400 kN
Table 43.2: Nonlinear Case Results
176
Ref. [43]
SOF.
M1 [kN m]
38.2
38.62
M2 [kN m]
22.5
22.52
M3 [kN m]
58.1
58.02
M4 [kN m]
58.8
58.79
δ [mm]
65.3
65.44
VERiFiCATiON MANUAL | SOFiSTiK 2014
BE40: Portal Frame
Linear Case
Nonlinear Case
Figure 43.2: Bending Moments
Table 43.3: Linear Case Results Ref. [Sect.43.2]
SOF.
H1 = H2 [kN]
5.54
5.52
M1 = M2 [kN m]
9.23
9.18
M3 = M4 [kN m]
18.46
18.43
Linear Case
Nonlinear 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 [43]
Literature
Schneider. Bautabellen fur ¨ 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
p
p 1.0
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 · 106 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
180
Conforming Element
Non-Conforming Element
4
0.4525
0.5917
8
0.8214
0.9056
16
0.9701
1.0082
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.
Normalised Displacement /
1.2
1
0.8
0.6
0.4
Analytical Solution Conforming Element Non-Conforming Element
0.2
0
0
2
4
6
8 10 Number of Elements
12
14
16
18
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: Softwaregestutze 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 dynamic 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: International 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 =
K=
SOFiSTiK 2014 | VERiFiCATiON MANUAL
p · r 4 (5 + μ) 64 · K (5 + μ) E · h3 12(1 − μ2 )
(45.1)
(45.2)
183
BE42: Thick Circular Plate
S =
1.2 · p · r 2
(45.3)
4·G·h
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 = My =
45.3
p · r2 16
· (3 + μ) = 4928.125
[kNm/ m]
(45.4)
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 , ϕy , 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/ m2 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/ cm2
h = 0.5 − 2.5 m
p = 1000 kN/ m2
G = 1300 kN/ cm2
r =5m
μ = 0.154
D = 10 m
x
y
Figure 45.2: FEM model
The maximum bending moment is calculated at the middle of the plate, as M = My = 4932.244
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[kNm/ m] with a deviation of 0.08 %. Table 45.2: Results h [m]
h/ D
Analytical z [mm]
SOF. z [mm]
|er | [%]
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 calculated results are in very good agreement with the analytical solution even for thicker plates.
45.5
Literature
[48] VDI 6201 Beispiel: Softwaregestutze Tragwerksberechnung - Beispiel Dicke Platte, Kategorie 1: ¨ Mechanische Grundlagen. Verein Deutscher Ingenieure e. V. ¨ [49] F. U. Mathiak. Ebene Flachentragwerke Teil II, Grundlagen der Plattentheorie. Hochschule Neubrandenburg. 2011.
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BE42: Thick Circular Plate
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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].
y p
h
A
D
d A0
p
h
L/ 2
L/ 2
t
Figure 46.1: Problem Description
46.2
Reference Solution
The maximum stress σA,,m resulting from a load p, at the edge of the hole can be determined at points A and A0 across a vertical cut, visualised in Fig. 46.1, as follows [51] [52]: σA,,m = Kt · σ,nom
(46.1)
where P = p · D = 1000
σ,nom =
SOFiSTiK 2014 | VERiFiCATiON MANUAL
P t · (D − d)
= 33.33
[kN]
[N/ mm2 ]
(46.2)
(46.3)
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BE43: Panel with Circular Hole
Kt = 3.000 − 3.140 · (d/ D) + 3.667 · (d/ D)2 − 1.527 · (d/ D)3 ,
46.3
(0 < d/ D < 1)
(46.4)
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 · 105 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
[44]
[310]
[168]
[390]
(a) Structured Meshing
[44]
[424]
[140]
[804]
(b) Unstructured Meshing
Figure 46.2: FEM Models
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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. 75
σA,,m [N/ mm2 ]
70 65 60 55 50 Analytical: 74.43 [ N/ mm2 ] Structured Mesh - Conforming Element Structured Mesh - Non-conforming Element Unstructured Mesh - Non-Conforming Element
45 40
0
100
200
300 400 500 600 Number of Degrees of Freedom nDOF
700
800
900
Figure 46.3: Convergence Diagram
16.54
74.75
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
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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 behaviour 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: Softwaregestutze Tragwerksberechnung - Beispiel Scheibe mit kreisformigem ¨ Loch - Konvergenztest fur ¨ 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.
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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
p B A h = 2 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 0 ) analysis using effective material parameters. G and ν 0 are effective soil parameters, while B represents the Skempton’s B-parameter. Self-weight is not taken into consideration.
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BE44: Undrained Elastic Soil Layer Subjected to Strip Loading
Table 47.1: Model Properties Material
Geometry
Loading
G, ν 0 = 0.3
p
B = 0.998
h = 4
ρ = 0.0 kg/ m3
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
analytical, drained fem, drained analytical, undrained fem, undrained
0 G·y p·h
0.1
0.2
0.3
0
1
2
3
4
/ Figure 47.3: Vertical displacement y of the surface
The drained and undrained horizontal and vertical total stresses (σ = σ 0 + pe ) 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.
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0
0
0.2
0.2 0.4
0.4
y/ h
y/ h 0.6
0.6
0.8
0.8
1
0
−0.2
−0.4
−0.6
−0.8
1 −0.6
−1
−0.7
−0.8
−0.9
−1
σy / p
σ / p (a) Horizontal stress σ
(b) Vertical stress σy
Figure 47.4: Stresses beneath footing center
Pore excess pressure (pe ) distribution for the undrained condition along the center line (A − C) is shown in Fig. 47.5. 0 0.2 0.4
y/ h 0.6 0.8 1 −0.4
−0.6
−0.8
−1
pe / p Figure 47.5: Excess pore pressure pe 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 [53]
Literature
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.
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BE44: Undrained Elastic Soil Layer Subjected to Strip Loading
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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 p0 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. p0
y
G, ν 0 , ν , ρ
h z
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 pe can be written as [55]: ∂pe ∂2 pe = c , (48.1) ∂t ∂z 2 where: c = k · Es / γ
coefficient of consolidation,
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BE45: One-Dimensional Soil Consolidation
Es
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 φcs ) plastic flow. The mobilized friction angle φm in Equation 51.1 is computed according to sin φm =
σ10 − σ30
(51.2)
2c · cot φ − σ10 − σ30
At failure, when φm ≡ φ, also the dilatancy angle reaches its final value ψm ≡ ψ. Accordingly, from Equation 51.1 the critical state friction angle can be derived as sin φcs =
sin φ − sin ψ
(51.3)
1 − sin φ sin ψ
It has been recognized that in some cases the Rowe’s model for dilatancy angles (Eq. 51.1) can overestimate the contractive behavior of the soil at low mobilized friction angles, φm < φcs . 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 sin φm / sin φ sin ψm =
sin φm − sin φcs 1 − sin φm sin φcs
·
sin φm sin φ
.
(51.4)
Wehnert [60] proposed a model based on a lower cut-off value ψ0 for the modification of the Rowe’s formulation from Eq. 51.1 at low mobilized friction angles sin ψ0 ; 0 < ψm ≤ ψRoe m sin ψm = . (51.5) sin φm − sin φcs Roe ; ψm < ψm ≤ ψ 1 − sin φm sin φcs This dilatancy model obviously requires a specification of an additional parameter, ψ0 .
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ψm [ ◦ ] Rowe Soreide Wehnert, ψ0 = −3◦
20
Constant ψ = 10 φcs 0 ψ0
5
15
25
φ = 35
φm [ ◦ ]
-10
-20
Figure 51.2: Comparison of models for mobilized dilatancy angle ψm 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 MohrCoulomb 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 ψ0 at low stress ratios, is used. The undrained calculation is conducted in the form of effective stresses with effective shear parameters (c0 , φ0 ) 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, σc = 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/ m2
Es,reƒ = 16.0 MN/ m2
H = 0.09 m
Phase I:
νr = 0.25
E50,reƒ = 12.0 MN/ m2
D = 0.036 m
σ1 = σ3 = σc =
γ = 0.0 MN/ m3
m = 0.75
= 200, 300 kP
c0 = 0.01 kN/ m2
Rƒ = 0.9
Phase II:
φ0 = 34◦
K0 = 0.44
σ3 = σc = 200, 300 kP
ψ = 2◦
B = 0.9832
σ1 = σ > σc
ψ0 = −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
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BE48: Triaxial Consolidated Undrained (CU) Test
failure condition (MC failure) based on the used shear parameters, c0 and φ0 , 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 ψ0 . Comparing the results of the development of the deviatoric stress q and the excess pore water pressure pe 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 ψ0 = −4◦2 . Due to the presence of the negative mobilized dilatancy angle (ψm < 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 φ0m becomes larger than the phase transition angle φ0ƒ which is approximately equal to the critical state friction angle φ0cs (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 (ψm = ψ0 for 0 ≤ ψm ≤ ψRoe , Eq. 51.5), the pore water under-pressures m are overestimated. Next the SOFiSTiK results obtained using the same soil model and dilatancy formulation as in [60] (HSWehnert) 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. used value ψ0 = −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]. 2 The
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BE48: Triaxial Consolidated Undrained (CU) Test
Hostun-RF Sand, σc = 200 kN/ m2
51.3.1 400
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]
350
q [kN/ m2 ]
300 250 200 150 100 50 0
0
50
100
150
200
250
300
p, p0 [kN/ m2 ] Figure 51.3: Effective stress path curve (q-p)
400
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp.1 [60] Exp.2 [60]
350
q [kN/ m2 ]
300 250 200 150 100 50 0
0
5
10
15
ϵ1 [%] Figure 51.4: Deviatoric stress - axial strain curve (q-ϵ1 )
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BE48: Triaxial Consolidated Undrained (CU) Test
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp. 1 [60] Exp. 2 [60]
200
pe [kN/ m2 ]
150
100
50
0
5
0
15
10
ϵ1 [%] Figure 51.5: Excess porewater pressure - axial strain curve (pe -ϵ1 )
Hostun-RF Sand, σc = 300 kN/ m2
51.3.2 400
MC failure TSP ESP, MC ESP, HS-Rowe ESP, HS-Cons ESP, HS-Soreide ESP, HS-Wehnert ESP, Wehnert [60] ESP, Exp. 1 [60]
350
q [kN/ m2 ]
300 250 200 150 100 50 0
0
50
100
150
200
p,
p0
250
300
350
400
450
[kN/ m2 ]
Figure 51.6: Effective stress path curve (q-p)
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450
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp.1 [60]
400 350
q [kN/ m2 ]
300 250 200 150 100 50 0
0
5
15
10
20
ϵ1 [%] Figure 51.7: Deviatoric stress - axial strain curve (q-ϵ1 )
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert [60] Exp. 1 [60]
300
pe [kN/ m2 ]
250
200
150
100
50
0
0
5
10
15
ϵ1 [%] Figure 51.8: Excess porewater pressure - axial strain curve (pe -ϵ1 )
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BE48: Triaxial Consolidated Undrained (CU) Test
HS-Rowe HS-Cons HS-Soreide HS-Wehnert
ψm [ ◦ ]
0
−10
−20
−30 0
5
10
15
20
φm
25
30
35
[ ◦]
Figure 51.9: Mobilised dilatancy angle - friction angle curve (ψm -φm )
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 fur ¨ ¨ Stuttgart: P. A. Vermeer, 2006. Geotechnik, Universitat [61] P.W. Rowe. “The stress-dilatancy relation for static equilibrium of an assembly of particles in contact”. In: Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences 269.1339 (1962), pp. 500–527. [62] O. K. Soreide. “Mixed hardening models for frictional soils”. PhD thesis. NTNU Norges teknisknaturvitenskapelige universitet, 2003.
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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]. σ1
D
H σ3
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,
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BE49: Triaxial Drained Test
σc = 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/ m2
Es,reƒ = 16.0 MN/ m2
H = 0.09 m
Phase I:
νr = 0.25
E50,reƒ = 12.0 MN/ m2
D = 0.036 m
σ1 = σ3 = σc =
γ = 0.0 MN/ m3
m = 0.75
= 100, 300 kP
c0 = 0.01 kN/ m2
Rƒ = 0.9
Phase II:
φ0 = 34◦
K0 = 0.44
σ3 = σc = 100, 300 kP
ψ = 2◦
B = 0.9832
σ1 = σ > σc
ψ0 = −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, c0 and φ0 , 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 (HSWehnert) with the reference numerical results from Wehnert [60], we can notice that the stress paths p-q are captured exactly for both σc -stress levels. Accordingly, the deviatoric stress q versus the axial strain ε1 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.
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Hostun-RF Sand, σc = 100 kN/ m2
52.3.1 600
MC failure MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1
500
q [kN/ m2 ]
400
300
200
100
0
0
100
200
300
400
500
600
p, p0 [kN/ m2 ] Figure 52.2: Effective stress path curve (q-p)
400
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp.1
350
q [kN/ m2 ]
300 250 200 150 100 50 0
0
5
10
15
20
ϵ1 [%] Figure 52.3: Deviatoric stress - axial strain curve (q-ϵ1 )
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BE49: Triaxial Drained Test
0.000
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1
0.200 0.400 0.600
ϵ [%]
0.800 1.000 1.200 1.400 1.600 1.800 2.000
5
0
15
10
20
ϵ1 [%] Figure 52.4: Volumetric strain - axial strain curve (ϵ -ϵ1 )
Hostun-RF Sand, σc = 300 kN/ m2
52.3.2 800
MC failure MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1
700
q [kN/ m2 ]
600 500 400 300 200 100 0
0
100
200
300
400
500
600
700
p, p0 [kN/ m2 ] Figure 52.5: Effective stress path curve (q-p)
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800
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp.1
700
q [kN/ m2 ]
600 500 400 300 200 100 0
0
5
10
15
20
ϵ1 [%] Figure 52.6: Deviatoric stress - axial strain curve (q-ϵ1 )
0.000
MC HS-Rowe HS-Cons HS-Soreide HS-Wehnert Wehnert Exp. 1
0.200 0.400
ϵ [%]
0.600 0.800 1.000 1.200 1.400 1.600 1.800 0
5
10
15
20
ϵ1 [%] Figure 52.7: Volumetric strain - axial strain curve (ϵ -ϵ1 )
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HS-Rowe HS-Cons HS-Soreide HS-Wehnert
ψm [ ◦ ]
0
−10
−20
−30
0
5
10
15
20
25
30
35
φm [ ◦ ] Figure 52.8: Mobilised dilatancy angle - friction angle curve (ψm -φm )
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 [60]
222
Literature
M. Wehnert. Ein Beistrag zur dreainerten und undrainerten Analyse in der Geotechnik. Institut fur ¨ ¨ Stuttgart: P. A. Vermeer, 2006. Geotechnik, Universitat
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 Under 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 r0 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. p(t) p(¯t ) r0
p0
0
1.5
3
¯t = t · cs / r0
(b) Pressure load (a) Circular cavity embedded in a fullplane
Figure 53.1: Problem Definition
53.2
Reference Solution
This problem is essentially a one dimensional problem which has an analytical solution [63]. The forcedisplacement 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.
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The dynamic-stiffness coefficient for this particular problem has an analytical expression and it reads (2) 2πG0 λ−1−F HF+1 (λ0 ) 0 , (53.2) S∞ (0 ) = · 2(1 − ν) − 2ν + 2(1 − ν) (2) 1 − 2ν λ HF (λ0 ) where G0
shear modulus,
ν
Poisson’s ratio,
ρ
mass density,
cs =
p
G/ ρ p cp = cs (2 − 2ν)/ (1 − 2n)
shear wave velocity,
0 = ωr0 / cp
dimensionless frequency,
λ = 2/ (2 − α)
coefficient,
α
non-homogeneity parameter of elasticity (α = 0 for the homogeneous case),
P-wave velocity,
(2)
Hk
F=
v t
the second kind Hankel function of the order k, (λ − 1)2 − λ2
ν(α + 1) − 1 1−ν
order of the Hankel function.
The static-stiffness coefficient K ∞ is used to non-dimensionlize displacement response K∞ =
2πG0 1 − 2ν
h i q · 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 dynamicstiffness 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 planestrain 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
cs , ρ, ν = 0.3
r0
P(t) = 2πr0 p(t)
Δt = 0.04 · r0 / cp
P0 = 2πr0 p0
M, N, θ = 1.4
G0 = ρc2s
Load and the finite element model of the structure are depicted in Fig. 53.2. The structure is comprised
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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 Scaled 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. 1
analytic constant M = 40, N = 10 M = 20, N = 15
r · K ∞ / P0
0.8 0.6 0.4 0.2 0 −0.2
0
2
4
6
8
10
12
14
16
18
20
t · cp / r0 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].
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53.5
Literature
[63]
M.H. Bazyar. “Dynamic Soil-Structure Interaction Analysis Using the Scaled Boundary FiniteElement 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).
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BE51: Pushover Analysis: Performance Point Calculation by EC8 Procedure
54 BE51: Pushover Analysis: Performance Point Calculation 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 El. Demand Diagram Performance Point Capacity Diagram Sp Demand Diagram
Sdp
Sd
Figure 54.1: Determination of the performance point PP (Sdp , Sp )
54.2
Reference Solution
The reference solution is provided in [66]. Assuming that the elastic demand diagram (5% elastic response spectrum in ADRS format1 ) and the capacity diagram are known, it is possible to determine the performance point PP (Sdp , Sp ) (Fig. 54.1). The procedure comprises of a series of trial calculations (trial performance points PPt (Sdp,t , Sp,t )), in which the equivalent inelastic single degree of freedom system (SDOF), represented by the capacity 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 PPt . 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 Sd format
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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. T ∗ = Ty
S
S Tc
Tc T ∗ = Ty
Se PE
PE Se Sp
PY
PP
μ=1 μ=1
μ>1
Sdy
Sde
Sdp
PY
Sp Sy
PP μ>1
Sdp = Sde
Sdy
Sd
(a) Short period range, T ∗ < TC
Sd
(b) Medium and long range, T ∗ > TC
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 (Sdy , Sy ) 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 ∗ = Ty from the known PY (Sdy , Sy ): T ∗ = Ty = 2π ·
v uS t dy Sy
;
(54.1)
2. Calculate the elastic spectral response PE (Sde , Se ) of the idealized equivalent SDOF system with the period T ∗ = Ty from the given 5%-damped elastic response spectrum (Fig. 54.2); 3. Calculate the yield strength reduction factor Ry : Ry =
Se ;
Sy
(54.2)
4. Calculate ductility μ: T (Ry − 1) · C + 1 T∗ μ= Ry
for
T ∗ < TC
for
T∗
;
(54.3)
≥ TC
5. Determine the performance point PP (Sdp , Sp ) from the inelastic design spectrum: Sdp = μ · Sdy = μ · Sp =
228
Se (T ∗ ) Ry
.
Sde Ry
,
(54.4a) (54.4b)
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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. ϕcnod = 1
;
=1
;
m=1,
(54.5)
where ϕcnod 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] Sd =
cnod
= cnod , ϕcnod · Vb S = 2 = Vb , ·m
(54.6a) (54.6b)
where Vb is the base shear and cnod is the control node displacement. Since the original system is a SDOF system, Vb 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 Sd , 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 Point
Sd [mm], S [m/ s2 ]
Elastic Demand
5%-Damped Elastic Response Spectrum
A
(61, 3.83)
g = {0.60g, 0.30g, 0.16g}
B
(∞, 3.83)
SA = 1.0, SB = 2.5, k1 = 1.0 TB = 0.15s, TC = 0.60s, TD = 3.00s
2 Not
that the point A is nothing else but the point PY (Sdy , Sy ).
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B
A
P [kN]
3.0
2.0
1.0
200.0
150.0
100.0
50.0
O 0.0
0.0
u [mm]
P
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) 0 ≤ T ≤ TB : S = SA + SB
T TB
· (SB − SA )
TB ≤ T ≤ TC : S = SB
SA
TC ≤ T ≤ TD : S = SB ·
TB
TC
TD
TC
k 1
T
T
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. SaS[m/sec2]
TbS=S0.2
TS=S0.5
TcS=S0.6
SPLS1
SPLS2 Ty
15.00
10.00
TS=S1.0 El.SDemand,S0.60g
5.00
PY
PP Capacity
Tp TS=S1.5 Demand,Sμ =S 2.91 TS=S2.0 TdS=S3.0 TS=S4.0
200.000
150.000
100.000
50.000
0.000
0.00
SdS[mm]
Figure 54.5: Capacity-Demand-Diagram (g = 0.60g)
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SaL[m/sec2]
TbL=L0.2
TL=L0.5
TcL=L0.6
TL=L1.0 Ty
8.00
Tp
SPLL1
SPLL2
6.00
PY
4.00
PP
TL=L1.5 Capacity
TL=L2.0 El.LDemand,L0.30g
2.00
Demand,Lμ =L 1.45 TdL=L3.0 TL=L4.0 SdL[mm]
200.000
150.000
100.000
50.000
0.000
0.00
Figure 54.6: Capacity-Demand-Diagram (g = 0.30g)
SaL[m/sec2]
TbL=L0.2
TL=L0.5 TcL=L0.6
TL=L1.0
Ty Tp SPLL1
4.00
TL=L1.5
SPLL2 Capacity
PY
PP TL=L2.0
2.00
TdL=L3.0 TL=L4.0
El.LDemand,L0.15g Demand,Lμ =L 1.00
SdL[mm]
200.000
150.000
100.000
50.000
0.000
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
Ryp
Ty
Sdy
Sdp
Sp
[−]
[ −]
[s]
[mm]
[ mm]
[ m/ s2 ]
SOF.
2.9
2.9
0.79
61
177
3.83
Ref. [66]
2.9
2.9
0.79
61
177
3.83
0.0
0.0
0.0
0.0
0.0
0.0
SOF.
1.5
1.5
0.79
61
89
3.83
Ref. [66]
1.5
1.5
0.79
61
89
3.83
0.0
0.0
0.0
0.0
0.0
0.0
1.0
1.0
0.79
44
44
2.78
[g]
0.60
|e|
0.30
|e|
[%]
[%]
SOF.
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Table 54.2: (continued) g [g] 0.15
Ref. [66] [%]
|e| μ Ryp Ty Sdy , Sdp Sp
μ
Ryp
Ty
Sdy
Sdp
Sp
[−]
[ −]
[s]
[ mm]
[ mm]
[ m/ s2 ]
1.0
1.0
0.79
44
44
2.76
0.0
0.0
0.0
0.0
0.0
0.7
displacement ductility factor reduction factor due to ductility at performance point period associated with yielding point spectral displacements at yielding and performance point 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 procedure 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 engineering and structural dynamics 28.9 (1999), pp. 979–993.
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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 5th order wave theory. In Fig. 55.1 the properties of a wave can be visualised. c z
η
H x
SWL
d L
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 C2 (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.
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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]. 60 50 40 30
η [ƒt ]
20 10 0 − 10 − 20 Reference SOFiSTiK
− 30 − 40
0
20
40
60
80
100 θ [o ]
120
140
160
180
200
Figure 55.2: Wave profile
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120
elevation from seabed (z − d) [ƒ t]
100
80
60
40
20 Reference SOFiSTiK
0
0
5
10
15
20
25 30 [ƒ t/ s]
35
40
45
50
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.
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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].
Dp
c z
η
H x
SWL
T L
d
Lp
d
h
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
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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
Dp = 6 m
H = 17.5 m
Lp = 54 m
T = 15 s
Cm = 2.0
SWL = 0 m
Cd = 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 m and the calculated depth criterion d/ L = 0.138 indicates that the examined case falls into finite water.
10 8 6 4
η [m ]
2 0 −2 −4 -5.000
−6 Reference SOFiSTiK
−8 − 10
0
2
4
6
8 T [s]
10
12
14
16
Figure 56.2: Wave profile
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5,000 4,000 3,000 2,000
F [kN]
1,000 0 −1,000 −2,000 −3,000 Reference SOFiSTiK
−4,000 −5,000
0
2
4
6
8
10 T [s]
12
14
16
18
20
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. Blumel. WaveLoads - A computer program to calculate wave ¨ loading on vertical and inclined tubes. ISEB - Fluid Mechanics Institute, University of Hannover. 2005.
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Part III
Design Code Benchmark Examples
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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 mEd and the required reinforcement is determined. b mEd
h d s1
s1
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) εc
Ac
Fc
d
z
Fs
As εs
Figure 57.2: Stress and Strain Distributions in the Design of Cross-sections
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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 σ
ƒtk,c = 525N/ mm2
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε εd = 25 0 /00 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
mEd = 25 kNm/ m
B 500A
d = 17.0 cm b = 1.0 m
Table 57.2: Results
s1 [cm2 / m]
244
SOF.
General Chart [74]
ω−Table [74]
kd −Table [74]
3.334
3.328
3.334
3.333
VERiFiCATiON MANUAL | SOFiSTiK 2014
DCE-EN1: Design of Slab for Bending
Design Process1
57.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: γc = 1.50 Steel: γs = 1.15
(NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
ƒck = 25 MP ƒcd = cc · ƒck / γc = 0.85 · 25/ 1.5 = 14.17 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects
3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Design Load: MEd = mEd · b = 25 kNm; NEd = 0 MEds = MEd − NEd · zs1 = 25 kNm Design with respect to General Design Chart Bending with axial force for rectangular cross-sections: μEds =
MEds b · d2 · ƒcd
=
25 · 10−3 1.0 · 0.172 · 14.17
= 0.061
ε = 25 · 10−3; ζ = 0.97 → σs1d = 456.52 MP 1 MEds s1 = · + NEd = 3.328 cm2 / m σs1d ζ·d
Tab. 9.1 [74]: General Chart for up to C50/ 60 - Cross-section without compression reinforcement
Design with respect to ω− (or μs − ) Design Table for rectangular cross-sections: μEds =
MEds b · d2 · ƒcd
=
25 · 10−3 1.0 · 0.172 · 14.17
= 0.061
ω = 0.0632 (interpolated) and σsd = 456.52 MP s1 =
1 σsd
Tab. 9.2 [74]: ω−Table for up to C50/ 60 - Rectangular section without compression reinforcement
· (ω · b · d · ƒcd + NEd ) = 3.334 cm2 / m
Design with respect to kd − Design Table for rectangular crosssections: kd = p
d MEds / b
=p
17 25/ 1.0
= 3.40
ks = 2.381, κs = 0.952 (interpolated values) MEds NEd s1 = ks · + · κs = 3.333 cm2 / m d σs1d
Tab. 9.3 [74]: kd −Table for up to C50/ 60 - Rectangular section without compression reinforcement
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 National 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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. Muller, and F. Lobisch. Bemessungshilfsmittel fur ¨ ¨ Betonbauteile nach Eurocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund 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
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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 MEd and the required reinforcement is determined. b
As2
h
d
MEd
d2
As1
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) εc
σcd
d2 As2
σs2d
d
εs2
Fs2d Fc
zs2 z zs1
As1 d1
εs1
σs1d
Fs1d
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 σ
ƒtk,c = 525N/ mm2
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε εd = 25 0 /00 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
MEd = 135 kNm
B 500A
d = 35.0 cm d2 = 5.0 cm b = 25 cm
Table 58.2: Results SOF.
General Chart [74]
ω−Table [74]
kd −Table [74]
As1 [cm2 / m]
10.73
10.73
10.77
10.79
As2 [cm2 / m]
2.47
2.47
2.52
2.43
248
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DCE-EN2: Design of a Rectangular CS for Bending
Design Process1
58.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: γc = 1.50
(NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Steel: γs = 1.15 ƒck = 20 MP ƒcd = cc · ƒck / γc = 0.85 · 20/ 1.5 = 11.33 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP Design Load:
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects
3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
NEd = 0 MEds = MEd − NEd · zs1 = 135 kNm
Design with respect to General Design Chart Bending with axial force for rectangular cross-sections: μEds =
MEds b · d2 · ƒcd
=
135 · 10−3 0.25 · 0.352 · 11.33
= 0.389
μEds > μEds,m = 0.296
5.4: (NA.5): Linear elastic analysis
→ compression reinforcement required
ξ = height of comression zone / d ≤ 0.45 for C12/ 15 − C50/ 60
from design chart for μEds,m = 0.296 and d2 / d = 0.143 : εs1 = 4.30 · 10−3 ; εs2 = −2.35 · 10−3 ; ζ = z/ d = 0.813
Tab. 9.1 [74]: General Chart for up to C50/ 60 - Section with compression reinforcement
for εs1 = 4.30 · 10−3 → σs1d = 436.8 MP for εs2 = −2.35 · 10−3 → σs1d = −434.9 MP MEds,m = μEds,m · b · d2 · ƒcd = 102.7 kNm ΔMEds = MEds − MEds,m = 135 − 102.7 = 32.3 kNm 1 MEds,m ΔMEds As1 = · + + NEd = 10.73 cm2 σs1d ζ·d d − d2 As2 =
1
·
ΔMEds
|σs2d | d − d2
= 2.47 cm2
Design with respect to ω− (or μs − )Table for rectangular crosssections: μEds =
MEds b · d2 · ƒcd
=
135 · 10−3 0.25 · 0.352 · 11.33
= 0.389
Because the internal force determination is done on the basis of a linear
5.4: (NA.5): Linear elastic analysis ξ = height of comression zone / d ≤ 0.45 for C12/ 15 − C50/ 60
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 National Annex) [72], [73], unless otherwise specified.
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249
DCE-EN2: Design of a Rectangular CS for Bending
Tab. 9.2 [74]: ω−Table for up to C50/ 60 - Rectangular section with compression reinforcement
elastic calculation, then ξm = 0.45 is chosen. Referring to the design table with compression reinforcement and for d2 / d = 0.15: ω1 = 0.4726; As1 =
As2 =
1 ƒyd ƒcd ƒyd
ω1 = 0.1104
· (ω1 · b · d · ƒcd + NEd ) = 10.77 cm2
· (ω2 · b · d) = 2.52 cm2
Design with respect to kd − Design Table for rectangular crosssections: kd = p Tab. 9.3 [74]: kd −Table for up to C50/ 60 - Rectangular section with compression reinforcement
5.4: (NA.5): Linear elastic analysis ξ = height of comression zone / d ≤ 0.45 for C12/ 15 − C50/ 60
d MEds / b
=p
35 135/ 0.25
= 1.51
Not able to read values from kd −table for simply reinforced rectangular cross-sections → compression reinforcement is required Because the internal force determination is done on the basis of a linear elastic calculation, then ξm = 0.45 is chosen. Referring to the kd −table with compression reinforcement: ks1 = 2.740; ks2 = 0.575 (interpolated values for kd = 1.51) ρ1 = 1.021; ρ2 = 1.097 (interpolated values for d2 / d = 0.143 and ks1 = 2.740) As1 = ρ1 · ks1 ·
As2 = ρ2 · ks2 ·
250
MEds d MEds d
+
NEd σs1d
= 10.79 cm2
= 2.43 cm2
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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. Muller, and F. Lobisch. Bemessungshilfsmittel fur ¨ ¨ Betonbauteile nach Eurocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund 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 MEd and the required reinforcement is determined.
hƒ MEd h
d zs As1
As1
d1 b 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
εc
zs As1 εs1 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 σ
ƒtk,c = 525N/ mm2
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε εd = 25 0 /00 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 1992-11:2004 (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
MEd = 425 kNm
B 500A
d = 60.0 cm d1 = 5.0 cm b = 30 cm beƒ ƒ = 258 cm
254
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DCE-EN3: Design of a T-section for Bending
Table 59.1: (continued) Material Properties
Geometric Properties
Loading
hƒ = 18 cm
Table 59.2: Results
As1 [cm2 / m]
SOFiSTiK 2014 | VERiFiCATiON MANUAL
SOF.
ω−Table [74]
kd −Table [74]
15.90
15.74
15.85
255
DCE-EN3: Design of a T-section for Bending
Design Process1
59.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects
3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Concrete: γc = 1.50 Steel: γs = 1.15 ƒck = 20 MP ƒcd = cc · ƒck / γc = 0.85 · 20/ 1.5 = 11.33 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP Design Load: NEd = 0 MEds = MEd − NEd · zs1 = 425 kNm Design with respect to ω− (or μs − )Table for T-beams:
Tab. 9.4 [74]: ω−Table for up to C50/ 60 - T-beam
μEds =
MEds beƒ ƒ · d2 · ƒcd
=
425 · 10−3 2.58 · 0.602 · 11.33
= 0.040
Referring to the design table for T-beams for: μEds = 0.040 and hƒ d
=
0.18 0.60
= 0.30;
beƒ ƒ b
=
2.58 0.30
= 8.6
→ ω1 = 0.039 As1 =
1 ƒyd
· ω1 · beƒ ƒ · d · ƒcd + NEd = 15.74 cm2
Design with respect to kd − Design Table for T-beams: Alternatively, the kd −Tables can be applied, demonstrated that the neutral line lies inside the flange. Tab. 9.3 [74]: kd −Table for up to C50/ 60 - Rectangular section without compression reinforcement
kd = p
d MEds / b
60 425/ 2.58
= 4.67
Referring to the table for kd = 4.67 and after interpolation → ks = 2.351;
kd −Table is applicable since the neutral line lies inside the flange
=p
ξ = 0.060 ;
κs = 0.952
= ξ · d = 0.060 · 60 = 3.6 cm ¡ hƒ = 18 cm MEds NEd As1 = ks · + · κs = 15.85 cm2 d σs1d 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 National 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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. Muller, and F. Lobisch. Bemessungshilfsmittel fur ¨ ¨ Betonbauteile nach Eurocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund 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 MEd and a compressive force NEd and the required reinforcement is determined. b
As,tot / 2
h
d
MEd
d2 d1 = d2
As,tot / 2
NEd
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) εc
σcd
d2 As2
σs2d
d
εs2
Fs2d Fc
zs2 z zs1
As1 d1
εs1
σs1d
Fs1d
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 σ
ƒtk,c
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε
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 respect 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 mentioned 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
MEd = 382 kNm
B 500A
d = 45.0 cm
NEd = −1785 kN
d1 = d2 = 5.0 cm b = 30 cm
Table 60.2: Results
As,tot [cm2 / m]
260
SOF.
Interaction Diagram [74]
35.03
35.19
VERiFiCATiON MANUAL | SOFiSTiK 2014
DCE-EN4: Design of a Rectangular CS for Bending and Axial Force
Design Process1
60.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: γc = 1.50 Steel: γs = 1.15
(NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
ƒck = 30 MP ƒcd = cc · ƒck / γc = 0.85 · 30/ 1.5 = 17.0 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects
3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Design Load: NEd = −1785 kN MEd = 382 kNm MEd 382 ed = 0.428 < 3.5 = = h NEd · h −1785 · 0.50 → 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: μEd =
νEd =
MEd b · h2 · ƒcd NEd b · h2 · ƒcd
=
=
382 · 10−3 0.30 · 0.502 · 17.0 −1785 · 10−3 0.30 · 0.50 · 17.0
= 0.30
= −0.70
Tab. 9.6 [74]: μ − ν Interaction diagram for concrete C12/ 15 − C50/ 60 - Rectangular cross-section with double symmetric reinforcement.
from design chart for d1 / h = 0.05/ 0.5 = 0.10: ωtot = 0.60 As,tot = ωtot · As1 = As2 =
b·h ƒyd / ƒcd
As,tot 2
= 35.19 cm2
= 17.6 cm2
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 National Annex) [72], [73], unless otherwise specified.
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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 accuracy.
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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. Muller, and F. Lobisch. Bemessungshilfsmittel fur ¨ ¨ Betonbauteile nach Eurocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund Bautechnik-Verein E.V. 2011.
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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 MEdy , MEdz and a compressive force NEd . MEdz each As,tot / 4 d1 = d2
NEd h
MEdy
b b1 = b2 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)
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DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force
εc
σcd
Fs2d
d2 As2
d
zs2
εs2
σs2d
Fc z
zs1 As1 εs1
d1
σs1d
Fs1d
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 σ
ƒtk,c
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε
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
MEdy = 500 kNm
B 500A
b1 = b2 = 5.0 cm
MEdz = 450 kNm
d1 = d2 = 5.0 cm
NEd = −1600 kN
b = 40 cm
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Table 61.2: Results
As,tot [cm2 / m]
SOFiSTiK 2014 | VERiFiCATiON MANUAL
SOF.
Interaction Diagram [74]
115.9
113.1
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DCE-EN5: Design of a Rectangular CS for Double Bending and Axial Force
Design Process1
61.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects
3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Concrete: γc = 1.50 Steel: γs = 1.15 ƒck = 30 MP ƒcd = cc · ƒck / γc = 0.85 · 35/ 1.5 = 19.8 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP Design Load: NEd = −1600 kN MEdy = 500 kNm MEdz = 450 kNm
Design with respect to Interaction diagram for Double Bending with axial force for rectangular cross-sections: μEdy =
μEdz =
νEd = Tab. 9.7 [74]: μ − ν Interaction diagram for concrete C12/ 15 − C50/ 60 - Rectangular cross-section with all-round symmetric reinforcement.
MEd b · h2 · ƒcd MEd b · h2 · ƒcd NEd
b · h2 · ƒcd
=
=
=
500 · 10−3 0.40 · 0.502 · 19.8 450 · 10−3 0.40 · 0.502 · 19.8 −1600 · 10−3
0.40 · 0.50 · 19.8
= 0.252
= 0.284
= −0.403
from design chart → ωtot = 1.24 for: • d1 / h = d2 / h = 0.05/ 0.5 = 0.10 • b1 / b = b2 / b = 0.05/ 0.4 = 0.08 ≈ 0.10 • ν = −0.4 • μ1 = m μEdy ; μEdz = 0.284 • μ2 = mn μEdy ; μEdz = 0.252 As,tot = ωtot ·
b·h ƒyd / ƒcd
= 113.1 cm2
As,tot/ 4 = 28.28 cm2
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 National Annex) [72], [73], unless otherwise specified.
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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 reproduced 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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. Muller, and F. Lobisch. Bemessungshilfsmittel fur ¨ ¨ Betonbauteile nach Eurocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund Bautechnik-Verein E.V. 2011.
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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 VEd and the required shear reinforcement is determined. b
As,tot / 2 h
d
z VEd
As,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) Fcd V α d
N
θ
z
V s
V(cot θ − cot α)
M
Ftd
shear reinforcement Figure 62.2: Shear Reinforced Members
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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 σ
ƒtk,c
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε
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 cm2 / m. Also in order to demonstrate that the correct value of VRd,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
VEd = 343.25 kN
B 500A
b = 30 cm d = 45.0 cm As,tot = 38.67 cm2
Table 62.2: Results As,tot [cm2 / m]
270
Design Code
SOF.
Ref.
DIN EN [72]
12.84
12.84
EN [77]
12.18
12.18
VERiFiCATiON MANUAL | SOFiSTiK 2014
DCE-EN6: Design of a Rectangular CS for Shear Force
Design Process1
62.4 Material:
Concrete: γc = 1.50
(NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Steel: γs = 1.15 ƒck = 30 MP ƒcd = cc · ƒck / γc = 0.85 · 30/ 1.5 = 17.0 MP
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects
ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP
3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Design Load: VEd = 343.25 kN Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 z = mx d − cV, − 30 mm; d − 2 cV,
(NDP) 6.2.3 (1): Inner lever arm z
z = mx {384; 378} = 384 mm 1.0 ≤ cot θ ≤ VRd,cc =
1.2 + 1.4 σcd / ƒcd 1 − VRd,cc / VEd
1/ 3 c · 0.48 · ƒck
· 1 − 1.2
≤ 3.0 σcd ƒcd
(NDP) 6.2.3 (2): Eq. 6.7aDE
· b · z
(NDP) Eq. 6.2.3 (2): 6.7bDE c = 0.5
VRd,cc = 0.5 · 0.48 · 30 1/ 3 · (1 − 0) · 0.3 · 0.384
(NDP) 6.2.3 (2): σcd = NEd / Ac
VRd,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
As,req / s = VEd / (ƒyd · z · cot θ) = 12.84 cm2 / m VRd,m = b · z · ν1 · ƒcd / (cot θ + tn θ) VRd,m = 0.3 · 0.384 · 0.75 · 17 / (1 + 1) = 734.4 kN
Design with respect to EN 1992-1-1:2004 [77]:
6.2.3 (3): Eq. 6.8 ƒyd = ƒyk / γs = 435 MP
(NDP) 6.2.3 (3): Eq. 6.9 Maximum shear force VRd,m occurs for θ = 45◦ : cot θ = tn θ = 1 ν1 = 0.75 · ν2 = 0.75, ν2 = 1 for ≤ C50/ 60
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) As,req / s = VEd / (ƒyd · z · cot θ) = 12.18 cm2 / m
6.2.3 (2): Eq. 6.7N The angle θ should be limited by Eq. 6.7N
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 National Annex) [72], [73], unless otherwise specified. 3 The sections mentioned in the margins refer to EN 1992-1-1:2004 [77], unless otherwise specified.
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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 fur ¨ 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
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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 VEd and the required reinforcement is determined. beƒ ƒ
hƒ VEd h
d zs As1
As1
d1 b 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) Fcd V α d
N
θ
z
V s
V(cot θ − cot α)
M
Ftd
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 σ
ƒtk,c
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε
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 1992-11:2004 [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
VEd = 450 kN
B 500A
d = 53.0 cm d1 = 7.0 cm b = 30 cm beƒ ƒ = 180 cm hƒ = 15 cm As1 = 15 cm2
The intermediate steps of calculating the required reinforcement are also validated in this example. First we calculate the design value for the shear resistance VRd,c for members not requiring shear reinforcement. It gives a value of VRd,c = 93.76 kN. Checking the results in AQB, we can see that SOFiSTiK outputs also Vrd1,c = 93.76. Just to test this result, if we input a shear force of VEd = 93.75 just below the value for VRd,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 VEd = 93.77 just larger than VRd,c , then AQB will start increasing cot θ and the minimum reinforcement will be printed. If we continue increasing VEd , 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 VRd,c = 93.76, would be to keep cot θ = 1.0
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DCE-EN7: Design of a T-section for Shear
and now with VEd = 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 VEd can be calculated as 642.23 kN. This can be found in AQB results as the Vrd2,c = 642.23 for the case of cot θ = 1.0. Giving as an input a shear force just above this value VEd = 642.24 triggers a warning ”Shear design not possible”. Next step is the validation of VRd,m . When the design shear force VEd exceeds VRd,m then cot θ must be decreased so that VEd = VRd,m . The reference result for VRd,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 VEd = 442.91 and 442.93 kN, respectively. Also the minimum reinforcement is calculated exactly by AQB with a value of 2.63 cm2 / m. Table 63.2: Results SOF.
Ref.
As,req / s [cm2 / m]
8.90
8.90
As,mn / s [cm2 / m]
2.63
2.63
93.76
93.76
VRd,m [kN]
442.92
442.92
VEd,m [kN]
642.23
642.23
VRd,c [kN]
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DCE-EN7: Design of a T-section for Shear
Design Process1
63.4 Material: (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects 3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Concrete: γc = 1.50 Steel: γs = 1.15 ƒck = 30 MP ƒcd = cc · ƒck / γc = 0.85 · 30/ 1.5 = 17.0 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP Design Load: VEd = 450.0 kN Design with respect to EN 1992-1-1:2004 [77]:2
6.2.3 (1): Inner lever arm z 6.2.2 (1): Design value for shear resistance VRd,c for members not requiring design shear reinforcement
z = 0.9 · d = 0.9 · 450 = 405 mm VRd,c = CRd,c · k · (100 · ρ1 · ƒck ) 1/ 3 + k1 · σcp · b · d CRd,c = 0.18/ γc = 0.12 v t 200 = 1.6143 < 2.0 k =1+ d ρ1 =
As
= 0.0094 < 0.02 b d VRd,c = 0.12 · 1.6143 · (100 · 0.0094 · 30)
1/ 3
+ 0 · 0.3 · 0.53
VRd,c = 93.76 kN VEd > VRd,c → shear reinforcement is required 6.2.3 (2): Eq. 6.7N: The angle θ should be limited by Eq. 6.7N 6.2.3 (3): Eq. 6.9
6.2.2 (6): Eq. 6.6N
1.0 ≤ cot θ ≤ 2.5 → start with cot θ = 2.50 VRd,m = b · z · ν · ƒcd / (cot θ + tn θ) ƒck = 0.528 ν = 0.6 · 1 − 2500 VRd,m = 0.3 · 0.477 · 0.528 · 17 / (2.5 + 0.4) = 442.92 kN VEd > VRd,m : increase θ so that VEd = VRd,m ⇒ cot θ = 2.44 As,req / s = VEd / (ƒyd · z · cot θ) = 8.90 cm2 / m ρ,mn = 0.08 ·
9.2.2 (5): Eq. 9.4
As,mn / s = ρ,mn · b · sin α = 2.63 cm2 / m
6.2.2 (6): Eq. 6.5
VEd ≤ 0.5 · b · d · ν · ƒcd = 642.23 kN
p
ƒck / ƒyk = 0.08 ·
p
9.2.2 (5): Eq. 9.5N
30 / 500 = 0.0008764
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 otherwise specified.
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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
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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 VEd and a compressive force NEd and and the required reinforcement is determined. b
As,tot / 2 h
d
z NEd VEd
As,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) Fcd V α d
N
θ
z
V s
V(cot θ − cot α)
M
Ftd
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
σ ƒyk ƒyd = ƒyk / γs
A Idealised
B
B Design
ε
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
VEd = 343.25 kN
B 500A
b = 30 cm
NEd = 500.0 kN
d = 45.0 cm As,tot = 38.67 cm2 cV, = 3.6 cm
Table 64.2: Results
As / s [cm2 / m] VRd,c [kN] cot θ
280
SOF.
Ref.
11.27
11.27
132.71
132.71
1.82
1.82
VERiFiCATiON MANUAL | SOFiSTiK 2014
DCE-EN8: Design of a Rectangular CS for Shear and Axial Force
Design Process1
64.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: γc = 1.50
(NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Steel: γs = 1.15 ƒck = 30 MP ƒcd = cc · ƒck / γc = 0.85 · 30/ 1.5 = 17.0 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects 3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Design Load: VEd = 343.25 kN NEd = −500.0 kN z = mx d − cV, − 30 mm; d − 2 cV,
(NDP) 6.2.3 (1): Inner lever arm z
z = mx {384; 378} = 384 mm VRd,c = CRd,c · k · (100 · ρ1 · ƒck )
1/ 3
+ 0.12 · σcp · b · d
with a minimum of νmn + 0.12 · σcp · b · d
(NDP) 6.2.2 (1): Eq. 6.2b
CRd,c = 0.15/ γc = 0.1 v v t 200 t 200 =1+ = 1.6667 < 2.0 k =1+ d 450 ρ1 =
As,tot / 2 b d
(NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance VRd,c for members not requiring design shear reinforcement
(NDP) 6.2.2 (1): CRd,c = 0.15/ γc
= 0.01432 < 0.02
VRd,c,mn = νmn + 0.12 · σcp · b · d νmn = (0.0525/ γc ) · k 3/ 2 · ƒck
1/ 2
= 0.41249
(NDP) 6.2.2 (1): Eq. 6.3aDE: νmn for d ≤ 600 mm
VRd,c,mn = 109.68 kN σcp = NEd / Ac < 0.2 · ƒcd
(NDP) 6.2.2 (1): Eq. 6.2 σcp > 0 for compression
σcp = −500 · 10−3 / 0.15 · 106 = −3.3333 N/ mm2 < 3.4 VRd,c = 0.1 · 1.6667 · (1.432 · 30) 1/ 3 + 0.112 · 3.3333 · 0.3 · 0.45 = 132.71 kN VEd > VRd,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 National Annex) [72], [73], unless otherwise specified.
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(NDP) 6.2.3 (2): Eq. 6.7aDE
(NDP) 6.2.3 (2): Eq. 6.7bDE c = 0.5 (NDP) 6.2.3 (2): σcd = NEd / Ac
6.7DE: σcd > 0 for compression
1.0 ≤ cot θ ≤
VRd,cc =
1.2 + 1.4 σcd / ƒcd 1 − VRd,cc / VEd
1/ 3 c · 0.48 · ƒck
· 1 − 1.2
≤ 3.0 σcd ƒcd
· b · z
σcp = NEd / Ac σcd = −500 · 10−3 / 0.15 · 106 = −3.3333 N/ mm2 3.3333 1/ 3 VRd,cc = 0.5 · 0.48 · 30 · 1 − 1.2 · 0.3 · 0.384 17.0 VRd,cc = 65.6948 kN
(NDP) 6.2.3 (2): The angle θ should be limited by Eq. 6.7DE
6.2.3 (3): Eq. 6.8 ƒyd = ƒyk / γs = 435 MP (NDP) 6.2.3 (3): Eq. 6.9 Maximum shear force VRd,m occurs for θ = 45◦ : cot θ = tn θ = 1 (NDP) ν1 = 0.75 · ν2 = 0.75, ν2 = 1 for ≤ C50/ 60
282
cot θ =
1.2 + 1.4 · 3.3333 / 17.0 1 − 65.6948 / 343.25
= 1.823
As,req / s = VEd / (ƒyd · z · cot θ) = 11.27 cm2 / m VRd,m = b · z · ν1 · ƒcd / (cot θ + tn θ) VRd,m = 0.3 · 0.384 · 0.75 · 17 / (1 + 1) = 734.4 kN
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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012.
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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 VEd and torsion TEd and the required shear and torsion reinforcement is determined. b As,tot / 2
h
d TEd VEd
As,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)
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DCE-EN9: Design of a Rectangular CS for Shear and Torsion
z
Ak centre-line
cover TEd teƒ / 2 V
teƒ 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 σ
ƒtk,c
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε
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
VEd = 175.0 kN
B 500A
b = 30 cm
TEd = 35.0 kNm
d = 65.0 cm As,tot = 26.8 cm2
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Table 65.2: Results SOF.
Ref.
As / s (T) [cm2 / m]
3.35
3.35
As (T) [cm2 ]
5.37
5.37
13.60
13.59
1303.04
1303.03
124.95
124.95
As,tot / s [cm2 / m] VRd,m [kN] TRd,m [kNm]
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Design Process1
65.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: (NA) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Tab. 3.1: Strength for concrete (NDP) 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects (NDP) 3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Concrete: γc = 1.50 Steel: γs = 1.15 ƒck = 30 MP ƒcd = cc · ƒck / γc = 0.85 · 35/ 1.5 = 19.833 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP Design Load: VEd = 175.0 kN, TEd = 35.0 kN
(NDP) 6.3.2 (5): Eq. (NA.6.31.1): For approximately rectangular solid sections no shear and torsion reinforcement 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 (NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance VRd,c for members not requiring design shear reinforcement (NDP) 6.2.2 (1): CRd,c = 0.15/ γc
TEd ≤ 35 >
VEd · b 4.5 135 · 0.3 4.5
= 9.0
→ Eq. NA.6.31.1 is not fulfilled 4.5 · TEd ≤ VRd,c VEd · 1 + VEd · b VRd,c = CRd,c · k · (100 · ρ1 · ƒck )
1/ 3
+ 0.12 · σcp · b · d
CRd,c = 0.15/ γc = 0.1 v t 200 = 1.5773 < 2.0 k =1+ d ρ1 =
As
= 0.01276 < 0.02 b d VRd,c = 0.1 · 1.5773 · (100 · 0.01276 · 35)
1/ 3
+ 0 · 0.3 · 0.65
VRd,c = 109.13 kN 4.5 · 35 = 700 > 109.13 175 · 1 + 135 · 0.3 → requirement of Eq. NA.6.31.2 is not met Torsional reinforcement 6.3.1 (3): Solid sections may be modelled by equivalent thin-walled sections (Fig. 65.2)
teƒ ƒ ,1 = teƒ ƒ ,2 = 2 · 50 = 100 mm (so = s = ss = 50 mm) Ak = (h − s − so ) · b − teƒ ƒ ,1 = 100 mm
Ak : 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 National Annex) [72], [73], unless otherwise specified.
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Ak = (700 − 50 − 50) · (300 − 100) = 120000 mm2 = 0.12 m2 k = 2 · [(700 − 50 − 50) + (300 − 100)] = 1600 mm = 1.6 m
k : circumference of area Ak
Simplifying, the reinforcement for torsion may be determined alone under the assumption of cot θ = 1.0, θ = 45◦ and be added to the independently calculated shear force reinforcement.
6.3.2 (2): The effects of torsion and shear may be superimposed, assuming the same value for θ
As,req / s = TEd · tn θ / (ƒyd · 2Ak )
(NDP) 6.3.2 (3): Eq. (NA.6.28.1)
As,req / s = 350 · 1.0 / (435 · 2 · 0.12) = 3.35 cm2 / m (T) As,req = TEd · k · cot θ / (ƒyd · 2Ak ) = 5.37 cm2 (T)
6.3.2 (3): Eq. 6.28
Torsional resistance moment TRd,m = 2 · ν · ƒcd · Ak · teƒ ƒ , · sin θ · cos θ sin θ · cos θ = 0.5 since θ = 45◦
6.3.2 (4): Eq. 6.30 (NDP): ν = 0.525
TRd,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: 2 2 VEd TEd + ≤1 TRd,m VRd,m z = mx d − cV, − 30 mm; d − 2 cV, cV, = so − Do / 2 = 50 − 28/ 2 = 36 mm
(NDP) 6.3.2 (4): Eq. (NA.6.29.1) for solid cross-sections (NDP) 6.2.3 (1): Inner lever arm z so : offset of reinforcement Do : bar diameter
z = mx {584; 578} = 584 mm VRd,m = b · z · ν1 · ƒcd / (cot θ + tn θ) VRd,m = 0.3 · 0.584 · 0.75 · 19.833 / (1 + 1) = 1303.03 kN
35 124.95
2
+
175
2
1303.04
(NDP) 6.2.3 (3): Eq. 6.9 Maximum shear force VRd,m occurs for θ = 45◦ : cot θ = tn θ = 1 (NDP) 6.2.3 (3):ν1 = 0.75 · ν2 = 0.75, ν2 = 1 for ≤ C50/ 60
= 0.0965 < 1
Shear reinforcement As,req / s = VEd / (ƒyd · z · cot θ) = 6.89 cm2 / m
6.2.3 (3): Eq. 6.8 ƒyd = ƒyk / γs = 435 MP
Total required reinforcement Required torsional reinforcement: 2 · As / s = 2 · 3.35 = 6.7 cm2 / m (double-shear connection) Total reinforcement: As,tot / s = 6.7 + 6.89 = 13.59 cm2 / m
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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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund Bautechnik-Verein E.V. 2011.
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DCE-EN10: Shear between web and flanges of T-sections
66 DCE-EN10: Shear between web and flanges of Tsections 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. beƒ ƒ hƒ VEd h
d
As1
As1
d1 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)
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DCE-EN10: Shear between web and flanges of T-sections
Fd Fd
beƒ ƒ sƒ
θƒ
Δ
hƒ
longitudinal bar anchored beyond this point
Fd + ΔFd Asƒ
compressive struts
Fd + ΔFd
b
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 σ
ƒtk,c = 525N/ mm2
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε εd = 25 0 /00 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.
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250
135
29
39.4
106
95.6
104
104 1
1
1
1
40
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
Pg = 500 kN/ m
B 500A
hƒ = 29 cm, h = 106 cm d1 = 7.0 cm b = 40 cm beƒ ƒ , = 104 cm, beƒ 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.
500.0
500.0
1000.0
1500.0
2000.0
2500.0
-2500.0
-2000.0
-1500.0
-1000.0
-500.0
2250.0
4000.0
5250.0
6000.0
6250.0
6000.0
5250.0
4000.0
2250.0
Figure 66.5: Loaded Structure, Resulting Moment and Shear Force
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Table 66.2: Results At beam 1001
SOF.
Ref.
As1 [cm2 ]
39.44
39.44
Asƒ / sƒ [cm2 / m]
10.46
10.45
VRd,c [kN] at = 0.0 m
73.41
73.41
VRd,c [kN] at = 1.0 m
154.37
154.51
73.41
73.41
1366.65
1367.16
cot θ
1.65
1.65
z [m] at = 1.0 m
1.25
1.25
748.99
748.80
VRd,c,mn [kN] VRd,m / s [kN]
VEd [kN]
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DCE-EN10: Shear between web and flanges of T-sections
Design Process1
66.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: γc = 1.50
(NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Steel: γs = 1.15 ƒck = 25 MP ƒcd = cc · ƒck / γc = 0.85 · 25/ 1.5 = 14.17 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects 3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
σsd = 456.52 MP Design Load for beam 1001: MEds = 2250 kNm μEds =
MEds beƒ ƒ · d2 · ƒcd
=
2250 · 10−3 2.5 · 1.282 · 14.17
= 0.03876
ω = 0.03971 and ξ = 0.9766 (interpolated) As1 =
1 σsd
· (ω · b · d · ƒcd + NEd ) = 39.44 cm2
Tab. 9.2 [74]: ω−Table for up to C50/ 60 without compression reinforcement NEd = 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: VEd = ΔFd =
ΔMEds z
·
beƒ ƒ , beƒ ƒ
=
6.2.4 (3): Eq. 6.20
2250 1.04 · = 748.8 kN 1.25 2.5
VRd,c = CRd,c · k · (100 · ρ1 · ƒck )
1/ 3
+ 0.12 · σcp · b · d
with a minimum of νmn + 0.12 · σcp · b · d CRd,c = 0.15/ γc = 0.1
(NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance VRd,c for members not requiring design shear reinforcement (NDP) 6.2.2 (1): Eq. 6.2b (NDP) 6.2.2 (1): CRd,c = 0.15/ γc
for beam 1001 at = 0.0 m: v v t 200 t 200 =1+ = 1.4057 < 2.0 k =1+ d 1215 ρ1 =
As 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 National Annex) [72], [73], unless otherwise specified.
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VRd,c = 0 VRd,c,mn = νmn + 0.12 · σcp · b · d (NDP) 6.2.2 (1): Eq. 6.3bDE
νmn = (0.0375/ γc ) · k 3/ 2 · ƒck
1/ 2
= 0.20833
VRd,c,mn = 73.41 kN → VRd,c = 73.41 kN VEd > VRd,c → shear reinforcement is required for beam 1001 at = 1.0 m: v v t 200 t 200 k =1+ =1+ = 1.3953 < 2.0 d 1280 As
=
39.44
= 0.01062 < 0.02 b d 29 · 128 VRd,c = 0.1 · 1.3953 · (100 · 0.01062 · 25)
ρ1 =
1/ 3
+ 0 · 0.29 · 1280
VRd,c = 154.51 kN VEd > VRd,c → shear reinforcement is required
(NDP) 6.2.3 (2): Eq. 6.7aDE
(NDP) Eq. 6.2.3 (2): 6.7bDE c = 0.5 (NDP) 6.2.4 (4) (NDP) 6.2.3 (2): σcd = NEd / Ac NEd = 0
1.0 ≤ cot θ ≤ VRd,cc =
1.2 + 1.4 σcd / ƒcd 1 − VRd,cc / VEd
1/ 3 c · 0.48 · ƒck
· 1 − 1.2
≤ 3.0 σcd ƒcd
· b · z
b = hƒ = 0.29 m and z = Δ = 1 m VRd,cc = 0.5 · 0.48 · 25
1/ 3
· (1 − 0) · 0.29 · 1.0
VRd,cc = 203.5kN cot θ = (NDP) 6.2.3 (2): The angle θ should be limited by Eq. 6.7DE
6.2.3 (3): Eq. 6.8 ƒyd = ƒyk / γs = 435 MP
1.2 + 0 1 − 203.5 / 748.8
= 1.647
cot θƒ = 1.647 Asƒ ,req / sƒ = VEd / (ƒyd · z · cot θ) Asƒ ,req / sƒ = 748.8 · 10−3 / (435 · 1.0 · 1.647) Asƒ ,req / sƒ = 10.45 cm2 / m
(NDP) 6.2.4 (4): Eq. 6.22 Maximum shear force VRd,m (NDP) 6.2.3 (3): ν = ν1 = 0.75 for ≤ C50/ 60 sin θƒ · cos θƒ = 0.4436
296
VEd ≤ ν · ƒcd · sin θƒ · cos θƒ · hƒ · Δ VRd,m = 0.75 · 14.17 · 0.4436 · 0.29 · 1.0 VRd,m = 1367.16 kN
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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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. Muller, and F. Lobisch. Bemessungshilfsmittel fur ¨ ¨ Betonbauteile nach Eurocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund Bautechnik-Verein E.V. 2011.
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DCE-EN11: Shear at the interface between concrete cast
67 cast
DCE-EN11: Shear at the interface between concrete
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. beƒ ƒ hƒ
VEd
h
d Shear section
zs As1
As1
d1 b 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
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DCE-EN11: Shear at the interface between concrete cast
anchorage
new concrete
NEd
b
h2 ≤ 10d 45◦ ≤ α ≤ 90◦ α ≤ 30◦
VEd
b
h1 ≤ 10d old concrete anchorage 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 σ
ƒtk,c
ƒyk ƒyd = ƒyk / γs
B
A
Idealised
B
Design
ε
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
Vz = 800 kN
B 500A
hƒ = 29 cm
My = 2250 kNm
d1 = 7.0 cm b = 40 cm , beƒ ƒ = 250 cm As1 = 1.0 cm2 zs = 95.56 cm
300
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DCE-EN11: Shear at the interface between concrete cast
Table 67.2: Results s [cm2 / m]
SOF.
Ref.
state
7.06
7.07
state only V
4.91
4.90
state V + M
4.99
4.99
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Design Process 1
67.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: (NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
Tab. 3.1: Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects 3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Concrete: γc = 1.50 Steel: γs = 1.15 ƒck = 25 MP ƒcd = cc · ƒck / γc = 0.85 · 25/ 1.5 = 14.17 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP σsd = 456.52 MP τ=
T b
=
V·S y · b
where S is the static moment of the separated area S = h · b · (zs − h / 2) = 0.18058 m3 τ=
0.8 · 0.18058 0.16667 · 0.4
T = The shear section with a length of 0.4 m is split into two equal parts with b = 0.2 m
= 2.1669 MP
0.8 · 0.18058 0.16667
= 0.86676 MN/ m = 866.76 kN/ m
T = 866.76 / 2 = 433.38 kN/ m State :
The associated design shear flow VEd is:
Design Load: VEd = T = 433.38 kN/ m
(NDP) 6.2.2 (1): Eq. 6.2a: Design value for shear resistance VRd,c for members not requiring design shear reinforcement
Ed = τ = 2.1669 MP VRd,c = CRd,c · k · (100 · ρ1 · ƒck ) Rd,c = CRd,c · k · (100 · ρ1 · ƒck ) ρ1 =
As b d
1/ 3
1/ 3
+ 0.12 · σcp · b · d
+ 0.12 · σcp
= 0.0 → Rd,c = 0.0
with a minimum of (NDP) 6.2.2 (1): Eq. 6.2b
VRd,c,mn = νmn + 0.12 · σcp · b · d Rd,c,mn = νmn + 0.12 · σcp
(NDP) 6.2.2 (1): Eq. 6.3bDE
νmn = (0.0375/ γc ) · k 3/ 2 · ƒck
1/ 2
= 0.20833 MP
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 National Annex) [72], [73], unless otherwise specified.
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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 · ƒctd + μ · σn + ρ · ƒyd · (1.2 · μ · sin α + cos α) and Rd ≤ 0.5 · ν · ƒcd Rd,m = 0.5 · ν · ƒcd = 4.9585 MP c = 0.50 and μ = 0.9 for indented surface ƒctd = αct · ƒctk;0.05 / γc
Rd = 0.51 +
s 0.2
s 0.2 · 1.0
Maximum shear stress Rd,m (NDP) 6.2.5 (1): ν = 0.70 for indented surface
6.2.5 (2): c, μ: factors depending on the roughness of the interface (NDP) 3.1.6 (2)P: Eq. 3.16 αct = 0.85 3.1.2 (3): Tab. 3.1 - Strength for concrete: ƒctk;0.05 = 1.8 MP:
ƒctd = 0.85 · 1.80 / 1.5 = 1.02 Rd = 0.5 · 1.02 + 0 +
(NDP) 6.2.5 (1): Eq. 6.25
· 435 · (1.2 · 0.9 · 1 + 0)
s : area of reinforcement crossb · ing interface / area of joint ρ=
· 469.56 = 2.1669
s = 7.07 cm2 / 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 VEd is: VEd = 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 · ƒctd + μ · σn + ρ · ƒyd · (1.2 · μ · sin α + cos α) Rd = 0.5 · 1.02 + 0 + Rd = 0.51 +
s 0.2
s 0.2 · 1.0
· 435 · (1.2 · 0.9 · 1 + 0)
· 469.56 = 1.66
s = 4.90 cm2 / m
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State shear force V and moment M: MEds = 2250 kNm μEds =
Tab. 9.2 [74]: ω−Table for up to C50/ 60 without compression reinforcement NEd = 0
(NDP) 6.2.3 (1): Inner lever arm z
MEds beƒ ƒ · d2 · ƒcd
2250 · 10−3
=
2.5 · 1.282 · 14.17
= 0.03876
ω = 0.03971 and ξ = 0.9766 (interpolated) 1
· (ω · b · d · ƒcd + NEd ) = 39.44 cm2 σsd z = mx d − cV, − 30 mm; d − 2 cV,
As1 =
z = mx {1160; 1190} = 1190 mm Design Load: The shear section with a length of 0.4 m is split into two equal parts with b = 0.2 m
T = V / z = 800 / 1.19 = 672.268 kN/ m T = 672.268 / 2 = 336.134 kN/ m VEd = 336.134 kN/ m and Ed = 336.134 / 0.2 = 1.68 MP Rd,c = CRd,c · k · (100 · ρ1 · ƒck )
(NDP) 6.2.2 (1): CRd,c = 0.15/ γc
1/ 3
+ 0.12 · σcp
CRd,c = 0.15/ γc = 0.1 v v t 200 t 200 =1+ = 1.3953 < 2.0 k =1+ d 1280 ρ1 =
As 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 · ƒctd + μ · σn + ρ · ƒyd · (1.2 · μ · sin α + cos α) Rd = 0.5 · 1.02 + 0 + Rd = 0.51 +
s 0.2
s 0.2 · 1.0
· 435 · (1.2 · 0.9 · 1 + 0)
· 469.56 = 1.68
s = 4.99 cm2 / m
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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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [74] K. Holschemacher, T. Muller, and F. Lobisch. Bemessungshilfsmittel fur ¨ ¨ Betonbauteile nach Eurocode 2 Bauingenieure. 3rd. Ernst & Sohn, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund Bautechnik-Verein E.V. 2011.
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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. Different loading conditions are examined, always consisting of a bending moment MEd , and in addition with or without a compressive or tensile axial force NEd . The crack width is determined. b
As0 h
d
MEd
z
NEd As
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 ϕ
h− B
c h
ε2 = 0
d A
C D
E
hc,eƒ B
ε1
5(c + ϕ/ 2)
Figure 68.2: Stress and Strain Distributions in the Design of Cross-sections
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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
NEd = 0 or ±300 kN
B 500A
d = 96.0 cm
MEd = 562.5 kNm
b = 30.0 cm ϕs = 25.0 mm, As = 24.50 cm2 ϕs0 = 12.0 mm, As0 = 2.26 cm2 k = 0.3 mm
Table 68.2: Results
Case
Load
As given [cm2 ]
Result
SOF.
Ref.
MEd , NEd = 0
24.50
As,req [cm2 ]
6.93
6.93
σs [MP]
207.85
207.85
MEd , NEd = 300
24.50
As,req [cm2 ]
10.39
10.39
MEd , NEd = −300
24.50
As,req [cm2 ]
4.04
4.04
V
MEd , NEd = 0
24.50
As [cm2 ]
passed with given reinforcement
σs [MP] V
MEd , NEd = 0
12.0
As [cm2 ]
440.57
440.53
not passed with given reinforcement
V
308
MEd , NEd = 0
→ new: 14.54
σs [MP]
436.30
436.28
24.50
k [mm]
0.13
0.13
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DCE-EN12: Calculation of crack widths
Design Process 1
68.4
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: γc = 1.50 Steel: γs = 1.15
(NDP) 2.4.2.4: (1), Tab. 2.1DE: Partial factors for materials
ƒck = 25 MP ƒcd = cc · ƒck / γc = 0.85 · 25/ 1.5 = 14.17 MP ƒyk = 500 MP ƒyd = ƒyk / γs = 500/ 1.15 = 434.78 MP
Tab. 3.1 : Strength for concrete 3.1.6: (1)P, Eq. (3.15): cc = 0.85 considering long term effects
3.2.2: (3)P: yield strength ƒyk = 500 MP 3.2.7: (2), Fig. 3.8
Design Load: MEd = 562.5 kNm NEd = 0.0 or ±300 kN
Minimum reinforcement areas
7.3.2: If crack control is required, a minimum amount of bonded reinforcement is required to control cracking in areas where tension is expected
• Case : MEd , NEd = 0.0 As,mn · σs = kc · k · ƒct,eƒ ƒ · Act
7.3.2 (2): Eq. 7.1: As,mn minimum area of reinforcing steel within tensile zone
ƒct,eƒ ƒ = ƒctm ≥ 3.0 MP ƒct,eƒ ƒ = 2.6 < 3.0 MP → ƒct,eƒ ƒ = 3.0 MP Act = b · heƒ ƒ = 0.3 · 0.5 = 0.15 m2 fct,eff
(NDP) 7.3.2 (2): ƒct,eƒ ƒ mean value of concrete tensile strength When the cracking time can’t be placed with certainty in the first 28 days then ƒct,eƒ ƒ ≥ 3.0 MP for normal concrete Tab. 3.1: ƒctm = 2.6 MP for C 25 7.3.2 (2): Act is the area of concrete within tensile zone, for pure bending of a rectangular section its height is h/ 2
heƒ ƒ = h/ 2
kc = 0.4 · 1 −
σc
7.3.2 (2): Eq. 7.2: kc for bending of rectangular sections
k1 · (h/ h∗ ) · ƒct,eƒ ƒ
≤1
σc = NEd / (b · h) = 0.0 → kc = 0.4 k = 0.8 for h ≤ 300 ϕs = ϕ∗ · ƒct,eƒ ƒ / 2.9 N/ mm2 s ϕ∗ = 25 · 2.9 / 3.0 = 24.16667 s σs =
q
k · 3.48 · 106 / ϕ∗ = 207.846 MP s
As,req = 0.4 · 0.8 · 3.0 · 0.15 · 104 / 207.846 = 6.928cm2
7.3.2 (2): Eq. 7.4: σc mean stress of the concrete 7.3.2 (2): h is the lesser value of b, h h = min {300, 1000} = 300 mm 7.3.2 (NA.5): Eq. NA.7.5.2: For determination of the reinforcement stress the maximum bar diameter has to be modified
(NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width
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 National Annex) [72], [73], unless otherwise specified.
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• Case : NEd = 300 kN 7.3.2 (2): Eq. 7.1: As,mn minimum area of reinforcing steel within tensile zone
As,mn · σs = kc · k · ƒct,eƒ ƒ · Act ƒct,eƒ ƒ = ƒctm ≥ 3.0 MP
Tab. 3.1: ƒctm = 2.6 MP for C 25 7.3.2 (2): Act is the area of concrete within tensile zone 7.3.2 (2): Eq. 7.4: σc mean stress of the concrete, tensile stress σc < 0 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 reinforcement stress the maximum bar diameter has to be modified (NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width 7.3.2 (2): Eq. 7.2: kc for bending with axial force of rectangular sections h∗ = 1.0 m for h ≥ 1.0 m k1 = 2 · h∗ / (3 · h) if NEd tensile force 7.3.2 (2): Eq. 7.1: Tensile stress σc < 0
ƒct,eƒ ƒ = 2.6 < 3.0 MP → ƒct,eƒ ƒ = 3.0 MP Act = b · heƒ ƒ = 0.15 m2 σc =
NEd b·h
=
300 · 103 300 · 1000
= 1.0 MP
k = 0.8 for h ≤ 300 ϕs = ϕ∗ · ƒct,eƒ ƒ / 2.9 N/ mm2 s ϕ∗ = 25 · 2.9 / 3.0 = 24.16667 s σs =
q
k · 3.48 · 106 / ϕ∗ = 207.846 MP s
kc = 0.4 · 1 −
σc k1 · (h/ h∗ ) · ƒct,eƒ ƒ
≤1
k1 = 2 · h∗ / (3 · h) = 2/ 3 1.0 kc = 0.4 · 1 + = 0.6 ≤ 1 (2/ 3) · 1 · 3.0 As,req = 0.6 · 0.8 · 3.0 · 0.15 · 104 / 207.846 = 10.39cm2 • Case : NEd = −300 kN
7.3.2 (2): Eq. 7.1: As,mn minimum area of reinforcing steel within tensile zone
As,mn · σs = kc · k · ƒct,eƒ ƒ · Act ƒct,eƒ ƒ = ƒctm ≥ 3.0 MP
Tab. 3.1: ƒctm = 2.6 MP for C 25 7.3.2 (2): Act is the area of concrete within tensile zone
ƒct,eƒ ƒ = 2.6 < 3.0 MP → ƒct,eƒ ƒ = 3.0 MP Act = b · heƒ ƒ The height of the tensile zone is determined through the stresses:
7.3.2 (2): Eq. 7.4: σc mean stress of the concrete, compressive stress σc > 0
σc =
NEd b·h
=
300 · 103 300 · 1000
= 1.0 MP
σ = ƒct,eƒ ƒ = 3.0 MP ⇒ heƒ ƒ =
3.0 · 50 3.0 + 1.0
= 37.5cm
Act = 0.3 · 0.375 = 0.1125 m2 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 reinforcement stress the maximum bar diameter has to be modified
310
k = 0.8 for h ≤ 300 ϕs = ϕ∗ · ƒct,eƒ ƒ / 2.9 N/ mm2 s ϕ∗ = 25 · 2.9 / 3.0 = 24.16667 s
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DCE-EN12: Calculation of crack widths
σs =
q
k · 3.48 · 106 / ϕ∗ = 207.846 MP s
kc = 0.4 · 1 −
(NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width
σc k1 · (h/ h∗ ) · ƒct,eƒ ƒ
≤1
7.3.2 (2): Eq. 7.2: kc for bending with axial force of rectangular sections
k1 = 1.5
kc = 0.4 · 1 −
1.0 1.5 · 1 · 3.0
h∗ = 1.0 m for h ≥ 1.0 m k1 = 1.5 if NEd compressive force
= 0.3111 ≤ 1
As,req = 0.3111 · 0.8 · 3.0 · 0.1125 · 104 / 207.846 = 4.04 cm2
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 (NDP) 7.3.2 (2): ƒct,eƒ ƒ mean value of concrete tensile strength
• Case V: NEd = 0.0 ƒct,eƒ ƒ = ƒctm ≥ 3.0 MP ƒct,eƒ ƒ = 2.6 < 3.0 MP → ƒct,eƒ ƒ = 3.0 MP ϕs = ϕ∗ · s
σs · As 4(h − d) · b · 2.9
· ϕs = 25 mm = ϕ∗ s
≥ ϕ∗ · s
Tab. 3.1: ƒctm = 2.6 MP for C 25 (NDP) 7.3.3: Eq. 7.7.1DE: The maximum bar diameters should be modified for load action
ƒct,eƒ ƒ 2.9
264.06 · 24.50
= ϕ∗ · 4.6476 s
4(100 − 96) · 30 · 2.9
As = 24.5 cm2 prescribed reinforcement
→ ϕ∗ = 5.3791 mm s σs =
q
k · 3.48 · 106 / ϕ∗ = 440.53 MP s
(NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width
σs = 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 σs are equal, as it can be seen in Case V below. and ϕs = 25 mm > ϕ∗ · s
ƒct,eƒ ƒ 2.9
= 5.3791 ·
3.0 2.9
= 5.5646
also control the steel stress with respect to the calculated strains εs = 0.440 + 1.913 · (0.50 − 0.04) = 1.31998 → σs = 0.00131998 · 200000 = 264.0 MP or from Tab. 7.2DE and for σs = 264.04 ≈ 260.0 MP → ϕ∗ = 15 mm → ϕs = ϕ∗ · s s ϕs = 15 ·
σs · As 4(h − d) · b · 2.9
264.06 · 24.50 4(100 − 96) · 30 · 2.9
≥ ϕ∗ · s
(NDP) 7.3.3 Tab. 7.2DE (a)
ƒct,eƒ ƒ 2.9
(NDP) 7.3.3: Eq. 7.7.1DE: The maximum bar diameters should be modified for load action
= 69.69 mm > 25 mm
→ crack width control is passed with given reinforcement.
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• Case V: NEd = 0.0, As = 12.0 cm2 In this case, the prescribed reinforcement is decreased from As = 24.5 cm2 to As = 12.0 cm2 in order to examine a case where the the crack width control is not passed with the given reinforcement. (NDP) 7.3.2 (2): ƒct,eƒ ƒ mean value of concrete tensile strength
ƒct,eƒ ƒ = ƒctm ≥ 3.0 MP
Tab. 3.1: ƒctm = 2.6 MP for C 25
ƒct,eƒ ƒ = 2.6 < 3.0 MP → ƒct,eƒ ƒ = 3.0 MP
(NDP) 7.3.3 Tab. 7.2DE (a)
from Tab. 7.2DE and for σs = 509.15 ≈ 510.0 MP
(NDP) 7.3.3: Eq. 7.7.1DE: The maximum bar diameters should be modified for load action
→ ϕ∗ ≈ 3.9 mm → ϕs = ϕ∗ · s s ϕs = 3.9 ·
σs · As 4(h − d) · b · 2.9
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 (NDP) 7.3.3 Tab. 7.2DE (a)
7.3.3: Eq. 7.7.1DE: The maximum bar diameters should be modified for load action
from Tab. 7.2DE and for σs = 436.43 MP → ϕ∗ ≈ 5.6 mm → ϕs = ϕ∗ · s s ϕs = 5.6 ·
σs · As 4(h − d) · b · 2.9
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. As = 14.54 cm2 we get a steel stress of σs = 436.36 MP which gives ϕs = 25 mm = ϕ∗ · s
436.36 · 14.54 4(100 − 96) · 30 · 2.9
→ ϕ∗ = 5.4849 mm s (NDP) 7.3.3 Tab. 7.2DE (a) where k = 0.3 m the prescribed maximum crack width
σs =
q
k · 3.48 · 106 / ϕ∗ s
σs =
p
0.3 · 3.48 · 106 / 5.4849 = 436.28 MP
Here we can notice that the stresses are equal leading to a usage factor of 1.0
7.3.4: Control of cracking with direct calculation
Control of cracking with direct calculation • Case V: NEd = 0.0
7.3.4 (1): αe is the ratio Es / Ecm 7.3.4 (1): Eq. 7.10: where A0p and Ac,eƒ ƒ are defined in 7.3.2 (3)
312
αe = Es / Ecm = 200000 / 31476 = 6.354 ρp,eƒ ƒ = (As + ξ21 · A0p ) / Ac,eƒ ƒ
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DCE-EN12: Calculation of crack widths
Ac,eƒ ƒ = hc,eƒ · b
7.3.2 (3): hc,eƒ see Fig. 7.1DE (d)
h / d1 = 100/ 4 = 25.00 → hc,eƒ / d1 = 3.25 Ac,eƒ ƒ = (3.25 · 4) · 30 = 13 · 30 = 390 cm2 A0p = 0.0 cm2
7.3.2 (3): where A0p is the area of pre or post-tensioned tendons within Ac,eƒ ƒ
ρp,eƒ ƒ = 24.50 / 390 = 0.06282 σs − kt · εsm − εcm =
ƒct,eƒ ƒ
· (1 + αe · ρp,eƒ ƒ )
ρp,eƒ ƒ
≥ 0.6 ·
Es 2/ 3
ƒct,eƒ ƒ = ƒctm = 0.30 · ƒck
σs Es
= 2.565 ≈ 2.6 MP
ƒct,eƒ ƒ = 2.6 < 3.0 MP → ƒct,eƒ ƒ = 3.0 MP 264.06 − 0.4 · εsm − εcm =
2.565
· (1 + 6.354 · 0.06282) 0.06282 200000
εsm − εcm = 1.2060 · 10−3 > 0.6 · sr,m = sr,m = sr,m ≤
3.6 · ρp,eƒ ƒ
3.6 · 0.06282 3.6 · ρp,eƒ ƒ
Tab. 3.1: ƒctm = 2.6 MP for C 25 or 2/ 3 ƒctm = 0.30 · ƒck = 2.565 MP kt : factor dependent on the duration of the load, kt = 0.4 for long term loading
= 0.79218 · 10−3
3.6 · ρp,eƒ ƒ
25 σs · ϕ
200000
(NDP) 7.3.2 (2): ƒct,eƒ ƒ mean value of concrete tensile strength, here no minimum value of ƒct,eƒ ƒ ≥ 3.0 MP is set
σs · ϕ
ϕ ≤
264.06
7.3.4 (1): Eq. 7.9: the difference of the mean strain in the reinforcement and in the concrete
(NDP) 7.3.4 (3): Eq. 7.11: sr,m is the maximum crack spacing
= 110.545 mm
= 611.25 mm
k = sr,m · (εsm − εcm ) = 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
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DCE-EN12: Calculation of crack widths
68.5
Conclusion
This example shows the calculation of crack widths. Various ways of reference calculations are demonstrated, 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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012.
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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. b tƒ r t Mz Vy y
h
z
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 Mz and a shear force Vy , with respect to EN 1993-1-1:2005 [78]. The calculation steps are
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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
Vy = 200 or 300 kN
b = 20.0 cm
Mz = 20 kNm
h = 19.0 cm tƒ = 1.0 cm, t = 0.65 cm r = 1.8 cm
Table 69.2: Results Case
Result
SOF.
Ref.
Vp,Rd,y [kN]
542.71
542.71
Vp,Rd,z [kN]
245.32
245.32
Util. level Vy
0.369
0.369
Util. level Mz
0.418
0.418
Util. level Vy
0.553
0.553
Util. level Mz,red
0.422
0.422
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DCE-EN13: Design of a Steel I-section for Bending and Shear
Design Process1
69.4 Material:
Structural Steel S 235 ƒy = 235 N/ mm2 Cross-sectional properties: Wp,y = 429.5 cm3
Tab. 3.1 : Nominal values of yield strength ƒy and ultimate tensile strength ƒ for hot rolled structural steel.
Plastic section modulus of HEA 200 w.r.t. y− and z−axis
Wp,z = 203.8 cm3
Where the shear force is present allowance should be made for its effect on the moment resistance. This effect may be neglected, where the shear force is less than half the plastic shear resistance.
6.2.8: Bending and shear
VEd ≤ 0.5 Vp,Rd
6.2.8 (2)
Vp,Rd =
AV · ƒy /
p 3 6.2.6 (2): Eq. 6.18: Vp,Rd the design plastic shear resistance
γM0
AVy = 2 · Aƒ nge (in the y−direction only contribution the two flanges )
AV the shear area w.r.t. y− and z−axis, respectively
AVy = 2 · tƒ · b = 40 cm2 AVz = A − 2 · b · tƒ + (t + 2 · r) · tƒ but not less than η · h · t h = h − 2 · tƒ = 17 cm AVz = 18.08 cm2 > 1 · 17 · 0.65 = 11.05 p 40 · 235 / 3 Vp,Rd,y = = 542.71 kN 1.00 p 18.08 · 235 / 3 Vp,Rd,z = = 245.32 kN 1.00
6.2.6 (3): The shear area AV may be taken as follows for rolled I-sections with load parallel to the web
6.2.6 (3): η may be conservatively taken equal to 1.0 6.1 (1): Partial factor γM0 = 1.00 is recommended
• Case : VEd = Vy = 200 kN, Mz = 20 kNm VEd Vp,Rd,y
= 0.369 < 0.5
→ no reduction of moment resistance due to shear Mz,V,Rd = Mz Mz,V,Rd
Wp,z · ƒy γM0
= 47.89 kNm
6.2.8 (5): Eq. 6.30: The reduced design plastic resistance moment (for ρ = 0)
= 0.418
1 The sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless otherwise specified.
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DCE-EN13: Design of a Steel I-section for Bending and Shear
• Case : VEd = Vy = 300 kN, Mz = 20 kNm VEd Vp,Rd,y 6.2.8 (3): The reduced moment resistance should be taken as the design resistance of the cross-section, calculated using a reduced yield strength (1 − ρ) · ƒy for the shear area
= 0.553 > 0.5
→ reduction of moment resistance due to shear 2 2 VEd − 1 = 0.01114 ρ= Vp,Rd,y Mz,V,Rd = (1 − ρ) ·
Wp,z · ƒy γM0
Mz,V,Rd = (1 − 0.01114) · 47.89 = 47.36 kNm Mz Mz,V,Rd
318
= 0.422
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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.
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DCE-EN13: Design of a Steel I-section for Bending and Shear
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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. b tƒ r t
y
h
z
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)
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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]. Bending Moment
Class 1: Fullly plastic moment is reached
Fully Plastic Moment
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 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 My = 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
My = 500 kNm
h = 462.0 mm tƒ = 17.0 mm t = 9.6 mm r = 10.2 mm
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DCE-EN14: Classification of Steel Cross-sections
Table 70.2: Results Case
Part
Result
Flange c/ t Web Flange Bending
Ref.
3.66
3.66
42.46
42.46
1
1
1
1
1
1
4
4
Class Web Flange
Compression
Class Web
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SOF.
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DCE-EN14: Classification of Steel Cross-sections
70.4
Design Process1
Material: Structural Steel S 275 Tab. 3.1 : Nominal values of yield strength ƒy and ultimate tensile strength ƒ for hot rolled structural steel. Tab. 5.2: Maximum width-to-thickness ratios for compression parts
Tab. 5.5.1(1): Classification of crosssection basis
ƒy = 275 N/ mm2 for maximum thickness ≤ 40 mm Æ ε = 235/ ƒy = 0.924 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. • Bending:
Tab. 5.2 (sheet 2): Outstand flanges
For the flange: c = b/ 2 − t / 2 − r
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
Tab. 5.2 (sheet 2): Outstand flanges part subject to compression
c / t ≤ 9 ε ≤ 8.32 → Flange classification: Class 1
Tab. 5.2 (sheet 1): Internal compression parts
For the web: c = h − 2tƒ − 2r c / t = (462 − 2 · 17 − 2 · 10.2) / 9.6 = 42.46
Tab. 5.2 (sheet 1): Internal compression parts - part subject to bending
c / t ≤ 72 ε ≤ 66.53 → Web classification: Class 1 Overall classification for bending: Class 1
5.5.2 (1): Classification
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.
• Compression: For the flange as above → Flange classification: Class 1 For the web as above : c / t = 42.46 Tab. 5.2 (sheet 1): Internal compression parts - part subject to compression Tab. 5.5.2 (8): A part which fails to satisfy the limits of Class 3 should be taken as Class 4 Tab. 5.5.2 (6): A cross-section is classified according to the highest least favourable class
5.5.2 (1): Classification
Class 3: c / t ≤ 42 ε ≤ 38.8 c / t = 42.46 > 38.8 → Web classification: Class 4 Overall classification for bending: Class 4 Class 4 cross-sections are those in which local buckling will occur before the attainment of yield stress in one or more parts of the crosssection. 1 The sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless otherwise specified.
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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.
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DCE-EN14: Classification of Steel Cross-sections
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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.
My
L b tƒ r t h
y
My
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)
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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=8m
My = 150 kNm
E = 210000 N/ mm2
UB 457x152x74 h = 462.0 mm b = 154.4 mm tƒ = 17.0 mm t = 9.6 mm r = 10.2 mm A = 94.48 cm2 z = 1046.5 cm4 T = 66.23 cm4 = 516297.12 cm6
Table 71.2: Results SOF.
Ref.
154.26
154.26
λLT
1.703
1.703
LT
1.907
1.907
χLT
0.321
0.321
MEd / Mb,Rd (BDK)
1.045
1.045
Mcr [kNm]
328
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DCE-EN15: Buckling Resistance of Steel Members
Design Process1
71.4 Material:
Structural Steel S 275 Tab. 3.1 : Nominal values of yield strength ƒy and ultimate tensile strength ƒ for hot rolled structural steel.
ƒy = 275 N/ mm2 for maximum thickness ≤ 40 mm Æ ε = 235/ ƒy = 0.924
Tab. 5.2: Maximum width-to-thickness ratios for compression parts
Design Load: MEd = 150kNm The cross-section is classified as Class 1, as demonstrated in Benchmark DCE-EN14. Wp,y · ƒy
Mc,Rd = Mp,Rd,y =
Mcr =
Mcr =
π
p
γM0
= 447.31 kNm
6.2.5 (2):Eq. 6.13: Bending resistance Mc,Rd for Class 1 cross-section 6.1 (1): γM0 = 1.00
v π 2 E Ez GT u t 1+ L GT L2
3.14 ·
p
v 2197.74 · 53.496 u t 8
6.3.2.2 (2): Mcr , the elastic critical moment for ltb is based on gross cross sectional properties
1+
3.142 · 108.42
6.3.2.2 (1): λLT non dimensional slenderness for lateral torsional buckling
h i 2 LT = 0.5 1 + αLT λLT − λLT,0 + βλLT
6.3.2.3 (1): LT for rolled sections in bending. Recommended values: λLT,0 = 0.4 (maximum value) β = 0.75 (minimum value)
h / b = 462/ 154.4 = 2.99 > 2 for rolled I-sections and h / b > 2 → buckling curve c for buckling curve c → αLT = 0.49 LT = 0.5 1 + 0.49 1.703 − 0.4 + 0.75 · 1.7032 = 1.907 1 LT +
r
2LT
2
= 0.321
6.3.2.3 (1): Table 6.5: Recommendation for the selection of ltb curve for cross-sections using Eq. 6.57 6.3.2.2 (2): Table 6.5: Recommendation values for imperfection factors αLT for ltb curves
6.3.2.3 (1): Eq. 6.57: χLT reduction factor for ltb
− β λLT
but χLT = 0.321 ≤ 1.0 and χLT = 0.321 ≤
Various empirical or approximate formulae exist for the determination of Mcr
53.496 · 82
Mcr = 154.26 kNm v v uW ƒ t y y t 447.31 λLT = = 1.703 = Mcr 154.26
χLT =
Tab. 5.5: Classification of cross-section
1 2
= 0.345
λLT Mb,Rd = χLT Wy MEd Mb,Rd
ƒy γM1
= 143.587 kNm
= 1.045 → Beam fails in LTB
6.3.2.1 (3): Eq. 6.55: Mb,Rd The design buckling resistance moment of laterally unrestrained beam For Class 1 section Wy = Wp,y 6.1 (1): γM1 = 1.00 recommended value
1 The sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless otherwise specified.
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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.
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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. b tƒ r t Mz Vy y
h
z
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)
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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 My , a shear force Vz 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 My,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
Vz = 500 kN
b = 30.0 cm
My = 150 kNm
h = 30.0 cm
N = −2000 kN
tƒ = 1.9 cm, t = 1.1 cm r = 2.7 cm A = 149.078 cm2
Table 72.2: Results Result
332
SOF.
Ref.
Util. level N
0.571
0.571
Util. level Nred
0.633
0.633
Util. level Vz
0.777
0.777
Util. level My
0.342
0.342
Util. level My,red
0.879
0.882
VERiFiCATiON MANUAL | SOFiSTiK 2014
DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force
Design Process1
72.4 Material:
Structural Steel S 235 ƒy = 235 N/ mm2
Tab. 3.1 : Nominal values of yield strength ƒy and ultimate tensile strength ƒ for hot rolled structural steel.
Cross-sectional properties: Wp,y = 1868.6 cm3
Plastic section modulus of HEB 300 w.r.t. y−
AVz = A − 2 · b · tƒ + (t + 2 · r) · tƒ but not less than η · h · t h = h − 2 · tƒ = 26.2 cm AVz = 47.428 cm2 > 1 · 26.2 · 1.1 = 28.82 p 47.428 · 235 / 3 = 643.49 kN Vp,Rd,z = 1.00 Mc,Rd =
Wp,y · ƒy γM0
= 439.12 kNm
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. Provided that the design value of the shear force VEd does not exceed the 50% of the design plastic shear resistance Vp,Rd no reduction of the resistances for bending and axial force need to be made.
6.2.6 (3): The shear area AV may be taken as follows for rolled I-sections with load parallel to the web
6.2.6 (3): η may be conservatively taken equal to 1.0 6.1 (1): Partial factor γM0 = 1.00 is recommended
6.2.5 (2): Eq. 6.13: The design resistance for bending for Class 1 crosssection 6.2.10 (1): Bending, shear and axial force
6.2.10 (2)
VEd ≤ 0.5 Vp,Rd VEd Vp,Rd,z
= 0.777 > 0.5
→ shear resistance exceeded Where VEd exceeds 50% of Vp,Rd , the design resistance of the crosssection to combinations of moment and axial force should be calculated using a reduced yield strength (1 − ρ) · ƒy for the shear area ρ=
2 VEd Vp,Rd,z
2 −1
Wp,y − My,V,Rd = My My,V,Rd My Mc,Rd
6.2.10 (3)
= 0.3069
ρ · A2 4t
γM0
· ƒy = 425.503 kNm
6.2.8 (5): Eq. 6.30: The reduced design plastic resistance moment allowing for shear force may alternitavely be obtained for I-sections with equal flanges and bending about major axis where A = h · t
= 0.353
= 0.342 (with no reduction)
1 The
sections mentioned in the margins refer to EN 1993-1-1:2005 [78] unless otherwise specified.
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DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force
6.2.3 (2): The design plastic resistance of gross cross-section NpR d 6.2.10 (3): The reduced plastic resistance ue to shear α = AVz / A
Np,Rd =
= 3503.3 kN
NV,Rd = 3161.249 kN
NV,Rd NEd Np,Rd
6.2.9.1 (4): For doubly symmetrical 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
γM0
NV,Rd = Np,Rd · (1 − α · ρ)
NEd
6.2.9.1 (1): Bending and axial force Class 1 and 2 cross-sections
A · ƒy
= 0.633 = 0.571 (with no reduction)
Where an axial force is present, allowance should be made for its effect on the plastic moment resistance. 0.25 NV,Rd NEd ≤ 0.5 h t ƒy (1 − ρ) γM0 790.31 kN NEd = 2000 > 234.71 kN → consideration of axial force in interaction
6.2.9.1 (5): Eq 6.36: The following approximations may be used for standard rolled I-sections with equal flanges
MN,y,Rd = Mp,y,Rd ·
1−n 1 − 0.5 α
and MN,y,Rd ≤ Mp,y,Rd n = NEd / NV,Rd = 0.6327 α = (1 − ρ) · (A − 2 · b · tƒ ) / A = 0.1631 MN,y,Rd = 425.503 · My,Ed MNV,y,Rd
334
1 − 0.6327 1 − 0.5 0.1631
= 170.164 kNm
= 0.882
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 fur ¨ 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.
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DCE-EN16: Design of a Steel I-section for Bending, Shear and Axial Force
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DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS
73 DCE-EN17: Stress Calculation at a Rectangular Prestressed 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.
h
d
My Np
zp Ap
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
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DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS
NP · zp / W2 M/ W2
NP / A
+
+
=
zp Ap NP · zp / W1
NP / A
M/ W1
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
σp ƒpk ƒp0,1k ƒpd = ƒp0,1k / γs
B
A
Idealised
B
Design
εp
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
Mg = 1250 kNm
Y 1770
b = 100.0 cm
Np = −3651.1 kN
d = 95.0 cm L = 20.0 m
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0.916 3572.3 0.917 3574.2 0.917 3576.1 0.918 3578.1 0.918 3580.2 0.919 3582.6 0.920 3585.6 0.920 3588.9 0.921 3592.4 0.922 3596.1 0.923 3600.0 0.924 3604.2 0.926 3608.5 0.927 3613.1 0.928 3618.0 0.929 3623.0 0.931 3628.3 0.932 3633.8 0.933 3639.5 0.935 3645.4 0.937 3651.6 0.938 3657.8 0.940 3663.8 0.941 3669.6 0.943 3675.1 0.944 3680.5 0.944 3678.8 0.942 3673.9 0.941 3669.2 0.940 3664.8 0.939 3660.6 0.938 3656.6 0.937 3652.8 0.936 3649.2 0.935 3645.9 0.934 3642.9 0.934 3640.4 0.933 3638.2 0.933 3636.3 0.932 3634.3
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0.944 3682.2
10011 1010 10012 1011 10013 1012 10014 1013 10015 1014 10016 1015 10017 1016 10018 1017 10019 1018 10020
20.00
19.00
18.00
17.00
16.00
0.000 -0.057
0.057 -0.057
0.115 -0.057
0.171 -0.055
0.223 -0.050
0.271 -0.045
0.313 -0.038
0.347 -0.031
0.374 -0.022
0.390 -0.011
0.396 0.000
0.390 0.011
0.374 0.022
Geometric Properties
15.00
14.00
13.00
12.00
2
11.00
10001 1001 10002 1002 10003 1003 10004 1004 10005 1005 10006 1006 10007 1007 10008 1008 10009 1009 10010
10.00
9.00
8.00
0.347 0.031
0.313 0.038
0.271 0.045
0.223 0.050
0.171 0.055
0.114 0.057
0.057 0.057
0.000 0.057
Material Properties
7.00
6.00
5.00
4.00
3.00
2.00
1
1.00
0.00
DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS
Table 73.1: (continued) Loading (at = 10 m)
Ap = 28.5 cm2
L
Figure 73.4: Simply Supported Beam
3
Figure 73.5: Tendon Geometry
1.000 = 1368 N/mm2
0.973 0.944 0.932
0.916
Figure 73.6: Prestress Forces and Stresses
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DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS
Table 73.2: Results Case
CS
Result
SOF.
Ref.
0
σc,b [MP]
−12.455
−12.454
My [kNm]
−1451.02
−1451.02
−4.882
−4.882
−201.02
−201.02
−11.866
−11.866
−1426.65
−1426.65
−4.626
−4.626
−176.65
−176.65
0 My [kNm]
1 My [kNm]
V
1 My [kNm]
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73.4
Design Process1
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: C 35/ 45
3.1: Concrete
Ecm = 34000 N/ mm2
3.1.2: Tab. 3.1: Ecm for C 35/ 45
Prestressing Steel: Y 1170
3.3: Prestressing Steel
Ep = 195000 N/ mm2
3.3.6 (3): Ep for wires
ƒpk = 1770 N/ mm2
3.3.2, 3.3.3: ƒpk Characteristic tensile strength of prestressing steel
ƒp0,1k = 1520 N/ mm2 3.3.2, 3.3.3: ƒp0,1k 0.1% proof-stress of prestressing steel, yield strength
Prestressing system: BBV L19 150 mm2 19 wires with area of 150 mm2 each, giving a total of Ap = 28.5 cm2
Cross-section: Ac = 1.0 · 1.0 = 1 m2 Diameter of duct ϕdct = 67 mm Ratio αE,p = Ep / Ecm = 195000 / 34000 = 5.74 Ac,netto = Ac − π · (ϕdct / 2)2 = 0.996 m2 Ade = Ac + Ap · αE,p = 1.013 m2
The force applied to a tendon, i.e. the force at the active end during tensioning, should not exceed the following value Pm = Ap · σp,m where σp,m = min 0.80ƒpk ; 0.90ƒp0,1k
5.10.2.1 (1)P: Prestressing force during tensioning - Maximum stressing force 5.10.2.1 (1)P: Eq. 5.41: Pm maximum stressing force (NDP) 5.10.2.1 (1)P: σp,m maximum stress applied to the tendon
Pm = Ap · 0.80 · ƒpk = 28.5 · 10−4 · 0.80 · 1770 = 4035.6 kN Pm = Ap · 0.90 · ƒp0,1k = 28.5 · 10−4 · 0.90 · 1520 = 3898.8 kN → Pm = 3898.8 kN and σp,m = 1368 N/ mm2 The value of the initial prestress force at time t = t0 applied to the concrete immediately after tensioning and anchoring should not exceed the following value Pm0 () = Ap · σp,m0 () 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 National Annex) [72], [73], unless otherwise specified.
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5.10.3 (2): Prestress force 5.10.3 (2): Eq. 5.43: Pm0 initial prestress force at time t = t0
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DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS
(NDP) 5.10.3 (2): σp,m0 () stress in the tendon immediately after tensioning or transfer
where σp,m0 () = min 0.75ƒpk ; 0.85ƒp0,1k Pm0 = Ap · 0.75 · ƒpk = 28.5 · 10−4 · 0.75 · 1770 = 3783.4 kN Pm0 = Ap · 0.85 · ƒp0,1k = 28.5 · 10−4 · 0.85 · 1520 = 3682.2 kN → Pm0 = 3682.2 kN and σp,m0 = 1292 N/ mm2 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 σp,m = 1368 N/ mm2 and Pm0 = 3682.2 kN can be visualised. Load Actions: Self weight per length: γ = 25 kN/ m → g1 = γ · A = 25 · 1 = 25 kNm
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 prestress
Safety factors at ultimate limit state Actions (unfavourable)
Safety factor at final state
•
permanent
γG = 1.35
•
prestress
γP = 1.00
Combination coefficients at serviceability limit state g1 = 25 kNm: for rare, frequent and quasi-permanent combination (for stresses) At = 10.0 m middle of the span: Mg = g1 · L2 / 8 = 1250 kNm Np = Pm0 ( = 10.0 m) = −3651.1 kN (from SOFiSTiK)
The concrete stresses may be determined for each construction stage under the total quasi-permanent combination σc {G + Pm0 + ψ2 · Q} In this Benchmark no variable load Q is examined
Calculation of stresses σc,b at = 10.0 m middle of the span: Position of the tendon: z = 0, 396 m • Case : Prestress at construction stage section 0 (P cs0) −σc Mp Np zp Pm0,=10 +σc Np = −3651.1 kN
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DCE-EN17: Stress Calculation at a Rectangular Prestressed Concrete CS
Mp1 = NP · z = −3651.1 · 0.396 = −1445.86 kNm Mp2 = NP · zs = −3651.1 · 0.001415 = −5.16 kNm Mp = −1445.86 − 5.16 = −1451.02 kNm = My σc,b = σc,b =
Np Ac,netto
+
−3651.1 0.996
zs the new position of the center of the cross-section for cs0 zp = z + zs Mp bending moment caused by prestressing
My W1,cs0 cross-section moduli for contruction stage 0 at the bottom left and right point
W1,cs0 +
−1451.02 0.165
= −12.454 MP
σc,b stress at the concrete at the bottom of the cross section
• Case : Prestress and self-weight at con. stage sect. 0 (P+G cs0) −σc Mg + Mp Np zp Pm0,=10 +σc Np = −3651.1 kN and Mg = 1250 kNm As computed above: Mp = −1451.02 kNm My = 1250 − 1451.02 = −201.02 kNm σc,b =
−3651.1 0.996
+
−201.02 0.165
= −4.882 MP
• Case : Prestress at con. stage sect. 1 (P cs1) Np = −3651.1 kN and Mp1 = −1445.86 kNm (as above) Mp2 = NP · zs = −3651.1 · (−0.005259) = 19.2 kNm Mp = −1445.86 + 19.2 = −1426.65 kNm = My σc,b = σc,b =
Np Ade
+
My W1,cs1 cross-section moduli for contruction stage 1
W1,cs1
−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) Np = −3651.1 kN and Mg = 1250 kNm As computed above: Mp = −1426.65 kNm My = 1250 − 1426.65 = −176.65 kNm σc,b =
−3651.1 1.013
+
−176.65 0.173
= −4.626 MP
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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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund 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.
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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 concrete, 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.
h
My Np
zp Ap
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.
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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 concrete 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 Benchmark 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
Mg = 1250 kNm
t0 = 28 dys
Y 1770
b = 100.0 cm
Np = −3651.1 kN
ts = 0 dys
RH = 80
L = 20.0 m
teƒ ƒ = 1000000 dys
Ap = 28.5 cm2
L
Figure 74.2: Simply Supported Beam
Table 74.2: Results Result
AQB
CSM+AQB
Ref.
εcs
−18.85 · 10−5
-
−18.85 · 10−5
ε
−31.58 · 10−5
-
−31.58 · 10−5
ϕ0
1.463
1.463
1.463
ϕ(t, t0 )
1.393
1.393
1.393
−66.94
−68.20
−68.79
190.8
194.4
196.05
Δσp,c+s [MP] ΔPc+s [kN]
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DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS
74.4
Design Process1
Design with respect to DIN EN 1992-1-1:2004 (NA) [72] [73]:2 Material: Concrete: C 35/ 45 Ecm = 34077 N/ mm2
3.1: Concrete 3.1.2: Tab. 3.1: Ecm , ƒck and ƒcm for C 35/ 45
ƒck = 35 N/ mm2 ƒcm = 43 N/ mm2 Prestressing Steel: Y 1170
3.3: Prestressing Steel
Ep = 195000 N/ mm2
3.3.6 (3): Ep for wires
ƒpk = 1770 N/ mm2
3.3.2, 3.3.3: ƒpk Characteristic tensile strength of prestressing steel
Prestressing system: BBV L19 150 mm2 19 wires with area of 150 mm2 each, giving a total of Ap = 28.5 cm2
Cross-section: Ac = 1.0 · 1.0 = 1 m2 Diameter of duct ϕdct = 67 mm Ratio αE,p = Ep / Ecm = 195000 / 34077 = 5.7223 Ac,netto = Ac − π · (ϕdct / 2)2 = 0.996 m2 Ade = Ac + Ap · αE,p = 1.013 m2
Load Actions: Self weight per length: γ = 25 kN/ m At = 10.0 m middle of the span: Mg = g1 · L2 / 8 = 1250 kNm Np = Pm0 ( = 10.0 m) = −3651.1 kN (from SOFiSTiK) Calculation of stresses at = 10.0 m midspan: Position of the tendon: zcp = 0, 396 m
Prestress and self-weight at con. stage sect. 0 (P+G cs0) Np = −3651.1 kN and Mg = 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 National Annex) [72], [73], unless otherwise specified.
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DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS
−σc Mg + Mp Np zp Pm0,=10 +σc
zs the new position of the center of the cross-section for cs0 zp = zcp + zs Mp bending moment caused by prestressing
Mp1 = NP · zcp = −3651.1 · 0.396 = −1445.86 kNm Mp2 = NP · zs = −3651.1 · 0.001415 = −5.16 kNm Mp = −1445.86 − 5.16 = −1451.02 kNm My = 1250 − 1451.02 = −201.02 kNm
σc,QP stress in concrete
σc,QP =
−3651.1 0.996
+
−201.02
= −4.882 MP
0.165
Calculation of creep and shrinkage at = 10.0 m midspan: t0 minimun age of concrete for loading ts age of concrete at start of drying shrinkage t age of concrete at the moment considered
3.1.4 (6): Eq. 3.8: εcs total shrinkage strain 3.1.4 (6): Eq. 3.9: εcd drying shrinkage strain
t0 = 28 days ts = 0 days t = teƒ ƒ + t0 = 1000000 + 28 = 1000028 days εcs = εcd + εc εcd (t) = βds (t, ts) · kh · εcd,0 The development of the drying shrinkage strain in time is strongly depends on βds (t, ts) factor. SOFiSTiK accounts not only for the age at start of drying ts but also for the influence of the age of the prestressing t0 . Therefore, the calculation of factor βds reads:
3.1.4 (6): Eq. 3.10: βds
3.1.4 (6): h0 the notional size (mm) of the cs h0 = 2Ac / = 500 mm
βds = βds (t, ts ) − βds (t0 , ts ) βds =
βds =
(t − ts ) (t − ts ) + 0.04 ·
q
h30
−
(t0 − ts ) (t0 − ts ) + 0.04 ·
(1000028 − 0) (1000028 − 0) + 0.04 ·
p
5003
−
q
h30
(28 − 0) (28 − 0) + 0.04 ·
p
5003
βds = 0.99955 − 0.05892 = 0.94063 3.1.4 (6): Tab. 3.3: kh coefficient depending on h0 Annex B.2 (1): Eq. B.11: εcd,0 basic drying shrinkage strain
kh = 0.70 for h0 ≥ 500 mm εcd,0 = 0.85 (220 + 110 · αds1 ) · exp −αds2 ·
Annex B.2 (1): Eq. B.12: βRH RH the ambient relative humidity (%) Annex B.2 (1): αds1 , αds1 coefficients depending on type of cement. For class N αds1 = 4, αds2 = 0.12
348
βRH = 1.55 1 −
RH RH0
3
= 1.55 1 −
80
ƒcm ƒcmo
3
100
εcd,0 = 0.85 (220 + 110 · 4) · exp −0.12 ·
43 10
· 10−6 · βRH
= 0.7564
· 10−6 · 0.7564
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DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS
εcd,0 = 2.533 · 10−4 εcd = 0.94063 · 0.70 · 2.533 · 10−4 = 0.0001668 εcd = 1.668 · 10−4 = 0.1668 ◦ /◦◦ εc (t) = βs (t) · εc (∞)
3.1.4 (6): Eq. 3.11: εc autogenous shrinkage strain
εc (∞) = 2.5 (ƒck − 10) · 10−6 = 2.5 (35 − 10) · 10−6
3.1.4 (6): Eq. 3.12: εc (∞)
εc (∞) = 6.25 · 10−5 = 0.0625 ◦ /◦◦ Proportionally to βds (t, ts), SOFiSTiK calculates factor βs as follows: βs = βs (t) − βs (t0 ) p p p p βs = 1 − e−0.2· t − 1 − e−0.2· t0 = e−0.2· t0 − e−0.2· t
3.1.4 (6): Eq. 3.13: βs
βs = 0.347 ε = εcd,0 + εc (∞) = 2.533 · 10−4 + 6.25 · 10−5 ε = −31.58 · 10−5
ε absolute shrinkage strain negative sign to declare losses
εc = 0.347 · 6.25 · 10−5 = 2.169 · 10−5 = 0.02169 ◦ /◦◦ εcs = 1.668 · 10−4 + 2.169 · 10−5 = −18.85 · 10−5 ϕ(t, t0 ) = ϕ0 · βc (t, t0 )
negative sign to declare losses
Annex B.1 (1): Eq. B.1: ϕ(t, t0 ) creep coefficient
ϕ0 = ϕRH · β(ƒcm ) · β(t0 ) 1 − RH/ 100 ϕRH = 1 + · α1 · α2 p 0.1 · 3 h0
Annex B.1 (1): Eq. B.3: ϕRH factor for effect of relative humidity on creep
16.8 p = 16.8/ 43 = 2.562 β(ƒcm ) = p ƒcm
Annex B.1 (1): Eq. B.4: β(ƒcm ) factor for effect of concrete strength on creep
α1 =
α2 =
α3 =
ϕRH
0.7
35 ƒcm
0.2
35 ƒcm
0.5
35 ƒcm
≤ 1 = 0.8658
Annex B.1 (1): Eq. B.2: ϕ0 notional creep coefficient
Annex B.1 (1): Eq. B.8c: α1 , α2 , α3 coefficients to consider influence of concrete strength
≤ 1 = 0.9597
≤ 1 = 0.9022
1 − 80/ 100 = 1+ · 0.8658 · 0.9597 = 1.1691 p 3 0.1 · 500
β(t0 ) =
1 0.1 + t00.20
t0 = t0,T ·
9
Annex B.1 (1): Eq. B.5: β(t0 ) factor for effect of concrete age at loading on creep
α +1
≥ 0.5
Annex B.1 (2): Eq. B.9: t0,T temperature adjusted age of concrete at loading adjusted according to expression B.10
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1.2 2 + t0,T
DCE-EN18: Creep and Shrinkage Calculation of a Rectangular Prestressed Concrete CS
Annex B.1 (3): Eq. B.10: tT temperature adjusted concrete age which replaces t in the corresponding equations
Annex B.1 (2): Eq. B.9: α a power which depends on type of cement For class N α = 0
tT =
Pn =1
t0,T = 28 · e−(4000/ [273+20]−13.65) = 28 · 1.0 = 28.0 ⇒ t0 = 28.0 · β(t0 ) =
Annex B.1 (1): Eq. B.7: βc (t, t0 ) coefficient to describe the development of creep with time after loading Annex B.1 (1): Eq. B.8: βH coefficient depending on relative humidity and notional member size
e−(4000/ [273+T(Δt )]−13.65) · Δt
9 2 + 28.01.2 1
0.1 + 28.00.20
βc (t, t0 ) =
+1
0
= 28.0
= 0.48844 0.3
(t − t0 ) (βH + t − t0 )
βH = 1.5 · 1 + (0.012 · RH)18 · h0 + 250 · α3 ≤ 1500 · α3 βH = 1.5 · 1 + (0.012 · 80)18 · 500 + 250 · 0.9022 βH = 1335.25 ≤ 1500 · 0.9022 = 1353.30 ⇒ βc (t, t0 ) = 0.9996 ϕ0 = 1.1691 · 2.562 · 0.48844 = 1.463
Annex B.1 (3): The values of ϕ(t, t0 ) given above should be associated with the tangent modulus Ec 3.1.4 (2): The values of the creep coefficient, ϕ(t, t0 ) is related to Ec , the tangent modulus, which may be taken as 1.05 · Ecm
ϕ(t, t0 ) = 1.463 · 0.9996/ 1.05 = 1.393 According to EN, the creep value is related to the tangent Young’s modulus Ec , where Ec being defined as 1.05 · Ecm . To account for this, SOFiSTiK adopts this scaling for the computed creep coefficient (in SOFiSTiK, all computations are consistently based on Ecm ).
ΔPc+s+r = Ap ·Δσp,c+s+r 5.10.6 (2): Eq. 5.46: ΔPc+s+r time dependent losses of prestress
Ep ϕ(t, t0 ) · σc,QP εcs · Ep + 0.8Δσpr + Ecm = Ap Ep Ap Ac 1+ 1+ z 2 [1 + 0.8ϕ(t, t0 )] Ecm Ac c cp
In this example only the losses due to creep and shrinkage are taken into account, the reduction of stress due to relaxation (Δσpr ) is ignored. Δσp,c+s = Δσp,c+s variation of stress in tendons due to creep and shrinkage at location , at time t
−0.1885 · 10−3 · 195000 + 5.7223 · 1.393 · (−4.882) 28.5 · 10−4 0.996 2 1 + 5.7223 1+ 0.396 [1 + 0.8 · 1.393] 0.996 0.08277
Δσp,c+s = −68.79 MP ΔPc+s = Ap · Δσp,c+s = 28.5 · 10−4 · 68.79 · 103 = 196.05 kN
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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 fur ¨ den Hochbau. BVPI, DBV, ISB, VBI. Ernst & Sohn, Beuth, 2012. [75] Beispiele zur Bemessung nach Eurocode 2 - Band 1: Hochbau. Ernst & Sohn. Deutschen Betonund Bautechnik-Verein E.V. 2011.
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