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8/22/2019 AD Validation Guide 2014 En
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Advance Design
Validation Guide
Version: 2014
Tests passed on: 12 September 2014
Number of tests: 470
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INTRODUCTION
Before being officially released, each version of GRAITEC software, including
Advance Design, undergoes a series of validation tests. This validation is performed in
parallel and in addition to manual testing and beta testing, in order to obtain the
"operational version" status. This document contains a description of the automatictests, highlighting the theoretical background and the results we have obtained using
the current software release.
Usually, a test is made of a reference (independent from the specific software version
tested), a transformation (a calculation or a data processing scenario), a result (given
by the specific software version tested) and a difference usually measured in
percentage as a drift from a set of reference values. Depending on the cases, the
used reference is either a theoretical calculation done manually, a sample taken from
the technical literature, or the result of a previous version considered as good by
experience.
Starting with version 2012, Graitec Advance has made significant steps forward in term ofquality management by extending the scope and automating the testing process.
While in previous versions, the tests were always about the calculation results which
were compared to a reference set, starting with version 2012, tests have been
extended to user interface behavior, import/export procedures, etc.
The next major improvement is the capacity to pass the tests automatically. These
current tests have obviously been passed on the “operational version”, but they are
actually passed on a daily basis during the development process, which helps improve
the daily quality by solving potential issues, immediately after they have been
introduced in the code.
In the field of structural analysis and design, software users must keep in mind that theresults highly depend on the modeling (especially when dealing with finite elements)
and on the settings of the numerous assumptions and options available in the
software. A software package cannot replace engineers experience and analysis.
Despite all our efforts in term of quality management, we cannot guaranty the correct
behavior and the validity of the results issued by Advance Design in any situation.
With this validation guide, we are providing a set of concrete test cases showing the
behavior of Advance Design in various areas and various conditions. The tests cover
a wide field of expertise: modeling, climatic load generation according to Eurocode 1,
combinations management, meshing, finite element calculation, reinforced concrete
design according to Eurocode 2, steel member design according to Eurocode 3, steel
connection design according to Eurocode 3, timber member design according toEurocode 5, seismic analysis according to Eurocode 8, report generation, import /
export procedures and user interface behavior.
We hope that this guide will highly contribute to the knowledge and the confidence you
are placing in Advance Design.
Manuel LIEDOT
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TABLE OF CONTENTS
1 FINITE ELEMENT METHOD .................................................................................................23
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2 CAD, RENDERING AND VISUALIZATION ........................................................................ 335
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3 CLIMATIC GENERATOR ....................................................................................................343
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4 COMBINATIONS................................................................................................................. 357
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5 CONCRETE DESIGN ..........................................................................................................365
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6 GENERAL APPLICATION.................................................................................................. 679
7 IMPORT / EXPORT ............................................................................................................. 685
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8 JOINT DESIGN....................................................................................................................689
9 MESH...................................................................................................................................691
10 OLD CLIMATIC GENERATOR............................................................................................693
11 REPORTS GENERATOR....................................................................................................695
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12 SEISMIC ANALYSIS ........................................................................................................... 703
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13 STEEL DESIGN...................................................................................................................709
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14 TIMBER DESIGN...............................................................................................................1071
15 XML TEMPLATE FILES ....................................................................................................1115
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1 Finite element method
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1.1 Slender beam with variable section (fixed-free) (01-0004SDLLB_FEM)
Passed
1.1.1 Description
1.1.2 Background
1.1.2.1 Model description
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Geometry
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1.1.2.2 Eigen mode frequencies
Reference solutions
Finite elements modeling
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Eigen mode shapes
1.1.2.3 Theoretical results
Solver Result name Result description Reference value
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1.1.3 Calculated results
Result name Result description Value Error
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1.2 Thin lozenge-shaped plate fixed on one side (alpha = 15 °) (01-0008SDLSB_FEM)
Passed
1.2.1 Description
1.2.2 Background
1.2.2.1 Model description
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Materials properties
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1.2.2.2 Eigen modes frequencies function by angle
Reference solution
Finite elements modeling
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Eigen mode shapes
1.2.2.3 Theoretical results
Solver Result name Result description Reference value
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1.2.3 Calculated results
Result name Result description Value Error
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1.3 Bimetallic: Fixed beams connected to a stiff element (01-0026SSLLB_FEM)
Passed
1.3.1 Description
1.3.2 Background
1.3.2.1 Model description
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Materials properties
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1.3.2.2 Deflection at B and D
Reference solution
Finite elements modeling
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1.3.2.3 Vertical reaction at A and C
Reference solution
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Finite elements modeling
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1.3.2.4 Bending moment at A and C
Reference solution
Finite elements modeling
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1.3.2.5 Theoretical results
Solver Result name Result description Reference value
1.3.3 Calculated results
Result name Result description Value Error
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1.4 Cantilever rectangular plate (01-0001SSLSB_FEM)
Passed
1.4.1 Description
1.4.2 Background
1.4.2.1 Model description
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Boundary conditions
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1.4.2.2 Displacement of the model in the linear elastic range
Reference solution
Finite elements modeling
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Deformed shape
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1.4.2.3 Theoretical results
Solver Result name Result description Reference value
1.4.3 Calculated results
Result name Result description Value Error
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1.5 Thin lozenge-shaped plate fixed on one side (alpha = 0 °) (01-0007SDLSB_FEM)
Passed
1.5.1 Description
1.5.2 Background
1.5.2.1 Model description
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Materials properties
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Boundary conditions
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Loading
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1.5.2.2 Eigen mode frequencies relative to the angle
Reference solution
Finite elements modeling
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Eigen mode shapes
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1.5.2.3 Theoretical results
Solver Result name Result description Reference value
1.5.3 Calculated results
Result name Result description Value Error
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1.6 Thin lozenge-shaped plate fixed on one side (alpha = 45 °) (01-0010SDLSB_FEM)
Passed
1.6.1 Description
1.6.2 Background
1.6.2.1 Model description
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Geometry
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1.6.2.2 Eigen mode frequencies relative to the angle
Reference solution
Finite elements modeling
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Eigen mode shapes
1.6.2.3 Theoretical results
Solver Result name Result description Reference value
1.6.3 Calculated results
Result name Result description Value Error
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1.7 Portal frame with lateral connections (01-0030SSLLB_FEM)
Passed
1.7.1 Description
1.7.2 Background
1.7.2.1 Model description
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Geometry
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1.7.2.2 Displacements at A
Reference solution
Finite elements modeling
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Displacements shape
1.7.2.3 Moments in A
Reference solution
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1.7.2.4 Theoretical results
Solver Result name Result description Reference value
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1.7.3 Calculated results
Result name Result description Value Error
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1.8 Circular plate under uniform load (01-0003SSLSB_FEM)
Passed
1.8.1 Description
1.8.2 Background
1.8.2.1 Model description
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Geometry
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Boundary conditions
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Loading
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1.8.2.2 Vertical displacement of the model at the center of the plate
Reference solution
Finite elements modeling
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Deformed shape
1.8.2.3 Theoretical results
Solver Result name Result description Reference value
1.8.3 Calculated results
Result name Result description Value Error
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1.9 Thin circular ring fixed in two points (01-0006SDLLB_FEM)
Passed
1.9.1 Description
1.9.2 Background
1.9.2.1 Model description
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Materials properties
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Boundary conditions
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Loading
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1.9.2.2 Eigen mode frequencies
Reference solutions
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Finite elements modeling
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Eigen mode shapes
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1.9.2.3 Theoretic results
Solver Result name Result description Reference value
1.9.3 Calculated results
Result name Result description Value Error
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1.10 Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM)
Passed
1.10.1Description
1.10.2Background
1.10.2.1Model description
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1.10.2.2Critical load on node 5
Reference solution
Finite elements modeling
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Deformed shape
1.10.2.3Theoretical results
Solver Result name Result description Reference value
1.10.3Calculated results
Result name Result description Value Error
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1.11 Vibration mode of a thin piping elbow in plane (case 1) (01-0011SDLLB_FEM)
Passed
1.11.1Description
1.11.2Background
1.11.2.1Model description
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Geometry
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Boundary conditions
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1.11.2.2Eigen mode frequencies
Reference solution
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Finite elements modeling
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Eigen mode shapes
1.11.2.3Theoretical results
Solver Result name Result description Reference value
1.11.3Calculated results
Result name Result description Value Error
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1.12 EDF Pylon (01-0033SFLLA_FEM)
Passed
1.12.1Description
1.12.2Background
1.12.2.1Model description
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Loading
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1.12.2.2Displacement of the model in the linear elastic range
Reference solution
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Software ANSYS 5.3 NE/NASTRAN 7.0
Finite elements modeling
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Deformed shape
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Buckling modal deformation (dominating mode)
1.12.2.3Theoretical results
Solver Result name Result description Reference value
1.12.3Calculated results
Result name Result description Value Error
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1.13 Tied (sub-tensioned) beam (01-0005SSLLB_FEM)
Passed
1.13.1Description
1.13.2Background
1.13.2.1Model description
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1.13.2.2Compression force in CE bar
Reference solution
Finite elements modeling
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Force diagrams
1.13.2.3Bending moment at point H
Reference solution
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Shape of the bending moment diagram
1.13.2.4Vertical displacement at point D
Reference solution
Finite elements modeling
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Deformed shape
1.13.2.5Theoretical results
Solver Result name Result description Reference value
1.13.3Calculated results
Result name Result description Value Error
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1.14 Vibration mode of a thin piping elbow in plane (case 2) (01-0012SDLLB_FEM)
Passed
1.14.1Description
1.14.2Background
1.14.2.1Model description
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Geometry
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1.14.2.2Eigen mode frequencies
Reference solution
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Finite elements modeling
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Eigen mode shapes
1.14.2.3Theoretical results
Solver Result name Result description Reference value
1.14.3Calculated results
Result name Result description Value Error
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1.15 Thin lozenge-shaped plate fixed on one side (alpha = 30 °) (01-0009SDLSB_FEM)
Passed
1.15.1Description
1.15.2Background
1.15.2.1Model description
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Geometry
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1.15.2.2Eigen mode frequencies relative to the angle
Reference solution
Finite elements modeling
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Eigen mode shapes
1.15.2.3Theoretical results
Solver Result name Result description Reference value
1.15.3Calculated results
Result name Result description Value Error
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1.16 Caisson beam in torsion (01-0037SSLSB_FEM)
Passed
1.16.1Description
1.16.2Background
1.16.2.1Model description
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Boundary conditions
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Loading
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1.16.2.2Displacement and stress at two points
Reference solution
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Finite elements modeling
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Deformed shape
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1.16.2.3Theoretical results
Solver Result name Result description Reference value
1.16.3Calculated results
Result name Result description Value Error
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1.17 Thin cylinder under a uniform radial pressure (01-0038SSLSB_FEM)
Passed
1.17.1Description
1.17.2Background
1.17.2.1Model description
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Boundary conditions
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1.17.2.2Stresses in all points
Reference solution
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Finite elements modeling
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1.17.2.3Cylinder deformation in all points
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1.17.2.4Theoretical results
Solver Result name Result description Reference value
1.17.3Calculated results
Result name Result description Value Error
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1.18 Vibration mode of a thin piping elbow in plane (case 3) (01-0013SDLLB_FEM)
Passed
1.18.1Description
1.18.2Background
1.18.2.1Model description
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Geometry
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1.18.2.2Eigen mode frequencies
Reference solution
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Finite elements modeling
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Eigen mode shapes
1.18.2.3Theoretical results
Solver Result name Result description Reference value
1.18.3Calculated results
Result name Result description Value Error
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1.19 Double fixed beam with a spring at mid span (01-0015SSLLB_FEM)
Passed
1.19.1Description
1.19.2Background
1.19.2.1Model description
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■
Units
Geometry
■
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Materials properties
■
■
Boundary conditions
■
■
■
■
Loading
■
■
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1.19.2.2Displacement of the model in the linear elastic range
Reference solution
EI EI EI EI
EI EI EI EI
EI EI EI EI
EI EI EI EI
■
■
■
M
R
P
M
R
v
v
v
v
v
EI
33
333
333
333
33
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U
U EI k
v
u
v
u
EI K
v v
v v v v
y)unnecessar (usually026826
244423222 vvvv
EI l
P v
EI
P v v v
Finite elements modeling
■
■
■
Deformed shape
1.19.2.3Theoretical results
Solver Result name Result description Reference value
1.19.3Calculated results
Result name Result description Value Error
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1.20 Beam on two supports considering the shear force (01-0041SSLLB_FEM)
Passed
1.20.1Description
1.20.2Background
1.20.2.1Model description
■
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Units
Geometry
Materials properties
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■
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Boundary conditions
■
►
►
■
Loading
■
■
1.20.2.2Vertical displacement of the model in the linear elastic range
Reference solution
cm017.1105.0912.043.16
3.012
22859384
30020246
5462228593848
30020246
448
33
6
x x
x
x x
x
GS
Pl
EI
Pl v
shear
y
flexion
z
Finite elements modeling
■
■
■
Deformed shape
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1.20.2.3Theoretical results
Solver Result name Result description Reference value
1.20.3Calculated results
Result name Result description Value Error
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1.21 Torus with uniform internal pressure (01-0045SSLSB_FEM)
Passed
1.21.1Description
1.21.2Background
1.21.2.1Model description
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Units
Geometry
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Materials properties
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Boundary conditions
■
■
Loading
■
■
1.21.2.2Stresses
Reference solution
Finite elements modeling
■
■
■
1.21.2.3Cylinder deformation
Reference solution
■
Finite elements modeling
■
■
■
1.21.2.4Theoretical results
Solver Result name Result description Reference value
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1.21.3Calculated results
Result name Result description Value Error
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1.22 Short beam on simple supports (on the neutral axis) (01-0017SDLLB_FEM)
Passed
1.22.1Description
1.22.2Background
■ Software
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Units
Geometry
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■
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Materials properties
■
■ ■
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Boundary conditions
■
►
►
■
Loading
■
■
1.22.2.1Eigen modes frequencies
Reference solution
E
Finite elements modeling
■
■
■
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Eigen mode shapes
1.22.2.2Theoretical results
Solver Result name Result description Reference value
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1.22.3Calculated results
Result name Result description Value Error
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1.23 Annular thin plate fixed on a hub (repetitive circular structure) (01-0022SDLSB_FEM)
Passed
1.23.1Description
1.23.2Background
1.23.2.1Model description
■
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Units
Geometry
■
■
■
Material properties
■
■
■
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Boundary conditions
■
■
Loading
■■
1.23.2.2Eigen modes frequencies
Reference solution
Finite elements modeling
■
■
■
1.23.2.3Theoretical results
Solver Result name Result description Reference value
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1.23.3Calculated results
Result name Result description Value Error
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1.24 Double fixed beam (01-0016SDLLB_FEM)
Passed
1.24.1Description
1.24.2Background
1.24.2.1Model description
■
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Units
Geometry
■
■
■
Materials properties
■
■
■
Boundary conditions
■
■
Loading
■
■
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1.24.2.2Displacement of the model in the linear elastic range
Reference solution
E EI
Pl v
Finite elements modeling
■
■
■
Deformed shape
1.24.2.3Eigen mode frequencies of the model in the linear elastic range
Reference solution
S
I E
Lf nn
Finite elements modeling
■
■
■
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Modal deformations
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1.24.2.4Theoretical results
Solver Result name Result description Reference value
1.24.3Calculated results
Result name Result description Value Error
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1.25 Thin circular ring hanged on an elastic element (01-0014SDLLB_FEM)
Passed
1.25.1Description
1.25.2Background
1.25.2.1Model description
■
■
■
Units
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Geometry
■
■
■
►
►
■►
►
►
Materials properties
■
■
■
Boundary conditions
■■
Loading
■
■
1.25.2.2Eigen mode frequencies
Reference solutions
Finite elements modeling■
■
■
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Eigen mode shapes
1.25.2.3Theoretical results
Solver Result name Result description Reference value
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1.25.3Calculated results
Result name Result description Value Error
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1.26 Short beam on simple supports (eccentric) (01-0018SDLLB_FEM)
Passed
1.26.1Description
1.26.2Background
1.26.2.1Model description
■
■
■
Units
Geometry
■
■
■
■
■
Materials properties
■
■
■
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Boundary conditions
■
►
►
■
Loading
■
■
1.26.2.2Eigen modes frequencies
Reference solution
Finite elements modeling
■
■
■
Eigen modes shape
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1.26.2.3Theoretical results
Solver Result name Result description Reference value
1.26.3Calculated results
Result name Result description Value Error
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1.27 Thin cylinder under a uniform axial load (01-0042SSLSB_FEM)
Passed
1.27.1Description
1.27.2Background
1.27.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
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Boundary conditions
■
►
►
■
Loading
■
■
1.27.2.2Stress in all points
Reference solution
Finite elements modeling
■
■
■
1.27.2.3Cylinder deformation at the free end
Reference solution
■
■
Finite elements modeling
■
■
■
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Deformation shape
1.27.2.4Theoretical results
Solver Result name Result description Reference value
1.27.3Calculated results
Result name Result description Value Error
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1.28 Spherical shell under internal pressure (01-0046SSLSB_FEM)
Passed
1.28.1Description
1.28.2Background
1.28.2.1Model description
■
■
■
Units
Geometry
■
■
■
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Materials properties
■
■
Boundary conditions
■
■
Loading
■
■
1.28.2.2Stresses
Reference solution
Finite elements modeling
■
■
■
1.28.2.3Cylinder deformation
Reference solution
■
Finite elements modeling
■
■
■
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Deformed shape
1.28.2.4Theoretical results
Solver Result name Result description Reference value
1.28.3Calculated results
Result name Result description Value Error
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1.29 Simply supported square plate (01-0036SSLSB_FEM)
Passed
1.29.1Description
1.29.2Background
1.29.2.1Model description
■
■
■
Units
Geometry
■
■
Materials properties
■
■
■
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Boundary conditions
■
►
►
■
Loading
■
■
1.29.2.2Vertical displacement at O
Reference solution
Finite elements modeling
■
■
■
Deformed shape
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1.29.2.3Theoretical results
Solver Result name Result description Reference value
1.29.3Calculated results
Result name Result description Value Error
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1.30 Square plate under planar stresses (01-0039SSLSB_FEM)
Passed
1.30.1Description
1.30.2Background
1.30.2.1Model description
■
■
■
1;1,
Units
Geometry
■
■
Materials properties
■
■
Boundary conditions
■
■
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Loading
■
■
1.30.2.2Displacement of the model in the linear elastic range
Reference solution
Finite elements modeling
■
■
■
Deformed shape
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1.30.2.3Theoretical results
Solver Result name Result description Reference value
1.30.3Calculated results
Result name Result description Value Error
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1.31 Fixed thin arc in planar bending (01-0027SSLLB_FEM)
Passed
1.31.1Description
1.31.2Background
1.31.2.1Model description
■
■
■
Units
Geometry
■
■
►
►
►
►
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Materials properties
Boundary conditions
■
■
Loading
■
►
►
►
■
1.31.2.2Displacements at B
Reference solution
■
■
■
Finite elements modeling
■
■■
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Results shape
1.31.2.3Theoretical resultsSolver Result name Result description Reference value
1.31.3Calculated results
Result name Result description Value Error
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1.32 Simply supported rectangular plate under a uniform load (01-0052SSLSB_FEM)
Passed
1.32.1Description
1.32.2Background
1.32.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
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Boundary conditions
■
■
Loading
■
■
1.32.2.2Vertical displacement and bending moment at the center of the plate
Reference solution
Finite elements modeling
■
■
■
1.32.3Theoretical Background
Solver Result name Result description Reference value
1.32.4Calculated results
Result name Result description Value Error
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1.33 Shear plate perpendicular to the medium surface (01-0055SSLSB_FEM)
Passed
1.33.1Description
1.33.2Background
1.33.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
Boundary conditions
■
■
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Loading
■
►
►
■
1.33.2.2Vertical displacement at C
Reference solution
Finite elements modeling
■
■
■
Deformed shape
1.33.2.3Theoretical results
Solver Result name Result description Reference value
1.33.3Calculated results
Result name Result description Value Error
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1.34 Triangulated system with hinged bars (01-0056SSLLB_FEM)
Passed
1.34.1Description
1.34.2Background
1.34.2.1Model description
■
■
■
Units
Geometry
■ ■
■
Materials properties
■
■
Boundary conditions
■
►
►
■
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Loading
■
►
►
►
■
1.34.2.2Tension force in BD bar
Reference solution
Finite elements modeling
■
■
■
1.34.2.3Vertical displacement at D
Reference solution
Finite elements modeling
■
■
■
Deformed shape
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1.34.2.4Theoretical results
Solver Result name Result description Reference value
1.34.3Calculated results
Result name Result description Value Error
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1.35 A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0061SSLSB_FEM)
Passed
1.35.1Description
1.35.2Background
1.35.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
Boundary conditions
■
■
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Loading
■
■
1.35.2.2Vertical displacement at C
Reference solution
Finite elements modeling
■
■
■
1.35.2.3Theoretical results
Solver Result name Result description Reference value
1.35.3Calculated results
Result name Result description Value Error
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1.36 Thin cylinder under a hydrostatic pressure (01-0043SSLSB_FEM)
Passed
1.36.1Description
1.36.2Background
1.36.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
Boundary conditions
■
■
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Loading
■
■
1.36.2.2Stresses
Reference solution
Finite elements modeling
■
■
■
1.36.2.3Cylinder deformation
Reference solution
■
■
Finite elements modeling■
■
■
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Deformation shape
1.36.2.4Theoretical results
Solver Result name Result description Reference value
1.36.3Calculated results
Result name Result description Value Error
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1.37 Slender beam on three supports (01-0025SSLLB_FEM)
Passed
1.37.1Description
1.37.2Background
1.37.2.1Model description
■
■
■
Units
Geometry
■
■
Materials properties
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Boundary conditions
■
►
►
►
■
Loading
■
■
1.37.2.2Bending moment at B
Reference solution
■
Finite elements modeling
■
■
■
Results shape
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1.37.2.3Reaction in B
Reference solution
■
Finite elements modeling
■
■
■
1.37.2.4Vertical displacement at B
Reference solution
■
Finite elements modeling
■
■
■
Results shape
1.37.2.5Theoretical results
Solver Result name Result description Reference value
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1.37.3Calculated results
Result name Result description Value Error
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1.38 Bending effects of a symmetrical portal frame (01-0023SDLLB_FEM)
Passed
1.38.1Description
1.38.2Background
1.38.2.1Model description
■
■
■
Units
Geometry
■
■
■
■
■
■
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Materials properties
■
■
■
Boundary conditions
■
■
Loading
■
■
1.38.2.2Eigen modes frequencies
Reference solution
Finite elements modeling
■
■
■
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Deformed shape
1.38.2.3Theoretical results
Solver Result name Result description Reference value
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1.38.3Calculated results
Result name Result description Value Error
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1.39 A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0058SSLSB_FEM)
Passed
1.39.1Description
1.39.2Background
1.39.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
Boundary conditions
■
■
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Loading
■
■
1.39.2.2Vertical displacement at C
Reference solution
Finite elements modeling
■
■
■
1.39.2.3Theoretical results
Solver Result name Result description Reference value
1.39.3Calculated results
Result name Result description Value Error
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1.40 A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (01-0066SSLSB_FEM)
Passed
1.40.1Description
1.40.2Background
1.40.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
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Boundary conditions
■
■
Loading
■■
1.40.2.2Vertical displacement at point C (center of the plate)
Reference solution
Finite elements modeling
■
■■
1.40.2.3Theoretical results
Solver Result name Result description Reference value
1.40.3Calculated results
Result name Result description Value Error
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1.41 Reactions on supports and bending moments on a 2D portal frame (Rafters) (01-0077SSLPB_FEM)
Passed
1.41.1Description
1.41.2Background
1.41.2.1Model description
■
■
■
1.41.2.2Moments and actions on supports M.R. calculation on a 2D portal frame.
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1.41.2.3Theoretical results
Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the chords
Solver Result name Result description Reference value
1.41.3Calculated results
Result name Result description Value Error
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1.42 A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0057SSLSB_FEM)
Passed
1.42.1Description
1.42.2Background
1.42.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
Boundary conditions
■
■
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Loading
■
■
1.42.2.2Vertical displacement at C
Reference solution
Finite elements modeling
■
■
■
1.42.2.3Theoretical results
Solver Result name Result description Reference value
1.42.3Calculated results
Result name Result description Value Error
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1.43 Slender beam on two fixed supports (01-0024SSLLB_FEM)
Passed
1.43.1Description
1.43.2Background
1.43.2.1Model description
■
■
■
Units
Geometry
■
■
Materials properties
■
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Boundary conditions
■
■
Loading
■
►
►
►
►
►
■
1.43.2.2Shear force at G
Reference solution
■
Finite elements modeling
■
■
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Results shape
1.43.2.3Bending moment in G
Reference solution
■
Finite elements modeling
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Results shape
1.43.2.4Vertical displacement at G
Reference solution
■
Finite elements modeling
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Results shape
1.43.2.5Horizontal reaction at A
Reference solution
■
Finite elements modeling
■
■
■
1.43.2.6Theoretical results
Solver Result name Result description Reference value
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1.43.3Calculated results
Result name Result description Value Error
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1.44 Slender beam of variable rectangular section with fixed-free ends (ß=5) (01-0085SDLLB_FEM)
Passed
1.44.1Description
1.44.2Background
1.44.2.1Model description
■
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Units
Geometry
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Materials properties
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Boundary conditions
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Loading
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1.44.2.2Reference results
Reference values
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1.44.2.3Theoretical results
TheoreticalFrequency
Result name Result description Reference value
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1.44.3Calculated results
Result name Result description Value Error
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1.45 Vibration mode of a thin piping elbow in space (case 3) (01-0069SDLLB_FEM)
Passed
1.45.1Description
1.45.1.1Model description
■
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Units
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Geometry
■
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Materials properties
■
■
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Boundary conditions
■
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Loading
■
■
1.45.1.2Eigen modes frequencies
Reference solution
■
Finite elements modeling
■
■
■
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Eigen mode shapes
1.45.1.3Theoretical results
Reference
Solver Result name Result description Reference value
1.45.2Calculated results
Result name Result description Value Error
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1.46 Fixed thin arc in out of plane bending (01-0028SSLLB_FEM)
Passed
1.46.1Description
1.46.2Background
1.46.2.1Model description
■
■
■
Units
Geometry
■
■
►
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Materials properties
■
■
Boundary conditions
■■
Loading
■
■
1.46.2.2Displacements at B
Reference solution
Finite elements modeling
■
■
■
1.46.2.3Moments at = 15°
Reference solution
■
■
Finite elements modeling
■
■
■
1.46.2.4Theoretical results
Solver Result name Result description Reference value
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1.46.3Calculated results
Result name Result description Value Error
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1.47 Fixed/free slender beam with eccentric mass or inertia (01-0096SDLLB_FEM)
Passed
1.47.1Description
1.47.2Background
1.47.2.1Model description
■
■
■■
1.47.2.2Problem data
Units
Geometry
■
■
■
■
■
■
■
■
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Materials properties
■
■
■
■■
Boundary conditions
Loading
1.47.2.3Reference frequencies
Reference solutions
Hz 1.636
Hz 1.642
Hz 13.460
Hz 13.590
Uncertainty about the reference solutions
Finite elements modeling
■
■
■
■
Modal deformations
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1.47.2.4Theoretical results
Solver Result name Result description Reference value
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1.47.3Calculated results
Result name Result description Value Error
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1.48 Membrane with hot point (01-0099HSLSB_FEM)
Passed
1.48.1Description
1.48.2Background
■ NAFEMS, Test T1
Analysis
■
1.48.2.1Problem data
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Units
Geometry / meshing
Materials properties
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
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1.48.3 yy stress at point A:
Reference solution:
Finite elements modeling
■
■
■
1.48.3.1Theoretical results
Solver Result name Result description Reference value
1.48.4Calculated results
Result name Result description Value Error
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1.49 Double cross with hinged ends (01-0097SDLLB_FEM)
Passed
1.49.1Description
1.49.2Background
■
■
1.49.2.1Problem data
Units
Geometry
■
■
■
■
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Materials properties
■
■
Boundary conditions
■■
Loading
1.49.2.2Reference frequencies
Finite elements modeling
■
■
Modal deformations
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1.49.2.3Theoretical results
Solver Result name Result description Reference value
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1.49.3Calculated results
Result name Result description Value Error
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1.50 Stiffen membrane (01-0040SSLSB_FEM)
Passed
1.50.1Description
1.50.2Background
1.50.2.1Model description
■
■
■
1;1,
Units
Geometry
■
■
■
Materials properties
■
■
Boundary conditions
■
■
Loading
■
■
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1.50.2.2Results of the model in the linear elastic range
Reference solution
MPa96.17 N/cm1796;1
MPa98.8 N/cm898;0
0;1
for181
MPa55.11 N/cm1155;1
MPa77.5 N/cm577;0
0;1
for
MPa49.38 N/cm3849;1
MPa24.19 N/cm1924;0
0;1
for121
3410.97510.367410.2
6000
21
1
1
2
3
2
xy1
2
xy1
xy1
1
2
yy1
2
yy1
yy1
11
2
xx1
2
xx1
xx1
21
4
66
222
b
u E
a
u E
cm
a
ES
ba
eabE
F
K
F u
A xy
xx yy
A xx
A
Finite elements modeling
■
■
■
Deformed shape
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1.50.2.3Theoretical results
Solver Result name Result description Reference value
1.50.3Calculated results
Result name Result description Value Error
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1.51 A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (01-0062SSLSB_FEM)
Passed
1.51.1Description
1.51.2Background
■ Software
■
■
1.51.2.1Model description
Units
Geometry
■
■
■
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Materials properties
■
■
■
Boundary conditions
■
■
Loading
■
■
1.51.2.2Vertical displacement at point C (center of the plate)
Reference solution
Finite elements modeling
■
■
■
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1.51.2.3Theoretical results
Solver Result name Result description Reference value
1.51.3Calculated results
Result name Result description Value Error
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1.52 A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (01-0063SSLSB_FEM)
Passed
1.52.1Description
1.52.2Background
1.52.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
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Boundary conditions
■
■
Loading
■■
1.52.2.2Vertical displacement at point C (the center of the plate)
Reference solution
Finite elements modeling
■
■■
1.52.2.3Theoretical results
Solver Result name Result description Reference value
1.52.3Calculated results
Result name Result description Value Error
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1.53 Beam on elastic soil, hinged ends (01-0034SSLLB_FEM)
Passed
1.53.1Description
1.53.2Background
1.53.2.1Model description
■
■
■
Units
Geometry
■
■
Materials properties
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Boundary conditions
■
►
►
■
Loading
■
►
►
►
►
■
1.53.2.2Displacement and support reaction at A
Reference solution
■
■
Finite elements modeling
■
■
■
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Deformed shape
1.53.2.3Displacement and bending moment at D
Reference solution
■
■
Finite elements modeling
■
■
■
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Bending moment diagram
1.53.2.4Theoretical results
Solver Result name Result description Reference value
1.53.3Calculated results
Result name Result description Value Error
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1.54 Beam on 3 supports with T/C (k = -10000 N/m) (01-0102SSNLB_FEM)
Passed
1.54.1Description
1.54.2Background
1.54.2.1Model description
■ internal GRAITEC
■
■
Units
Geometry
■
■
Materials properties
■
■
Boundary conditions
■
►
►
►
■
Loading
■
■
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1.54.2.2References solutions
Displacements
Mz Moments
Finite elements modeling
■
■
■
Deformed shape
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Moment diagram
1.54.2.3Theoretical results
Solver Result name Result description Reference value
1.54.3Calculated results
Result name Result description Value Error
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1.55 Design of a concrete floor with an opening (03-0208SSLLG_BAEL91)
Passed
1.55.1Description
1.55.2Background
1.55.2.1Model description
Calculation model: 2D
►
►
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■
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Slab geometry
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Support positions
Positions of punctual loads
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Global loading overview
Load Combinations
Code Numbers Type Title
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1.55.2.2Effel Structure Results
SLS max displacements (load combination 101)
Mx bending moment for ULS load combination
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My bending moment for ULS load combination
Mxy bending moment for ULS load combination
1.55.2.3Effel RC Expert Results
Main hypothesis
■
■
■
■
■
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Axi reinforcements
Ayi reinforcements
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Axs reinforcements
Ays reinforcements
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1.55.2.4Theoretical results
Solver Resultname
Result description Referencevalue
1.55.3Calculated results
Result name Result description Value Error
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1.56 Spherical dome under a uniform external pressure (01-0050SSLSB_FEM)
Passed
1.56.1Description
1.56.2Background
1.56.2.1Model description
■
■
■
Units
Geometry
■
■
■
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Materials properties
■
■
Boundary conditions
■■
Loading
■
■
1.56.2.2Horizontal displacement and exterior meridian stress
Reference solution
Finite elements modeling
■
■
■
Deformed shape
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1.56.2.3Theoretical results
Solver Result name Result description Reference value
1.56.3Calculated results
Result name Result description Value Error
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1.57 A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (01-0064SSLSB_FEM)
Passed
1.57.1Description
1.57.2Background
1.57.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
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Boundary conditions
■
■
Loading
■■
1.57.2.2Vertical displacement at point C (the center of the plate)
Reference solution
Finite elements modeling
■
■■
1.57.2.3Theoretical results
Solver Result name Result description Reference value
1.57.3Calculated results
Result name Result description Value Error
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1.58 System of two bars with three hinges (01-0002SSLLB_FEM)
Passed
1.58.1Description
1.58.2Background
1.58.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
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Boundary conditions
■
■
Loading
■
■
1.58.2.2Displacement of the model in C
Reference solution
Finite elements modeling
■
■
■
Displacement shape
1.58.2.3Bars stresses
Reference solutions
Finite elements modeling
■
■
■
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1.58.2.4Shape of the stress diagram
1.58.2.5Theoretical results
Solver Result name Result description Reference value
1.58.3Calculated results
Result name Result description Value Error
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1.59 Rectangular thin plate simply supported on its perimeter (01-0020SDLSB_FEM)
Passed
1.59.1Description
1.59.2Background
1.59.2.1Model description
■
■
■
Units
Geometry
■
■
■
■
►
►
►
►
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Materials properties
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
1.59.2.2Eigen modes frequencies
Reference solution
Finite elements modeling
■■
■
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Eigen mode shapes
1.59.2.3Theoretical results
Solver Result name Result description Reference value
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1.59.3Calculated results
Result name Result description Value Error
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1.60 Simply supported square plate under a uniform load (01-0051SSLSB_FEM)
Passed
1.60.1Description
1.60.2Background
1.60.2.1Model description
■
■
■
Units
Geometry
■
■
Materials properties
■
■
Boundary conditions
■
■
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Loading
■
■
1.60.2.2Vertical displacement and bending moment at the center of the plate
Reference solution
Finite elements modeling
■
■
■
1.60.2.3Theoretical result
Solver Result name Result description Reference value
1.60.3Calculated results
Result name Result description Value Error
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1.61 Plane portal frame with hinged supports (01-0089SSLLB_FEM)
Passed
1.61.1Description
1.61.2Background
1.61.2.1Model description
■
■
■
Units
Geometry
■
■
■
■
■
■
Materials properties
■
■
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Boundary conditions
Loading
■
■
■
■
1.61.2.2Calculation method used to obtain the reference solution
■
■
■
■
■
■
■
■
1.61.2.3Reference values
1.61.2.4Theoretical results
Solver Result name Result description Reference value
1.61.3Calculated results
Result name Result description Value Error
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1.62 Double hinged thin arc in planar bending (01-0029SSLLB_FEM)
Passed
1.62.1Description
1.62.2Background
1.62.2.1Model description
■
■
■
Units
Geometry
■
■
►
►
►
►
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Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
■
1.62.2.2Displacements at A, B and C
Reference solution
■
■
Finite elements modeling
■■
■
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Displacements shape
1.62.2.3Theoretical results
Solver Result name Result description Reference value
1.62.3Calculated results
Result name Result description Value Error
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1.63 A 3D bar structure with elastic support (01-0094SSLLB_FEM)
Passed
1.63.1Description
1.63.1.1Model description
■ Internal GRAITEC;
Analysis
■
Units
Geometry
■
■■
Materials properties
■
■
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Boundary conditions
■
■
Loading
■■
1.63.1.2Theoretical results
System solution
2
22 B
H L
■
1L
2x= where
.12
1.1
2
1)(.).1()(
ji ji uuuu L
xu
L
x xu
)(
)(
)(
)(
0000
0101
0000
0101
)(
)(
11
11
4
1
4
14
1
4
12
=
2
2
1
2
1
2
12
1
1
1
1
101
j
j
i
i
j
i
L T
e
T
v
v
u
v
u
L
ES
u
u
L
ES d
L
ES
d L
ES dx B B ES dV B H Bk
e
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L
Hsinand
2cos
L
B
)(
)(
)(
)(
22
2222
22
2222
:)D
H(arctan=5,1nodes1elementfor
5
5
1
1
22
22
22
22
31
V
U
V
U
H HB
H HB
HB B HB B
H HB
H HB
HB B HB B
L
ES K
■
)(
)(
)(
)(
0000
0101
0000
0101
)(
)(
11
11 :systemcoordinatelocalin the
4
K K sayWe
5
j
j
i
i
j
i
v
u
v
u
K u
u K k
)(
)(
)(
)(
1010
0000
1010
0000
':90=6,5nodes5elementfor
6
6
5
5
5
V
U
V
U
K K
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■
■
■
■
1759
1759
11
11 and
200
200
002
002
5
1
5
1
5
5
1
1
5
5
1
1
N
N
u
u
L
ES
V
U
V
U
L
B
L
H L
H
L
B L
B
L
H L
H
L
B
v
u
v
u
5
1
5
1
5
5
1
1
5
5
1
1
11
11 and
cossin00
sincos00
00cossin00sincos
N
N
u
u
L
ES
V
U
V U
v
u
vu
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Reference values
7182/1015
Finite elements modeling
■
■
■
Deformed shape
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Normal forces diagram
1.63.1.3Reference values
Solver Result name Result description Reference value
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1.63.2Calculated results
Result name Result description Value Error
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1.64 Cantilever beam in Eulerian buckling with thermal load (01-0092HFLLB_FEM)
Passed
1.64.1Description
1.64.2Background
1.64.2.1Model description
■ Euler theory;
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
Boundary conditions
■■
Loading
■
■
)
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1.64.2.2Displacement of the model in the linear elastic range
Reference solution
98696.0100000
98696
986964P 2
2
critical
N L
EI
Finite elements modeling
■
■
■
Deformed shape
1.64.2.3Theoretical results
Solver Result name Result description Reference value
1.64.3Calculated results
Result name Result description Value Error
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1.65 Beam on elastic soil, free ends (01-0032SSLLB_FEM)
Passed
1.65.1Description
1.65.2Background
1.65.2.1Model description
■
■
■
Units
Geometry
■
■
Materials properties
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Boundary conditions
■
►
►
■
Loading
■
■
1.65.2.2Bending moment and displacement at C
Reference solution
■
■
Finite elements modeling
■
■
■
Bending moment diagram
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1.65.2.3Displacements at A
Reference solution
■
■
Finite elements modeling
■
■
■
1.65.2.4Theoretical results
Solver Result name Result description Reference value
1.65.3Calculated results
Result name Result description Value Error
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1.66 Vibration mode of a thin piping elbow in space (case 2) (01-0068SDLLB_FEM)
Passed
1.66.1Description
1.66.2Background
1.66.2.1Model description
■
■
■
Units
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Geometry
■
■
■
■
■
■
■
■
■
■
►
►
►
►
►
Materials properties
■
■
■
Boundary conditions
■
►
►
►
■
Loading
■
■
1.66.2.2Eigen modes frequencies
Reference solution
■
Finite elements modeling
■
■
■
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Eigen mode shapes
1.66.2.3Theoretical results
Solver Result name Result description Reference value
1.66.3Calculated results
Result name Result description Value Error
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1.67 Thin square plate fixed on one side (01-0019SDLSB_FEM)
Passed
1.67.1Description
1.67.2Background
1.67.2.1Model description
■
■
■
Units
Geometry
■
■
■
►
►
►
►
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Materials properties
■
■
■
Boundary conditions
■
■
Loading
■
■
1.67.2.2Eigen modes frequencies
Reference solution
Finite elements modeling
■
■
■
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Eigen mode shapes
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1.67.2.3Theoretical results
Solver Result name Result description Reference value
1.67.3Calculated results
Result name Result description Value Error
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1.68 Fixed/free slender beam with centered mass (01-0095SDLLB_FEM)
Passed
1.68.1Description
1.68.2Background
■ Software
■
■
■
1.68.2.1Test data
Units
Geometry
■
■
■
■
■
■
■■
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Materials properties
■
■
■
Boundary conditions
■
■
Loading
1.68.2.2Reference results
Reference frequency
Uncertainty about the reference frequencies
■
■
Finite elements modeling
■
■
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Modal deformations
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1.68.2.3Theoretical results
Solver Result name Result description Reference value
1.68.3Calculated results
Result name Result description Value Error
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1.69 Slender beam of variable rectangular section (fixed-fixed) (01-0086SDLLB_FEM)
Passed
1.69.1Description
1.69.2Background
1.69.2.1Model description
■
■
■
Units
Geometry
■
■
■
►
►
■
►
►
►
Materials properties
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
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1.69.2.2Reference results
Calculation method used to obtain the reference solution
Reference values
Eigen modes
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1.69.2.3Theoretical results
Solver Result name Result description Reference value
1.69.3Calculated results
Result name Result description Value Error
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1.70 Truss with hinged bars under a punctual load (01-0031SSLLB_FEM)
Passed
1.70.1Description
1.70.2Background
■ Software
■
■
1.70.2.1Model description
Units
Geometry
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Materials properties
Boundary conditions
■
■
Loading
■
■
1.70.2.2Displacements at C and D
Reference solution
Finite elements modeling
■■
■
Displacements shape
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1.70.2.3Theoretical results
Solver Result name Result description Reference value
1.70.3Calculated results
Result name Result description Value Error
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1.71 Beam on 3 supports with T/C (k = 0) (01-0100SSNLB_FEM)
Passed
1.71.1Description
1.71.2Background
1.71.2.1Model description
■ internal GRAITEC
■
■
Units
Geometry
■
■
Materials properties
■
■
Boundary conditions
■
►
►
►
■
Loading
■
■
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1.71.2.2References solutions
Displacements
Mz Moments
Finite elements modeling
■
■
■
Deformed shape
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Moment diagrams
1.71.2.3Theoretical results
Solver Result name Result description Reference value
1.71.3Calculated results
Result name Result description Value Error
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1.72 Linear element in simple bending - without compressed reinforcement (02-0162SSLLB_B91)
Passed
1.72.1Description
1.72.1.1Model description
■ J. Perchat
■
■
Units
■
■■
■
Geometry
■
■
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
►
►
►
■
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Reinforced concrete calculation hypothesis:
■
■
Span Concrete Reinforcement Application Concrete Cracking
1.72.1.2Reinforcement calculation
Reference solution
Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8
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Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8
Finite elements modeling
■
■
■
1.72.1.3Theoretical results
Reference
Solver Result name Result description Reference value
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1.72.2Calculated results
Result name Result description Value Error
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1.73 Non linear system of truss beams (01-0104SSNLB_FEM)
Passed
1.73.1Description
1.73.2Background
1.73.2.1Model description
■ internal GRAITEC
■
■
Units
Geometry■
■
Materials properties
Boundary conditions
■
►
►
■
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Loading
■
■
1.73.2.2References solutions
Displacements
Finite elements modeling
■
■
■
Deformed shape
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Normal forces
1.73.2.3Theoretical results
Solver Result name Result description Reference value
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1.73.3Calculated results
Result name Result description Value Error
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1.74 Simple supported beam in free vibration (01-0098SDLLB_FEM)
Passed
1.74.1Description
1.74.2Background
■ NAFEMS, FV5
Analysis type:
■
1.74.2.1Problem data
Units
Geometry
■
■
■
Materials properties
■
■
■
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Boundary conditions
■
►
►
■
Loading
1.74.2.2Reference frequencies
Finite elements modeling
■
■
Modal deformations
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1.74.2.3Theoretical results
Solver Result name Result description Reference value
1.74.3Calculated results
Result name Result description Value Error
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1.75 Study of a mast subjected to an earthquake (02-0112SMLLB_P92)
Passed
1.75.1Description
1.75.1.1Model description
■ internal GRAITEC
■
■
1.75.1.2Material strength model
Units
Geometry
■
■
■
■
■
■
Masses
■
■
Materials properties
■
■
■
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Boundary conditions
■
Loading
■
Finite elements modeling
1.75.1.3Seismic hypothesis in conformity with PS92 regulation
■
■
■
■
■
1.75.1.4Modal analysis
Eigen periods reference solution
Modal vectors
Normalizing relative to the mass
;
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Modal deformations
1.75.1.5Spectral study
Design spectrum
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Reference participation factors
Pseudo-acceleration
Reference modal displacement
Equivalent static forces
Displacement at the top of the mast
Shear force at the top of the mast
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Moment at the base
1.75.1.6Theoretical results
Reference
Solver Result name Result description Reference value
1.75.2Calculated results
Result name Result description Value Error
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1.76 Design of a Steel Structure according to CM66 (03-0206SSLLG_CM66)
Passed
1.76.1Description
1.76.2Background
1.76.2.1Model description
Calculation model: Simple
■
►
►
►
►
■
Model preview
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Structure’s load case
Code No. Type Title
1.76.2.2Effel Structure results
Displacement Envelope (“CMCD" load combinations)
Envelope of linear element forces
D DX DY DZEnv. Case No.
Max.
location (cm) (cm) (cm) (cm)
Envelope of forces on linear elements (“CMCFN” load combinations)
Envelope of linear element forces
Fx Fy Fz Mx My MzEnv. Case No. MaxSite
(T) (T) (T) (T*m) (T*m) (T*m)
Envelope of linear element stresses (“CMCFN” load combinations)
Envelope of linear element stresses
sxxMax sxyMax sxzMax sFxx sMxxMaxEnv. Case No. MaxSite
(MPa) (MPa) (MPa) (MPa) (MPa)
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1.76.2.3CM66 Effel Expertise results
Hypotheses
■
■
■
■
■
■
■
■
■
■
■
■
The results of the optimization given below correspond to an iteration of the finite elements calculation.
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Deflection verification
Ratio
Max values on the element
■
■
■
CM Stress diagrams
Work ratio
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Stresses
Max values on the element■
■
■
Buckling lengths
Lfy
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Lfz
Lateral-torsional buckling lengths
Ldi
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Lds
Optimization
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1.76.2.4Theoretical results
Solver Result name Result description Reference value
Solver Result name Result description Reference value
Solver Result name Result description Reference value
Solver Result name Result description Rate (%) Final section
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1.76.3Calculated results
Result name Result description Value Error
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1.77 Linear element in combined bending/tension - without compressed reinforcements - Partiallytensioned section (02-0158SSLLB_B91)
Passed
1.77.1Description
1.77.1.1Model description
■ J. Perchat
■
■
Units
■
■
■
■
Geometry
■
■
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
►
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►
►
■
Reinforced concrete calculation hypothesis:
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1.77.1.2Reinforcement calculation
Reference solution
Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8
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Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8
Finite elements modeling
■
■
■
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1.77.1.3Theoretical results
Solver Result name Result description Reference value
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1.77.2Calculated results
Result name Result description Value Error
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1.78 Thin cylinder under its self weight (01-0044SSLSB_MEF)
Passed
1.78.1Description
1.78.2Background
1.78.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■ ■
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Boundary conditions
■
►
►
■
Loading
■
■
1.78.2.2Stresses
Reference solution
Finite elements modeling
■
■
■
1.78.2.3Cylinder deformation
Reference solution
■
■
1.78.2.4Theoretical results
Solver Result name Result description Reference value
1.78.3Calculated results
Result name Result description Value Error
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1.79 Pinch cylindrical shell (01-0048SSLSB_FEM)
Passed
1.79.1Description
1.79.2Background
■ Software
■
■
1.79.2.1Model description
Units
Geometry
■■
■
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Materials properties
■
■
Boundary conditions
■
■
Loading
■
■
1.79.2.2Vertical displacement at point A
Reference solution
Finite elements modeling
■
■
■
1.79.2.3Theoretical result
Solver Result name Result description Reference value
1.79.3Calculated results
Result name Result description Value Error
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1.80 Spherical shell with holes (01-0049SSLSB_FEM)
Passed
1.80.1Description
1.80.2Background
■ Software
■
■
1.80.2.1Model description
Units
Geometry
■
■
■
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Materials properties
■
■
Boundary conditions
■
■
Loading
■
■
1.80.2.2Horizontal displacement at point A
Reference solution
Finite elements modeling
■
■
■
Deformed shape
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1.80.3Theoretical Background
Solver Result name Result description Reference value
1.80.4Calculated results
Result name Result description Value Error
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1.81 Simply supported rectangular plate loaded with punctual force and moments (01-0054SSLSB_FEM)
Passed
1.81.1Description
1.81.2Background
1.81.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
Boundary conditions
■
■
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Loading
■
►
►
►
►■
1.81.2.2Vertical displacement at C
Reference solution
Finite elements modeling
■
■
■
Deformed shape
1.81.2.3Theoretical results
Solver Result name Result description Reference value
1.81.3Calculated results
Result name Result description Value Error
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1.82 Simply supported rectangular plate under a uniform load (01-0053SSLSB_FEM)
Passed
1.82.1Description
1.82.2Background
1.82.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
Boundary conditions
■
■
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Loading
■
■
1.82.2.2Vertical displacement and bending moment at the center of the plate
Reference solution
Finite elements modeling
■
■
■
1.82.2.3Theoretical results
Solver Result name Result description Reference value
1.82.3Calculated results
Result name Result description Value Error
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1.83 A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0059SSLSB_FEM)
Passed
1.83.1Description
1.83.2Background
1.83.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
Boundary conditions
■
■
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Loading
■
■
1.83.2.2Vertical displacement at C
Reference solution
Finite elements modeling
■
■
■
1.83.2.3Theoretical resultsSolver Result name Result description Reference value
1.83.3Calculated results
Result name Result description Value Error
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1.84 A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0060SSLSB_FEM)
Passed
1.84.1Description
1.84.2Background
1.84.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
Boundary conditions
■
■
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Loading
■
■
1.84.2.2Vertical displacement at C
Reference solution
Finite elements modeling
■
■
■
1.84.2.3Theoretical results
Solver Result name Result description Reference value
1.84.3Calculated results
Result name Result description Value Error
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1.85 Short beam on two hinged supports (01-0084SSLLB_FEM)
Passed
1.85.1Description
1.85.2Background
1.85.2.1Model description
■
■
■
Units
Geometry
■
■
■
■
Materials properties
■
■
Boundary conditions
■
■
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Loading
1.85.2.2Reference results
Reference values
-1.25926 x 10-3
1.85.2.3Theoretical results
Solver Result name Result description Reference value
1.85.3Calculated results
Result name Result description Value Error
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1.86 Reactions on supports and bending moments on a 2D portal frame (Columns) (01-0078SSLPB_FEM)
Passed
1.86.1Description
1.86.2Background
1.86.2.1Model description
■ Design and
■
■
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1.86.2.2Moments and reactions on supports M.R. calculation on a 2D portal frame.
h H qh
M E B 2
²
1.86.2.3Theoretical results
Solver Result name Result description Reference value
1.86.3Calculated results
Result name Result description Value Error
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1.87 Vibration mode of a thin piping elbow in space (case 1) (01-0067SDLLB_FEM)
Passed
1.87.1Description
1.87.2Background
1.87.2.1Model description
■
■
■
Units
Geometry
■
■
■
■
■
■
■
■
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Materials properties
■
■
■
Boundary conditions
■
■
Loading
■
■
1.87.2.2Eigen modes frequencies
Reference solution
■
Finite elements modeling
■
■
■
Eigen mode shapes
1.87.2.3Theoretical results
Solver Result name Result description Reference value
1.87.3Calculated results
Result name Result description Value Error
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1.88 Beam on 3 supports with T/C (k -> infinite) (01-0101SSNLB_FEM)
Passed
1.88.1Description
1.88.2Background
1.88.2.1Model description
■ internal GRAITEC
■
■
Units
Geometry
■
■
Materials properties
■
■
Boundary conditions
■
►
►
►
■
Loading
■
■
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1.88.2.2References solutions
Displacements
Mz Moments
Finite elements modeling
■
■
■
Deformed shape
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Moment diagram
1.88.2.3Theoretical results
Solver Result name Result description Reference value
1.88.3Calculated results
Result name Result description Value Error
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1.89 Linear system of truss beams (01-0103SSLLB_FEM)
Passed
1.89.1Description
1.89.2Background
1.89.2.1Model description
■ internal GRAITEC
■
■
Units
Geometry■
■
Materials properties
Boundary conditions
■
►
►
■
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Loading
■
■
1.89.2.2References solutions
Displacements
N normal forces
Finite elements modeling
■
■■
Deformed shape
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Normal forces
1.89.2.3Theoretical results
Solver Result name Result description Reference value
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1.89.3Calculated results
Result name Result description Value Error
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1.90 Design of a 2D portal frame (03-0207SSLLG_CM66)
Passed
1.90.1Description
1.90.2Background
1.90.2.1Model description
Calculation model: 2D
►
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Model preview
Combinations
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1.90.2.2Effel Structure Results
Ridge displacements (combination 104)
Diagram of normal force envelope
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Envelope of bending moments diagram
1.90.2.3Effel Expert CM results
Main hypotheses
■
■
■
■
■
■
■
■
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Deflection verification
Ratio
CM Stress diagrams
Work ratio
Stresses
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Buckling lengths
Lfy
Lfz
Lateral-torsional buckling lengths
Ldi
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Lds
Optimization
1.90.2.4Theoretical results
Solver Result name Result description Reference value
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Solver Result name Result description Reference value
Solver Result name Result description Rate (%) Final section
1.90.3Calculated results
Result name Result description Value Error
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1.91 A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (01-0065SSLSB_FEM)
Passed
1.91.1Description
1.91.2Background
1.91.2.1Model description
■
■
■
Units
Geometry
■
■
■
Materials properties
■
■
■
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Boundary conditions
■
■
Loading
■■
1.91.2.2Vertical displacement at point C center of the plate)
Reference solution
Finite elements modeling
■
■■
1.91.2.3Theoretical results
Solver Result name Result description Reference value
1.91.3Calculated results
Result name Result description Value Error
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1.92 Double fixed beam in Eulerian buckling with a thermal load (01-0091HFLLB_FEM)
Passed
1.92.1Description
1.92.2Background
1.92.2.1Model description
■ Euler theory;
■
■
Units
Geometry
Materials properties
■
■
■
Boundary conditions
■
■
Loading
■
■
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1.92.2.2Displacement of the model in the linear elastic range
Reference solution
93.3724.117
925.29
k 724.117
2
P 2
2
critical
N L
EI
Finite elements modeling
■
■
■
Deformed shape of mode 1
1.92.2.3Theoretical results
Solver Result name Result description Reference value
1.92.3Calculated results
Result name Result description Value Error
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1.93 Verifying results of a steel beam subjected to dynamic temporal loadings (TTAD #14586)
Passed
1.93.1Description
1.93.2Background
The harmonic response of
1.93.2.1Model description
■■
■
Units
Geometry
■
■
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Materials properties
■
■
■
Boundary conditions
■
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Loading
■
►
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■
1.93.2.2Harmonic response from NE/Nastran V8
Reference solution
A
I E
L f n
n
2
2
2
Software NE/NASTRAN 8.0
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Finite elements modeling
■
■
■
1.93.2.3Reference results
Result name Result description Reference value
1.93.3Calculated results
Result name Result description Value Error
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1.94 Verifying the displacement results on linear elements for vertical seism (TTAD #11756)
Passed
1.94.1Description
1.95 Verifying Sxx results on beams (TTAD #11599)
Passed
1.95.1Description
1.96 Generating results for Torsors NZ/Group (TTAD #11633)
Passed
1.96.1Description
1.97 Verifying forces for triangular meshing on planar element (TTAD #11723)
Passed
1.97.1Description
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1.98 Verifying tension/compression supports on nonlinear analysis (TTAD #11518)
Passed
1.98.1Description
1.99 Verifying tension/compression supports on nonlinear analysis (TTAD #11518)
Passed
1.99.1Description
1.100 Generating a report with torsors per level (TTAD #11421)
Passed
1.100.1Description
1.101 Verifying the level mass center (TTAD #11573, TTAD #12315)
Passed
1.101.1Description
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1.102 Generating planar efforts before and after selecting a saved view (TTAD #11849)
Passed
1.102.1Description
1.103 Verifying stresses in beam with "extend into wall" property (TTAD #11680)
Passed
1.103.1Description
1.104 Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD#11854)
Passed
1.104.1Description
1.105 Verifying constraints for triangular mesh on planar elements (TTAD #11447)
Passed
1.105.1Description
1.106 Verifying diagrams for Mf Torsors on divided walls (TTAD #11557)
Passed
1.106.1Description
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1.107 Verifying forces results on concrete linear elements (TTAD #11647)
Passed
1.107.1Description
1.108 Verifying forces on a linear elastic support which is defined in a user workplane (TTAD#11929)
Passed
1.108.1Description
1.109 Verifying results on punctual supports (TTAD #11489)
Passed
1.109.1Description
1.110 Verifying the internal forces results for a simple supported steel beam
Passed
1.110.1Description
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1.111 Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontalforce (TTAD #13175)
Passed
1.111.1Description
1.112 Verifying nonlinear analysis results for frames with semi-rigid joints and rigid joints (TTAD#11495)
Passed
1.112.1Description
1.113 Verifying the main axes results on a planar element (TTAD #11725)
Passed
1.113.1Description
1.114 Verifying the display of the forces results on planar supports (TTAD #11728)
Passed
1.114.1Description
1.115 Verifying torsors on a single story coupled walls subjected to horizontal forces
Passed
1.115.1Description
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2 CAD, rendering and visualization
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2.1 System stability during section cut results verification (TTAD #11752)
Passed
2.1.1 Description
2.2 Verifying descriptive actors after creating analysis (TTAD #11589)
Passed
2.2.1 Description
2.3 Creating a circle (TTAD #11525)
Passed
2.3.1 Description
2.4 Verifying the display of punctual loads after changing the load case number (TTAD #11958)
Passed
2.4.1 Description
2.5 Verifying the display of a beam with haunches (TTAD #12299)
Passed
2.5.1 Description
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2.6 Modeling using the tracking snap mode (TTAD #10979)
Passed
2.6.1 Description
2.7 Verifying hide/show elements command (TTAD #11753)
Passed
2.7.1 Description
2.8 Verifying the coordinates system symbol (TTAD #11611)
Passed
2.8.1 Description
2.9 Verifying the grid text position (TTAD #11704)
Passed
2.9.1 Description
2.10 Verifying the "ghost display on selection" function for saved views (TTAD #12054)
Passed
2.10.1Description
2.11 Verifying drawing of joints in y-z plan (TTAD #12453)
Passed
2.11.1Description
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2.12 Verifying the snap points behavior during modeling (TTAD #11458)
Passed
2.12.1Description
2.13 Verifying holes in horizontal planar elements after changing the level height (TTAD #11490)
Passed
2.13.1Description
2.14 Creating a camera (TTAD #11526)
Passed
2.14.1Description
2.15 Verifying the local axes of a section cut (TTAD #11681)
Passed
2.15.1Description
2.16 Verifying the grid text position (TTAD #11657)
Passed
2.16.1Description
2.17 Generating combinations (TTAD #11721)
Passed
2.17.1Description
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2.18 Verifying rotation for steel beam with joint (TTAD #12592)
Passed
2.18.1Description
2.19 Turning on/off the "ghost" rendering mode (TTAD #11999)
Passed
2.19.1Description
2.20 Verifying the steel connections modeling (TTAD #11698)
Passed
2.20.1Description
2.21 Verifying the "ghost" display after changing the display colors (TTAD #12064)
Passed
2.21.1Description
2.22 Verifying the program behavior when launching the analysis on a model with overlapped loads(TTAD #11837)
Passed
2.22.1Description
2.23 Verifying the representation of elements with HEA cross section (TTAD #11328)
Passed
2.23.1Description
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2.24 Verifying the descriptive model display after post processing results in analysis mode (TTAD#11475)
Passed
2.24.1Description
2.25 Verifying the display of elements with compound cross sections (TTAD #11486)
Passed
2.25.1Description
2.26 Verifying the fixed load scale function (TTAD #12183).
Passed
2.26.1Description
2.27 Verifying the dividing of planar elements which contain openings (TTAD #12229)
Passed
2.27.1Description
2.28 Moving a linear element along with the support (TTAD #12110)
Passed
2.28.1Description
2.29 Verifying the program behavior when trying to create lintel (TTAD #12062)
Passed
2.29.1Description
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2.30 Creating base plate connections for non-vertical columns (TTAD #12170)
Passed
2.30.1Description
2.31 Verifying annotation on selection (TTAD #12700)
Passed
2.31.1Description
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3 Climatic generator
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3.1 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12835)
Passed
3.1.1 Description
3.2 EC1: Generating snow loads on two side by side buildings with gutters (TTAD #12806)
Passed
3.2.1 Description
3.3 EC1: generating wind loads on a square based lattice structure with compound profiles andautomatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12744)
Passed
3.3.1 Description
3.4 EC1: Generating wind loads on a square based structure according to UK standards (BS EN1991-1-4:2005) (TTAD #12608)
Passed
3.4.1 Description
3.5 EC1: generating Cf and Cp,net wind loads on an isolated roof with one slope (DEV2013#4.3)
Passed
3.5.1 Description
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3.6 EC1: generating snow loads on a 3 slopes 3D portal frame with parapets (NF EN 1991-1-3/NA)(TTAD #11111)
Passed
3.6.1 Description
3.7 EC1: verifying the snow loads generated on a monopitch frame (TTAD #11302)
Passed
3.7.1 Description
3.8 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11699)
Passed
3.8.1 Description
3.9 EC1: generating wind loads on a 2 slopes 3D portal frame using the Romanian national annex(TTAD #11687)
Passed
3.9.1 Description
3.10 EC1: generating snow loads on a 2 slopes 3D portal frame with roof thickness greater than theparapet height (TTAD #11943)
Passed
3.10.1Description
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3.11 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12841)
Passed
3.11.1Description
3.12 EC1: generating wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls (TTAD#11937)
Passed
3.12.1Description
3.13 EC1: Generating snow loads on a 4 slopes with gutters building. (TTAD #12719)
Passed
3.13.1Description
3.14 EC1: generating snow loads on a 2 slopes 3D portal frame (NF EN 1991-1-3/NA) (VT : 3.4 -Snow - Example A)
Passed
3.14.1Description
3.15 EC1: generating wind loads on a 55m high structure according to German standards (DIN EN1991-1-4/NA) (DEV2012 #3.12)
Passed
3.15.1Description
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3.16 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.3 - Wind- Example C)
Passed
3.16.1Description
3.17 EC1: Generating snow loads on a single slope with lateral parapets (TTAD #12606)
Passed
3.17.1Description
3.18 EC1: wind load generation on a signboard
Passed
3.18.1Description
3.19 EC1 NF: generating wind loads on a 3D portal frame with 2 slopes roof (TTAD #11932)
Passed
3.19.1Description
3.20 EC1: wind load generation on a high building with horizontal roof
Passed
3.20.1Description
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3.21 EC1: generating snow loads on a 3D portal frame with a roof which has a small span (< 5m)and a parapet (TTAD #11735)
Passed
3.21.1Description
3.22 EC1: wind load generation on simple 3D portal frame with 4 slopes roof (TTAD #11604)
Passed
3.22.1Description
3.23 EC1: generating wind loads on duopitch multispan roofs with pitch < 5 degrees (TTAD #11852)
Passed
3.23.1Description
3.24 EC1: wind load generation on a simple 3D structure with horizontal roof
Passed
3.24.1Description
3.25 EC1: wind load generation on a simple 3D portal frame with 2 slopes roof (TTAD #11602)
Passed
3.25.1Description
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3.26 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528)
Passed
3.26.1Description
3.27 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528)
Passed
3.27.1Description
3.28 EC1: generating wind loads on a 3D portal frame with one slope roof (NF EN 1991-1-4/NA) (VT :3.2 - Wind - Example B)
Passed
3.28.1Description
3.29 EC1: wind loads on a triangular based lattice structure with compound profiles and userdefined "n" (NF EN 1991-1-4/NA) (TTAD #12276)
Passed
3.29.1Description
3.30 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex(TTAD #11569)
Passed
3.30.1Description
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3.31 EC1: Verifying the geometry of wind loads on an irregular shed. (TTAD #12233)
Passed
3.31.1Description
3.32 EC1: generating snow loads on duopitch multispan roofs according to German standards (DINEN 1991-1-3/NA) (DEV2012 #3.13)
Passed
3.32.1Description
3.33 NV2009: generating wind loads and snow loads on a simple structure with planar support(TTAD #11380)
Passed
3.33.1Description
3.34 Generating the description of climatic loads report according to EC1 Romanian standards(TTAD #11688)
Passed
3.34.1Description
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3.35 EC1: generating snow loads on two side by side roofs with different heights, according toGerman standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)
Passed
3.35.1Description
3.36 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex(TTAD #11570)
Passed
3.36.1Description
3.37 EC1: Verifying the wind loads generated on a building with protruding roof (TTAD #12071,#12278)
Passed
3.37.1Description
3.38 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.1 - Wind- Example A)
Passed
3.38.1Description
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3.39 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11531)
Passed
3.39.1Description
3.40 EC1: generating wind loads on a triangular based lattice structure with compound profiles andautomatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12276)
Passed
3.40.1Description
3.41 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12808)
Passed
3.41.1Description
3.42 EC1: Generating snow loads on a 4 slopes with gutters building (TTAD #12716)
Passed
3.42.1Description
3.43 EC1: generating Cf and Cp,net wind loads on an isolated roof with double slope (DEV2013#4.3)
Passed
3.43.1Description
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3.44 EC1: generating snow loads on two close roofs with different heights according to Czechstandards (CSN EN 1991-1-3) (DEV2012 #3.18)
Passed
3.44.1Description
3.45 EC1: snow on a 3D portal frame with horizontal roof and parapet with height reduction (TTAD#11191)
Passed
3.45.1Description
3.46 EC1: generating wind loads on double slope 3D portal frame according to Czech standards(CSN EN 1991-1-4) (DEV2012 #3.18)
Passed
3.46.1Description
3.47 EC1: generating wind loads on an isolated roof with two slopes (TTAD #11695)
Passed
3.47.1Description
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3.48 EC1: generating wind loads on double slope 3D portal frame with a fully opened face(DEV2012 #1.6)
Passed
3.48.1Description
3.49 EC1: generating snow loads on a 2 slopes 3D portal frame with gutter (TTAD #11113)
Passed
3.49.1Description
3.50 EC1: generating snow loads on a 3D portal frame with horizontal roof and gutter (TTAD#11113)
Passed
3.50.1Description
3.51 EC1: generating snow loads on monopitch multispan roofs according to German standards(DIN EN 1991-1-3/NA) (DEV2012 #3.13)
Passed
3.51.1Description
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3.52 EC1: wind load generation on a building with multispan roofs
Passed
3.52.1Description
3.53 EC1: wind load generation on a high building with double slope roof using differentparameters defined per directions (DEV2013#4.2)
Passed
3.53.1Description
3.54 EC1: Generating snow loads on a 2 slope building with gutters and parapets. (TTAD #12878)
Passed
3.54.1Description
3.55 EC1: wind load generation on a high building with horizontal roof using UK annex(DEV2013#4.1) (TTAD #12608)
Passed
3.55.1Description
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3.56 EC1: generating Cf and Cp,net wind loads on an multibay canopy roof (DEV2013#4.3)
Passed
3.56.1Description
3.57 EC1: wind load generation on a high building with a horizontal roof using different CsCdvalues for each direction (DEV2013#4.4)
Passed
3.57.1Description
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4 Combinations
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4.1 Defining concomitance rules for two case families (TTAD #11355)
Passed
4.1.1 Description
4.2 Generating combinations for NEWEC8.cbn (TTAD #11431)
Passed
4.2.1 Description
4.3 Generating load combinations after changing the load case number (TTAD #11359)
Passed
4.3.1 Description
4.4 Generating load combinations with unfavorable and favorable/unfavorable predominant action(TTAD #11357)
Passed
4.4.1 Description
4.5 Performing the combinations concomitance standard test no.4 (DEV2012 #1.7)
Passed
4.5.1 Description
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4.6 Performing the combinations concomitance standard test no.7 (DEV2012 #1.7)
Passed
4.6.1 Description
4.7 Generating a set of combinations with Q group of loads (TTAD #11960)
Passed
4.7.1 Description
4.8 Generating combinations (TTAD #11673)
Passed
4.8.1 Description
4.9 Performing the combinations concomitance standard test no. 5 (DEV2012 #1.7)
Passed
4.9.1 Description
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4.10 Performing the combinations concomitance standard test no.8 (DEV2012 #1.7)
Passed
4.10.1Description
4.11 Generating the concomitance matrix after adding a new dead load case (TTAD #11361)
Passed
4.11.1Description
4.12 Performing the combinations concomitance standard test no.1 (DEV2010#1.7)
Passed
4.12.1Description
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4.13 Generating a set of combinations with seismic group of loads (TTAD #11889)
Passed
4.13.1Description
4.14 Performing the combinations concomitance standard test no.2 (DEV2012 #1.7)
Passed
4.14.1Description
4.15 Performing the combinations concomitance standard test no.6 (DEV2012 #1.7)
Passed
4.15.1Description
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4.16 Performing the combinations concomitance standard test no.10 (DEV2012 #1.7)
Passed
4.16.1Description
4.17 Performing the combinations concomitance standard test no.3 (DEV2012 #1.7)
Passed
4.17.1Description
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4.18 Generating a set of combinations with different Q "Base" types (TTAD #11806)
Passed
4.18.1Description
4.19 Generating the concomitance matrix after switching back the effect for live load (TTAD #11806)
Passed
4.19.1Description
4.20 Performing the combinations concomitance standard test no.9 (DEV2012 #1.7)
Passed
4.20.1Description
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4.21 Verifying combinations for CZ localization (TTAD #12542)
Passed
4.21.1Description
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5 Concrete Design
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5.1 C2 Test 16: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram
Passed
5.1.1 Description
5.1.2 Background
5.1.2.1 Model description■
■
■
■
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
Materials properties
■
■
■
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
400
8
81040
344
8
8103040
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5.1.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 17.3820
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
7.0
1
35
cm f
2.0
2
35
cm f
121
mmu
Ach 123
3140
192600*220
11.3488.0*17.3*2)(*)(*),(2123*1.0
100
501
1 003
t f t cm RH RH
Calculating the equivalence coefficient:
01
.
2996210
82022
10
822
3030
200000
683400
344
11311 0
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54.24
400
344*11.31
29962
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
Mpa f yk s 400*8,0
Neutral axis position calculation; Calculation of M tser :
MNmhbhd
hd
M f eff
f
f
e
stser 083,010,0*90.10*
10,0585,0
3
10,0585,0
*54.24*2
400**3*
*2
22
kNm M kNm M tser ser 83400
Concrete compressive stresses
MPah
d hb
M
f
f eff
ser m 31.8
)2
10,0585,0(*10,0*90.0
400,0
)2
(**
MPahd
d e
s
f
e
sm
c
61.1054.24
400
210,0585,0
54.24
40031.8
*585,0
2
*
MPa MPa cc 1261.10
The calculation of the tension reinforcement theoretical section As1
Calculation of the steel compressed reinforcement As2 :
mh x x f 1306,010,02306,012
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MPa x
xcc 01.6
2306,0
1306,0*61.1*
1
22
MN xbb N cweff c 282,0
2
01.6*1306,0*)18,090.0(
2*)( 2
22
Nm z N M cc 125,0441,0*282,0* 222
m z c 441,03
1306,010,0585,02
²06.7400*441,0
125,0
*2
22 cm
z
M A
sc
s
Notions of serviceability moment M 0 :
Theoretical steel reinforcement section :
²83.18435,0
400,0*46.20*
0
0 cm M
M A A ser s
s
Reference solution for minimal reinforcement area
d b
d b f
f
Max A
w
w
yk
eff ct
s
**0013.0
***26.0 ,
min,
MPa f f ctmeff ct 21.2,
²37.1
²10*37.1585.0*18.0*0013.0
²10*21.1585.0*18.0*500
21.2*26.0
max4
4
min, cm
m
m A s
Finite elements modeling
■
■
■
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SLS load combinations(kNm)
Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
5.1.2.3 Reference results
Result name Result description Reference value
5.1.3 Calculated results
Result name Result description Value Error
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5.2 EC2 Test 12: Verifying a rectangular concrete beam subjected to uniformly distributed load,without compressed reinforcement- Bilinear stress-strain diagram (Class XD3)
Passed
5.2.1 Description
5.2.2 Background
5.2.2.1 Model description
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
2058
4061525
1518
406153025
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5.2.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 725.2830
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
121 Mpa f cm 35
70
135
.
20
235
.
944.038
3535 7.07.0
1
cm f
984.0383535
2.02.0
2
cm f
mm
u
Ach 70.195
650280*2
650*280*220
37.2488.0*73.2*78.1)(*)(*),(78.1984.0*
70.195*1.0
100
501
1 003
t f t cm RH RH
Calculating the equivalence coefficient:
Ecar
Eqp
cm
se
M
M t
E
E
*),(1 0
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MPa f
E ck cm 32837
10
830*22
10
8*22
3.03.0
MPa E s 200000
75.2205
151*37.21*),(1 0
Ecar
Eqp
M
M t
76.16
205
151*37.21
32837
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
Mpa f ck bc 1830*6,0*6,0
Mpa f yk s 400*8,0
Neutral axis position calculation:
1
430.040018*76.16
15*76.16
*
*1
sce
ce
Moment resistance calculation:
1.
243057004300. 11
29703
2430570018243028050
3 1
2
1 11
297205
Calculation of reinforcement area with max constraint on steel and concrete
4300. 1
48503
43001570
3
.1 1
56104004850
2050 1
1
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Reference solution for minimal reinforcement area
00130
260
8962
402
1007257028000130
10402570280500
8962260
4
4
Finite elements modeling
■■
■
ULS and SLS load combinations (kNm)
Theoretical reinforcement area (cm2 )
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Minimum reinforcement area (cm2 )
5.2.2.3 Reference results
Result name Result description Reference value
5.2.3 Calculated results
Result name Result description Value Error
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5.3 EC2 Test 11: Verifying a rectangular concrete beam subjected to a uniformly distributed load,without compressed reinforcement- Bilinear stress-strain diagram (Class XD1)
Passed
5.3.1 Description
5.3.2 Background
5.3.2.1 Model description
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■
■
► 671651
25
►56225300300 3232
■
► 78434
151
500
►
MPa f
E ck cm 31476
10
825*22000
10
8*22000
3.03.0
Boundary conditions
■
►
►
■
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Loading
■
■
■
kNm M Ecq 94,1628
²80,5*75,38
kNm M Eqp 7.104
8
²80,5*75.24
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5.3.2.2 Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
9252825
816816
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
1.. 21 35
70
135
.
20
235
.
1.. 21
47176
6002502
600250220
702488092528918914717610
100
501
1 003
73.2163
104*70.21*),(1 0
Ecar
Eqp
M
M t
32.17
163
104*70.21
31476
200000
*),(1 0
Ecar
Eqp
cm
s
e
M
M t
E
E
Material characteristics:
Mpa f ck bc 1525*6,06,0
Mpa f yk s 400*8,0
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Neutral axis position calculation:
1
394.040015*32.17
15*32.17
*
*1 sce
ce
Moment resistance calculation:
1
204051903940. 11
17303
2040519015204025050
3 1
2
1 11
kNm M kNm M rb Ecq 17394.162
Calculation of reinforcement area with max constraint on steel and concrete
394,01
45103
394015190
3
.1 1
²03.9400*451.0
163.0
*,1 cm
z
M A
sc
ser ser s
Reference solution for minimal reinforcement area
d b
d b f
f
Max A
w
w
yk
eff ct
s
**0013.0
***26.0 ,
min,
MPa f f ctmeff ct 56.2,
²73.1
²10*69.1519.0*25.0*0013.0
²10*73.1519.0*25.0*500
56.226.0
max4
4
min, cm
m
m A s
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Finite elements modeling
■
■
■
ULS and SLS load combinations (kNm)
Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
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5.3.2.3 Reference results
Result name Result description Reference value
5.3.3 Calculated results
Result name Result description Value Error
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5.4 EC2: Verifying the longitudinal reinforcement area for a beam subjected to point loads
Passed
5.4.1 Description
5.5 EC2: Verifying the minimum reinforcement area for a simply supported beam
Passed
5.5.1 Description
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5.6 EC2 Test 45: Verifying a rectangular concrete beam supporting a balcony - Bilinear stress-strain diagram (Class XC1)
Passed
5.6.1 Description
5.6.2 Background
5.6.2.1 Model description
■
■
■
■
■
■
■
■
■
■
■
■
■
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Units
Geometry
■■
■
■
■
Boundary conditions
■
►
►
■
5.6.2.2 Reference results in calculating the concrete beam
The ULS load calculation:
■
■
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Load calculation:
■ Ed V
■ Ed M
■ Ed T
Shear and bending moment:
m KN P u /14.212*5,144.13*35,1
Maximum shear (ULS):
MN KN l P V u Ed 042,03,42
24*14,21
2*
Bending moment at the supports:
MNm KNml P
M u Ed 028,02,28
12
²4*14,21
12
²*
Maximum Moment at middle of span:
MNm KNml P
M u Ed 014,01,14
24
²4*14,21
24
²*
Torsion moment:
m MNmm KNmmtu /015,0/97,1425,2*5,159,8*35,1
MNml
mT tu Ed 03,02
4*015,0
2*
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5.6.2.3 Bending rebar
Span reinforcement (at bottom fiber)
007,067.16*²70.0*25,0
014,0cu
0088,0007,0*211*25,1 u
m z c 697,00088,0*4,01*70.0
²46.0²10*62.478,434*697,0
014,0 4 cmm Au
d b
d b f
f
Max A
w
w
yk
eff ct
s
**0013.0
***26.0 ,
min,
Mpa f f ctmeff ct 56.2,
²27.2
²27.270.0*25.0*0013.0**0013.0
²27.270.0*25.0*500
56.2*26.0***26.0
,
min, cm
cmd b
cmd b f
f
Max A
w
w
yk
eff ct
s
Reinforcement on supports (at top fiber)
014,0
67.16*²70.0*25,0
028,0
cu
018,0014,0*211*25,1 u
m z c 695,0018,0*4,01*70.0
²93.0²10*27.978,434*695,0
028,0 4 cmm Au
Shear reinforcement
MN V Ed 042,0˚ 1cot
Concrete rod verification:
²cot1
cotcot****max,
cd w Rd f v z bV
m z z c 695.0
Vertical frames:
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54.0250
251*6,0
v
MN tg
f v z b f v z bV cd cw
cd cw Rd 78.02
67.16*54.0*695.0*25.0
cot
***
²cot1
cotcot****max,
cot
***1max,
tg
b z f vV wucd
Rd
MN V MN V Rd Ed 78.0042.0 max,
Calculation of transverse reinforcement:
ml cm f z
tg V
s
A
yd u
Ed sw /²39.178.434*695.0
042.0
*
*.
From the minimum reinforcement percentage:
sin..min, ww sw b
s
A
0008.0500
25*08,0*08,0min,
yk
ck
w f
f
ml cm s
A sw /²225.0*0008.0
ml cm s
A sw /²2
MNmT Ed 03,0
Torsional shear stress:
k ief
Ed it
At
T
**2 ,
,
cmcmu
Acmc
t ief 375.9375.9)7525(2
)75*25(8*2
max,
²1025.0²1025)375.975(*)375.925( mcm Ak
561102500937502
0302
2
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Concrete verification:
cos*sin******2 ,max, ief k cd cw Rd t A f vT
54.0250
1*6,0
ck f
v
MN T Rd 085.070.0*70.0*09375.0*1025.0*67.16*54.0*2max,
0,1max,max,
Rd
Ed
Rd
Ed
V
V
T
T
0,1399.078.0042.0
085.003.0
cot***2
*
yd k
k Ed l
f A
uT A
mcmt ht bu ef ef k 625.15.162)]375.975()375.925[(*2)]()[(*2
²47.578.434*1025.0*2
625.1*03.0
cm Al
Torsion transversal reinforcement
ml cm f A
T
s
A
yd k
Ed
T
swT /²36.378.434*1025.0*2
03.0
cot***2
ml cm s
A
T
swT /²36,3
Finite elements modeling
■■
■
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ULS load combinations(kNm)
5.6.2.4 Reference results
Result name Result description Reference value
5.6.3 Calculated results
Result name Result description Value Error
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5.7 EC2 Test 6: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram
Passed
5.7.1 Description
5.7.2 Background
5.7.2.1 Model description
■
■
■
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
Materials properties
■
■
■
■
■ MPa f
f c
ck cd 67.10
5,1
25
■ MPa f f ck ctm 56.225*30.0*30.0 3/23/2
■ MPa f
f s
yk
yd 78.43415,1
500
Boundary conditions
■
►►
■
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Loading
■
■
kNm M Ed 8.9348
²8*85.116
5.7.2.2 Reference results in calculating the concrete beam moment
kNm f hd hb M cd f
f eff btu 967.16*215.0435.0*15.0*00.1*
2**
kNm M kNm M Ed btu 8.934900
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5.7.2.3 Reference reinforcement calculation:
Theoretical section 2:
kNm
h
d f hbb M f
cd f weff Ed
720
2
15.0
435.0*67.16*15.0*)20.01(2***)(2
²00.46
8.347*2
15.0435.0
720.0
*2
22 cm
f h
d
M A
yd
f
Ed
Theoretical section 1:
kNm M M M Ed Ed Ed 21572093521
341.067.16*²435.0*20.0
215.0
*²*
1 cd w
Ed cu
F d b
M
3720
544.0341.0*211*25.1)*21(1*25.1 cuu
md z uc 340.0)544.0*40.01(*435.0)*4.01(*1
²52.1478.347*340.0
215.0
*1
11 cm
f z
M A
yd c
Ed
Theoretical section 1:
Finite elements modeling
■
■
■
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ULS load combinations(kNm)
Theoretical reinforcement area(cm2 )
5.7.2.4 Reference results
Result name Result description Reference value
5.7.3 Calculated results
Result name Result description Value Error
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5.8 EC2 Test 24: Verifying the shear resistance for a rectangular concrete beam with verticaltransversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Passed
5.8.1 Description
5.8.2 Background
5.8.2.1 Model description
■
■
■
■
■
■
■
■
■
■
■
■
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Units
Geometry
■■
■
■
■
■
■ θ ˚
Boundary conditions
■
►
►
■
Loading:
l
a P V u Ed 1
KN V Ed 5,1893
11*25,284
Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to anunreduced shear, that will be used for further calculations.
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5.8.2.2 Reference results in calculating the maximum design shear resistance
1
.. 1
1.
2501601
1
45 90
1v
55.0250
201*6.0
2501*6,01
ck f
v
md z u 392.0435.0*9.0*9.0
MN V Rd 288.02
20.0*392.0*33.13*55.0max,
MN V MN V Rd Ed 288.01895.0 max,
Calculation of transversal reinforcement:■
■ ml cmtg
f z
tg V
s
A
ywd u
Ed sw /²13.11
15,1
500*392,0
45*18950,0
*
*.
■
KN l
a P V uu 75,94
3
1*25,284*
■
Finite elements modeling
■
■
■
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Advance Design gives the following results for Atz (cm2 /ml)
5.8.2.3 Reference results
Result name Result description Reference value
5.8.3 Calculated results
Result name Result description Value Error
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5.9 EC2 Test 10: Verifying a T concrete section, without compressed reinforcement - Inclinedstress-strain diagram
Passed
5.9.1 Description
5.9.2 Background
5.9.2.1 Model description
■
■
■
■
■
■
■
■
■ 30%2
■
■
■
■
■
Units
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Geometry
■
■
■
Materials properties
■
■
■
■
2051
30
90230300300 3232
■
78434151
500
Boundary conditions
■
►
►
■
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Loading
550628
61125
36008
6800
26558
6590
5.9.2.2 Reference results in calculating the concrete beam moment
310480420
2
20090200401
2
Reference reinforcement calculation:
Theoretical section 2:
3
2
1023
2
20090020200400401
2
466027727714321
3321 1086311020030635
28702090400
8631
1
33 1006255104804
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43502870211251211251.
5545034350
43501 0
.
.1 0 2
4660243600455027727714321
7491
024362
20090
2003
2
22
Theoretical section 1:
3321 1086311020030635
372028702090400
8631 1
37202870
43502870211251211251.
74304350400190.4011
4657024367430
8631
111
Theoretical section 1:
Reference solution for minimal reinforcement area
00130
260
902
425
1068490040000130
10425900400500
902260
4
4
Finite elements modeling
■
■
■
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ULS and SLS load combinations (kNm)
Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
5.9.2.3 Reference results
Result name Result description Reference value
5.9.3 Calculated results
Result name Result description Value Error
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5.10 EC2 Test36: Verifying a rectangular concrete column using the method based on nominalcurvature- Bilinear stress-strain diagram (Class XC1)
Passed
5.10.1Description
5.10.2Background
5.10.2.1Model description
■
■
■
■
►
►
■
►
► 3,02
■
►►
► 10 02
■
■
■
■
■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
MN N ED 99.03.0*5.14.0*35.1
00
2400
5002
4002 0
020201
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5.10.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 50
5.4935.0
5*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
EQP M
EQP M W QG M EQP *03.0
MNme N M qp Eqp 0098.002.0*)3.0*3.04.0(* 10
Ed M
ED M 0 W QG M oED *5.1*5.135.1
MNm L
H e N M W ed Eqp 48855.04
5*25.0*5.102.0*99.0
4**5.1* 10
02.048855.0
0098.0
0
0 Ed
Eqp
M
M
0,t
)(*)(*),( 00 t f t cm RH
MPa f
f cm
cm 73.2830
8.168.16)(
482.0301.0
1
1.0
1)(
20.020.0
0
0
t
t
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213
0
***1.0
1001
1
h
RH
RH
121 MPa f cm 357.0
1
35
cm f
2.0
2
35
cm f
mm
u
Ach 221
600350*2
600*350*2*20
MPa MPa f cm 3538 944.038
3535 7.07.0
1
cm f
984.0383535
2.02.0
2
cm f
752.1984.0*944.0*221*1.0
100
501
1***1.0
1001
1321
30
h
RH
RH
30.2482.0*73.2*752.1)(*)(*),( 00 t f t cm RH
The effective creep coefficient calculation:
046.002.0*30.2*, 0 Ed
EQP
ef M
M t
046.1046.011 ef
The necessity of buckling calculation (second order effect):
22 * e N M Ed
c
l
r e
2
02 *
1
0
1**
1
r K K
r r
).45,0(
1
0 d r
yd
1000
1739.2
200000
78.434
s
yd
yd E
f
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1
0
0161.0)30.0*45,0(
1000
1739.2
)*45,0(
1 md r
yd
1
bal u
ur
nnnn K
236.020*35.0*6.0
99.0
*
cd c
Ed
f A
N n
1un
166.020*35.0*6.0
78.434*10*08.16 4
cd c
yd s
f A
f A
166.1166.0.11 un
4,0bal n
214.14.0166.1
236.0166.1
bal u
ur
nn
nn K
1r K 1r K
K 1*1 ef K
15020035,0
ck f
5.4935.0
125120 h
l
17.0150
5.49
200
3035.0
15020035,0
ck f
008.1046.0*17.01*1 ef K
0162.00161.0*008.1*11
**1
0
r
K K r
r
mc
l
r e 03375.0
12
²5*0162.0*
1 2
02
MNme N M Ed 033.003375.0*990.0* 22
MNm M M M Ed Ed 522.0033.048855.020
MN N Ed 990.0
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MNm M Ed 522.0
Reinforcement calculation according the second order effects:
MN N Ed 990.0
MNm M Ed 522.0
Buckling checking:
Curvature calculation:
1
00161.0
1
mr
1
bal u
ur
nn
nn K
112.0n
60.020*35.0*60.0
78.434*10*02,58
*
* 4
cd c
yd s
f A
f A
60.160.011 un
4,0bal n
1124.140.060.1
112.060.1
r r K K
1 K 1.1 ef K
1
0
0161.01
**1 m
r K K
r r
Finite elements modeling
■
■
■
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Theoretical reinforcement area (cm2 )
Theoretical value (cm2 )
5.10.2.3Reference results
Result name Result description Reference value
5.10.3Calculated results
Result name Result description Value Error
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5.11 EC2 Test 17: Verifying a rectangular concrete beam subjected to a uniformly distributed load,without compressed reinforcement - Inclined stress-strain diagram (Class XD1)
Passed
5.11.1Description
5.11.2Background
σσ
5.11.2.1Model description
■
■
■
■
■
■
■
■
■
■ σ
■ σ
Units
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Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■ Mpa f ck 20■ Mpa f yk 400
■ ²42,9 cm A st
Boundary conditions
■
►
►
■
Loading
■
■
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5.11.2.2Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 17.3820
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
121 Mpa f cm 35
7.0
1
35
cm f
2.0
2
35
cm f
121
mm
u
Ach 86.142
500200*2
500*200*220
03.3488.0*17.3*96.1)(*)(*),(96.186.142*1.0
100
501
1 003
t f t cm RH RH
Ecar
Eqp
cm
se
M
M t
E
E
*),(1 0
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MPa f
E ck cm 29962
10
820*22
10
8*22
3.03.0
MPa E s 200000
03.41*03.31*),(1 0 Ecar
Eqp
M
M t
90.26
1*03.31
29962
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
152560601
320801
Checking inertia cracked or not:
ctm f
v
I f M ctm
cr
*
433
00208,012
50,0*20,0
12
*m
hb I
mh
v 25,02
Mpa f f ck ctm 21,220*30.0*30.0 3
2
3
2
MNmv
I f M ctmcr 018,0
25,000208,0*21,2*
Neutral axis position calculation:
0)'(**)(**²**2
1111 d x A xd A xb e sce st w
w
st scew st sce st sce
b
Ad Ad b A A A A x
)*'*(***2)²(*)(* 2
1
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0 sc A
w
st ew st e st e
b
Ad b A A x
)****2²*)(* 2
1
cm x 04,2320
)42,9/44(*90,26*20*2²42,9*²90.2642,9*90.261
Calculating the second moment:
443
1
3
1 001929,0)²230,044,0(*90.26*10*42,93
230,0*20,0)²(**
3
*m xd A
xb I e st
w
Stresses calculation:
Mpa Mpa x I
M c
ser c 1294,8230,0*001929,0
075,0
* 1
Mpa Mpa x
xd sce st 32057.219
230,0
230,044,0*94,8*90.26**
1
1
Finite elements modeling
■
■
■
SLS load combinations (kNm) and stresses (MPa)
Compressing stresses in concrete section σ c
σ
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Compressing stresses in the steel reinforcement section σ s
σ
5.11.2.3Reference results
Result name Result description Reference value
σ σ
σ σ
5.11.3Calculated results
Result name Result description Value Error
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5.12 EC2 Test 44: Verifying a rectangular concrete beam subjected to eccentric loading - Bilinearstress-strain diagram (Class X0)
Passed
5.12.1Description
5.12.2Background
5.12.2.1Model description
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
■
■
Boundary conditions
■
►
►
■
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Reference results in calculating the concrete beam
P P ba
be P T A
10
3
5
3**5,0**
P P ba
ae P T B51
52**5,0**
The ULS load calculation:
kN P u 70.203*5.112*35.1
KNm P T u Ed 21.67,20*10
3*
10
3
KNm P
T u Ed 14.4
5
7.20
5
Result ADVANCE Design 2012 - Moment: (in kNm)
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5.12.2.2Calculation in pure bending
Bending moment at ULS:
KN P u 7,203*5,112*35,1
MNm KNm P
M u Ed 0248,084,24
5
7,20*6
5
*6
Simple bending design:
110,020*²25.0*18,0
0248,0cu
146,0110,0*211*25,1 u
m z c 235,0146,0*4,01*25.0
²43,2²10*43,278,434*235,0
0248,0 4 cmm Au
5.12.2.3Torsion reinforcement
Torsion longitudinal reinforcement - Before application of the moment
cot***2
*
yd k
k Ed l
f A
uT A
))()((*2 ef ef k t ht bu
)(*)( ef ef k t ht b A
m
u
Act ef 07.005625.0;07.0max
3.018.0*2
30.0*18.0;035.0*2max;*2max
mcmuk 68.068))730()718((*2
²0253.0²253)730(*)718( mcm Ak
²92.178.434*0253.0*2
68.0*0062.0cm Al
Torsion longitudinal reinforcement - After application of the moment
²28.178.434*0253.0*2
68.0*00414.0cm Al
Torsion transversal reinforcement - Before application of the moment
ml cm f A
T
s
A
yd k
Ed
T
swT /²82.278.434*0253.0*2
0062.0
cot***2
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Torsion transversal reinforcement - After application of the moment
ml cm f A
T
s
A
yd k
Ed
T
swT /²88.178.434*0253.0*2
00414.0
cot***2
Concrete verification:
cos*sin******2 ,max, ief k cd cw Rd t A f vT
528.0250
1*6,0
ck f
v
MN T Rd 018.070.0*70.0*07.0*0253.0*20*528.0*2max,
²cot1
cotcot****max,
cd w Rd f v z bV
m z z c 235.0
1cot
528.0250
301*6,0
v
MN f v z bV cd cw Rd 223.02
1*20*528.0*235.0*18.0
²cot1
cotcot****max,
0,1max,max,
Rd
Ed
Rd
Ed
V
V
T
T
0,1400.0
223.0
0124.0
018.0
0062.0
Finite elements modeling
■
■
■
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ULS load combinations(kNm)
5.12.2.4Reference results
Result name Result description Reference value
5.12.3Calculated results
Result name Result description Value Error
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5.13 EC2 Test 14: Verifying a rectangular concrete beam subjected to a uniformly distributed load,with compressed reinforcement- Bilinear stress-strain diagram (Class XD1)
Passed
5.13.1Description
5.13.2Background
5.13.2.1Model description
■
■
■
■
■
3,02
Units
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Geometry
■
■
■
■■
■
Materials properties
■
■
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
■
mkN M cq ser .05.672
8
²3.6*60508
4
3.6*2530,
mkN M qp ser .11.436
8
²3.660*3.0508
4
3.6*25*3.030,
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5.13.2.2Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 92.2825
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
121 Mpa f cm 35
7.0
1
35
cm f
2.0
2
35
cm f
121
mm
u
Ach 267
800400*2
800*400*220
54.2488.0*92.2*78.1)(*)(*),(78.1267*1.0
100
501
1 003
t f t cm RH RH
Calculating the equivalence coefficient:
Ecar
Eqp
cm
se
M
M t
E
E
*),(1 0
MPa f
E ck cm 31476
10
825*22
10
8*22
3.03.0
MPa E s 200000
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65.2672
435*54.21*),(1 0
Ecar
Eqp
M
M t
82.16
672
435*54.21
31476
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
Mpa f ck bc 1515*6,0*6,0
Mpa f yk s 400*8,0
Neutral axis position calculation:
387.040018*82.16
15*82.16
*
*1
sce
ce
Moment resistance calculation:
1.
273.0707.0*387.0*11 d x
MNm x
d xb M cwrb 505.03
273.0707.0*15*273.0*4.0*5.0)
3(****
2
1 11
kNm M kNm M rb Ecq 505672
The calculation of the tension reinforcement theoretical section A1
m x
d z c 616.03
273.0707.0
31
²51.20400*616.0
505.0
*1 cm
z
M A
sc
rb
Stress calculation for steel reinforcement σ sc :
064.0707.0
045.0''
d
d
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MPace sc 78.210387.0
067.0387.0*15*82.16
'**
1
1
Calculation of the steel compressed reinforcement A’:
²96.1178.210*)045.0707.0(
505.0672.0*)'(
' cmd d
M M A sc
rb ser
Calculation of the steel tensioned reinforcement A2 :
²30.6400
78.210*96.11'*2 cm A A
s
sc
Calculation of the steel reinforcement :
²81.2630.651.2021 cm A A A
Reference solution for minimal reinforcement area
d b
d b f
f
Max A
w
w
yk
eff ct
s
**0013.0
***26.0 ,
min,
MPa f f ctmeff ct 56.2,
²76.3
²10*68.3707.0*40.0*0013.0
²10*76.3707.0*40.0*500
56.2*26.0
max4
4
min, cm
m
m A s
Finite elements modeling
■
■
■
ULS and SLS load combinations (kNm)
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Minimum reinforcement area (cm2 )
5.13.2.3Reference results
Result name Result description Reference value
5.13.3Calculated results
Result name Result description Value Error
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5.14 EC2 Test 43: Verifying a square concrete column subjected to a small rotation moment andsignificant compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1)
Passed
5.14.1Description
5.14.2Background
5.14.2.1Model description
■
■
■
■
►
►
►
■
►
►
■
■■
■
■
■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■
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5.14.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 60.11*20
37.8050.0
60.11*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
EQP M
Ed M
ieee 01
mei 03.0
me N
M eee i
Eqp
Eqp
i 125.030.0100*30.0150
7*30.015
0
0
01
kN N Eqp 180100*30.01501
MNmkNme N M Eqp Eqp 0225.050.22125.0*180* 111
MNm M Ed 041.01
0,t
)(*)(*),( 00 t f t cm RH
72.2830
8.168.16)(
cm
cm f
f
488.0281.0
11.0
1)( 20.020.0
0
0 t t
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213
0
***1.0
1001
1
h
RH
RH
MPa f cm 35 944.038
3535 7.07.0
1
cm f 984.0
38
3535 2.02.0
2
cm f
72.1984.0*944.0*
250*1.0
100
501
1250500500*2
500*500*2*230
RH mm
u
Ach
28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH
25.1041.0
0225.0*28.2*, 0
Ed
EQP
ef M
M t
The second order effects; The buckling calculation:
n
C B A ***20lim
071.020*²50.0
353.0
*
cd c
Ed
f A
N n
80.0
25.1*2.01
1
*2,01
1
ef
A
1.1*21 B
70.07,1 mr C
24.46
071.0
7.0*1.1*80.0*20lim
24.4637.80 lim
Calculation of the eccentricities and solicitations corrected for ULS :
■
■
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m N
M e
Ed
Ed 087.0353.0
03075.00
ml ei 03.0400
6.11
4000
The first order eccentricity: stresses correction:
MN N Ed 353.0
meee i 117.001
MNm N e M Ed Ed 041.0353.0*117.0*1
0* e N M Ed
mmmm
mmmm
mmhmm
e 207.16
20max
30
50020
max
30
20max0
Reinforcement calculation in the first order situation:
MNmh
d N M M Gua 112.02
50.045.0*353.0041.0)
2(*0
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494,0)45,0
50,0*4,01(*
45,0
50,0*8,0)*4,01(**8,0
d
h
d
h BC
055.020*²45.0*50.0
112.0
*²*
cd w
uacu
f d b
M
494.0055.0 BC cu
The calculation for the tensioned steel in pure bending:
055.0cu
071,0)055,0*21(1*25,1 u
md z uc 437,0)071,0*4,01(*45,0)*4,01(*
²89,578,434*437,0
112,0*
cm f z
M A yd c
ua
The calculation for the compressed steel in bending:
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5.14.2.3The nominal curvature method (second order effect):
The curvature calculation:
0
1**1r
K K r
r
1
0
0107.045.0*45,0
200000
78.434
)*45,0()*45,0(
1 md
E
f
d r s
yd
yd
1
bal u
ur
nn
nn K
0706.020*²50.0
353.0
*
cd c
Ed
f A
N n
0435.020*²50.0
78.434*10*5
*
* 4
cd c
yd s
f A
f A
0435.10435.011 un
4,0bal n
51.140.00435.1
0706.00435.1
r K
1r K 1r K
K 1*1 ef K
1
0
0107.00107.0*1*11
**1 m
r K K
r r
Calculation moment:
20 M M M Ed Ed
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Ed M 0
2 M
22 * e N M Ed
mc
l
r e 144.0
10
²60.11*0107.0*
1 2
02
MNme N M Ed 051.0144.0*353.0* 22
MNm M M M Ed Ed 09.0051.0041.020
Reinforcement calculation according the second order effects:
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2
min, 5*002.0²81.078.434
353.0*10.0*10,0cm Acm
f
N A c
yd
Ed s
Finite elements modeling
■
■
■
Theoretical reinforcement area(cm2 )
Theoretical value (cm2 )
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5.14.2.4Reference results
Result name Result description Reference value
5.14.3Calculated results
Result name Result description Value Error
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5.15 EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinearstress-strain diagram
Passed
5.15.1Description
5.15.2Background
5.15.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
Materials properties
■
■
■
■
■ MPa f
f c
ck cd 67.10
5,1
16
■ 90116300300 3232
■ MPa f
f s
yk
yd 8.34715,1
400
Boundary conditions
■
►
►
■
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Loading
■
kNm M Ed 89.5518
²7*105.70
kNm M Ecq 96.3998
²7*3.65
5.15.2.2Reference results in calculating the concrete beam moment
52367102
100595010090
2
5.15.2.3Reference reinforcement calculation:
Theoretical section 2:
■
MNm
hd f hbb M
f
cd f weff Ed
418.0
2
1.00595*67.10*1.0*)18.09.0(
2***)(2
■
²08.22
8.347*21.0595.0
418.0
*2
22 cm
f hd
M A
yd f
Ed
kNm M kNm M Ed btu 89.551523
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Theoretical section 1:
kNm M M M Ed Ed Ed 133418.0552.021
197.067.10*²595.0*18.0
133.0
*²*
1
cd w
Ed
cu F d b
M
3720
276.0197.0*211*25.1)*21(1*25.1 cuu
md z uc 529.0)276.0*40.01(*595.0)*4.01(*1
²25.78.347*529.0
133.0
*1
11 cm
f z
M A
yd c
Ed
Theoretical section 1:
Finite elements modeling
■
■
■
ULS and SLS load combinations(kNm)
Theoretical reinforcement area(cm2 )
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5.15.2.4Reference results
Result name Result description Reference value
5.15.3Calculated results
Result name Result description Value Error
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5.16 EC2 Test 3: Verifying a rectangular concrete beam subjected to uniformly distributed load,with compressed reinforcement- Bilinear stress-strain diagram
Passed
5.16.1Description
5.16.2Background
Simple Bending Design for
5.16.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
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Units
Geometry
■■
■
■
■
■
Materials properties
■
■■
■
■ 671651
25
■ 56225300300 3232
■ 78434151
500
■ 3147610
82522000
10
822000
3030
Boundary conditions
■
►
►
■
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Loading
■
■
6102555
8
145214
61067538
14150150
61030328
1494
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5.16.2.2Reference results in calculating the equivalent coefficient
■ 00
■ 9252825
816816
■ 48802810
1
10
1
2002000
0
φ
■ 2130
..10
1001
1
■ 121 35
■
70
135
.
20
235
.
■ 1338 21 Mpa Mpa f f ck cm
■ 6616342710
100501
16342712506502
125065022
30
■ 37524880925266100
■
01
.
Nm M Ecar
310*3675
Nm M Eqp
310*2303
8215
103975
1023037021
31476
200000 .
3
3
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5.16.2.3Reference results in calculating the concrete beam reduced moment limit
372.0lu
)500/372.0( S
Reference reinforcement calculation at SLU:
■
■
383,067,16*²125.1*65,0
10*25.5255
*²* 2
3
MPamm
Nm
f d b
M
cd w
Ed cu
372.0383.0 lucu
Calculation of the tension steel section (Section A1):
Nm f d b M cd wlu Ed
6
1 10*10.567.16*²125.1*65.0*372.0*²**
■ α 61803720211251211251.
■ 851061804011301.401
■
²35.13778.434*851.0
10*10.5
.
6
11 cm
MPam
Nm
f z
M A
yd lu
Ed
Compressed steel reinforcement reduction (Section As2):
00327.0045.0130.1*618.0130.1*618.0*1000
5.3)'*(
**1000
5,3 d d
d lu
lu
sc
MPa f yd sc yd sc 78.43400217.000327.0
²32.2
78.434)045.0130.1(
15.5255.5
)'(
12 cm
d d
M M A
sc
Ed Ed s
²32.2.22 cm f
A A yd
sc s
Total area to be implemented:
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Reference reinforcement calculation at SLS:
■ 6106753
8
14150150
■ 8215.
■
41901251400158215
158215
11.
1.1
61 10042154190650
2
1 1
2
1
9850
3
41901251
3
1
66 10012985010042
Theoretical section 1 (tensioned reinforcement only)
Nm M M rb
6
1 10*01.2
372.0125.1
419.011
d
x
m xd z c 985.03419.0125.1
31
²01.51400985.0
01.211 cm
z
M A
sc
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
Nm M M M rbcq ser
66
,2 10*665.110*)01.2675.3(
d
d d ce sc
*
'***
1
1
MPa sc 78.211125.1*372.0
045.0125.1*372.0*15*82.15
²92.71
78.211*045.0125.1
645.1
*)'(' 2 cm
d d
M A
sc
²08.38400
78.211*92.71*' cm A A
s
sc s
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Section area:
Reference reinforcement additional iteration for calculation at SLS:
MPa A
A s
ELU
ELS s 252400
42.141
18.89
md x sce
ce 55.0130.1*25215*82.15
15*82.15*
*
*1
N xb F cwc
6
1 10*67.215*55.0*65.0*2
1***
2
1
m x
d z c 94.03
55.0125.1
3
1
Nm x
d xb z F M cwccrb
611 10*52.2
3
55.0125.1*15*55.0*65.0*
2
1
3****
2
1*
Theoretical section 1 (tensioned reinforcement only)
Nm M M rb
6
1 10*52.2
488.0125.1
55.011
d
x
m x
d z c 94.03
548.0125.1
31
²38.10625294.0
52.211 cm
z
M A
sc
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Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
Nm M M M rbcq ser
66
,2 10*155.110*)52.2675.3(
d d d ce sc
*'***
1
1
MPa sc 73.217125.1*485.0
045.0125.1*485.0*15*82.15
²12.49
73.217*045.0125.1
155.1
*)'(' 2 cm
d d
M A
sc
²44.42252
73.217*12.49*' cm A A s
sc s
Section area:
Reference solution for minimal reinforcement area
d b
d b f
f
Max A
w
w
yk
eff ct
s
**0013.0
***26.0 ,
min,
MPa f f ctmeff ct 56.2,
²73.9
²10*51.9125.1*65.0*0013.0
²10*73.9125.1*65.0*500
56.2*26.0
max4
4
min, cm
m
m A s
Finite elements modeling
■
■
■
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ULS and SLS load combinations(kNm)
Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
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5.16.2.4Reference results
Result name Result description Reference value
5.16.3Calculated results
Result name Result description Value Error
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5.17 EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinearstress-strain diagram
Passed
5.17.1Description
5.17.2Background
5.17.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
Materials properties
■
■
■
■
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
kNm M cq ser 240
8
²8*1020,
kNm M qp ser 184
8
²8*10*3.020,
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5.17.2.2Reference results in calculating the concrete final value of the creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 17.3820
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
121 Mpa f cm 35
7.0
1
35
cm f
2.0
2
35
cm f
Mpa Mpa Mpa f f ck cm 35288
121
mmu
Ach 24.162
3920
318000*220
97.2488.0*17.3*92.1)(*)(*),(92.124.162*1.0
100
501
1 003
t f t cm RH RH
Calculating the equivalence coefficient:
Ecar
Eqp
cm
se
M
M t
E E
*),(1 0
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MPa f
E ck cm 29962
10
820*22
10
8*22
3.03.0
MPa E s 200000
28.3240
184*97.21*),(1 0
Ecar
Eqp
M
M t
89.21
240
184*97.21
2962
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
Mpa f ck bc 1220*6,0*6,0
Mpa f yk s 400*8,0
MNmhbhd
hd
M f eff
f
f
e
stser 122,010,0*20,1*
10,0669,0
3
10,0669,0
*77.43
400**3*
*2
22
kNm M kNm M tser ser 122240
Concrete compressive stresses
MPah
d hb
M
f
f eff
ser m 23,3
)2
10,0669,0(*10,0*20,1
240,0
)2
(**
MPah
d
d e
s
f
e
sm
c 96,489.21
400
2
10,0669,0
89.21
40023,3
*669,0
2
*
MPa MPa cc 1296.4
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The calculation of the tension reinforcement theoretical section As1
cmd x sce
ce 30.14669,0*40096,4*89.21
96,4*89.21
*
*1
MN xb N
c
eff c 426,02
96,4
*1430,0*20,12** 11
265062204260111 62203
1430066901
²65,10400*622,0
265,0
*1
11 cm
z
M A
sc
s
Calculation of the steel compressed reinforcement As2 :
043001001430012
49114300
043009641
1 1
2
2
02902
4910430030201
2
1 222
016055500290222
55503
0430010066902
7204005550
0160
12
22
Notions of serviceability moment M 0 :
9397206510210
249001602650210
Theoretical steel reinforcement section :
5892490
2400939
0
0
Reference solution for minimal reinforcement area
00130
260
212
612
10612669030000130
103126690300500
212260
4
4
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Finite elements modeling
■
■
■
SLS load combinations(kNm)
Theoretical reinforcement area (cm2 )
Minimum reinforcement area (cm2
)
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5.17.2.3Reference results
Result name Result description Reference value
5.17.3Calculated results
Result name Result description Value Error
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5.18 EC2 Test 13: Verifying a rectangular concrete beam subjected to a uniformly distributed load,without compressed reinforcement - Bilinear stress-strain diagram (Class XD1)
Passed
5.18.1Description
5.18.2Background
5.18.2.1Model description
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
kNm M Ecq 28.2318
²80.5*)3025(
kNm M Eqp 97.1428
²80.5*)30*3.025(
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5.18.2.2Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 92.2825
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
121 Mpa f cm 35
7.0
1
35
cm f
2.0
2
35
cm f
121
mm
u
Ach 176
600250*2
600*250*220
70.2488.0*92.2*89.1)(*)(*),(89.1176*1.0
100
501
1 003
t f t cm RH RH
Calculating the equivalence coefficient:
Ecar
Eqp
cm
se
M
M t
E E
*),(1 0
MPa f
E ck cm 31476
10
825*22
10
8*22
3.03.0
MPa E s 200000
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67.2231
143*69.21*),(1 0
Ecar
Eqp
M
M t
97.16
231
143*69.21
31476
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
Mpa f ck bc 1515*6,0*6,0
Mpa f yk s 400*8,0
Neutral axis position calculation:
1
389.040018*97.16
15*97.16
*
*1
sce
ce
Moment resistance calculation:
1.
202051903890. 11
MNm x
d xb M cwrb 171.03
202.0519.0*15*202.0*25.0*5.0)
3(****
2
1 11
kNm M kNm M rb Ecq 171231
The calculation of the tension reinforcement theoretical section A1
m x
d z c 452.03
202.0519.0
3
1
²46.9400*452.0
171.0
*1 cm
z
M A
sc
rb
Stress calculation for steel reinforcement σ sc :
087.0519.0
045.0''
d
d
MPace sc 75.197389.0
087.0389.0*15*97.16'**1
1
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Calculation of the steel compressed reinforcement A’:
²44.675.197*)045.0519.0(
171.0231.0
*)'(' cm
d d
M M A
sc
rb ser
Calculation of the steel tensioned reinforcement A2 :
²18.3400
75.197*44.6'*2 cm A A
s
sc
Calculation of the steel reinforcement :
²64.1218.346.921 cm A A A
²44.6' cm A
Reference solution for minimal reinforcement area
8962
²73.1²10*69.1519.0*25.0*0013.0
²10*73.1519.0*25.0*500
56.2*26.0
max4
4
min, cmm
m A s
Finite elements modeling
■
■
■
ULS and SLS load combinations(kNm)
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Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
5.18.2.3Reference results
Result name Result description Reference value
5.18.3Calculated results
Result name Result description Value Error
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5.19 EC2 Test 20: Verifying the crack openings for a rectangular concrete beam subjected to auniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram(Class XD1)
Passed
5.19.1Description
5.19.2Background
σσ
5.19.2.1Model description
■
■
■
■
■
■
■
■
■ σ
■ σ
■
■
Units
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Geometry
■
■
■
■■
■
Materials properties
■
■
■
■
■ 25
■ 500
■ 1630
Boundary conditions
■
►
►
■
Loading
■
►
►
►
►
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5.19.2.2Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 92.2825
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
Mpa f cm 35
121
7.0
1
35
cm f
2.0
2
35
cm f
121
mm
u
Ach 67.266
800400*2
800*400*220
55.2488.0*92.2*78.1)(*)(*),(78.167.266*1.0
100
501
1 003
t f t cm RH RH
),(1 0t
E
E
cm
se
MPa f
E ck cm 31476
10
825*22
10
8*22
3.03.0
MPa E s 200000
55.2),( 0 t
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56.22
55.21
31476
200000
),(1 0
t
E
E
cm
se
Material characteristics:
Mpa f ck bc 1525*6,06,0
Mpa f yk s 400*8,0
Neutral axis position calculation:
0..2
1111
2.. 2
1
0 sc A
3540
16307156224021630562216305622
.2.. 2
1
Calculating the second moment:
443
11
31
01450350071056221016303
350040
..3
Stresses calculation:
Mpa x I
M ser c 96.7350,0*
0145,0
330,0* 1
Mpa Mpa x
xd sce st 4007.184
350,0
350,071,0*96.7*56.22**
1
1
Maximum spacing of cracks:
2
, 06.015.0*4.0
4.02
8.0
15.03
)350.08.0(255.0)71.08.0(*5.2
min*40.0
2
3
)()(*5.2
min* m
h
xhd h
b A eff c
0500.006.0
10*16.30
4
,
,
eff c
seff p
A A
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mmnn
nneq 16
**
**
2211
2
22
2
11
eff pr
k k
ck s,
21
3max,
***425.0
*
113.251
25*4.3
25*4.3
3/23/2
3
c
k
mmk k
ck seff p
r 162050.0
)10*16(*5.0*8.0*425.0051.0*113.2
***425.0*
3
,
213max,
Calculation of average strain:
41.631187
200000
cm
se
E
E
44
,
,
,
10*54.5*6,010*88.7200000
)050.0*41.61(*050.0
56.2*4.071.184).1(**
s
s
s
eff pe
eff p
eff ct
t s
cm sm E E
f k
Calculation of crack widths:
mm sw cm smr k 128.0)1088.7(*162.0)(* 4
max,
Finite elements modeling
■
■
■
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SLS load combinations (kNm) and stresses (MPa)
σ
σ
σ
σ
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5.19.2.3Reference results
Result name Result description Reference value
σ σ
σ σ
5.19.3Calculated results
Result name Result description Value Error
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5.20 EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement – Inclinedstress-strain diagram
Passed
5.20.1Description
5.20.2Background
5.20.2.1Model description
■
■
■
■
■
■ 8,02 ■
■
■■
■
■
■
■
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Units
Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■
■
671651
25
MPa f
E ck cm 31476
10
82522000
10
822000
3.03.0
■
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Boundary conditions
■
►
►
■
Loading
■
■
■
■
kNm M Ed 10024
30.6*75.184
8
²30.6*05.175
kNm M Ecq 7054
30.6*60
8
²30.6*123
kNm M Eqp 6374
30.6*55
8
²30.6*111
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5.20.2.2Reference results in calculating the equivalent coefficient
)(*)(*),( 00 t f t cm RH
925.2825
8.168.16)(
cm
cm f
f
φ
213
0
***1.0
1001
1
h
RH
RH
121 MPa f CM 35
7.0
1
35
cm f
2.0
2
35
cm f
Mpa Mpa f f ck cm 338
121
78.167.266*1.0
100
501
167.266)800400(*2
800*400*2*230
RH mm
u
Ach
54.2488.0*92.2*78.1)(*)(*),( 00 t f t cm RH
94.20
705
637*54..21
31476
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
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5.20.2.3Reference results in calculating the concrete beam reduced moment limit
MPa f ck 50
)*62.7969.165(*)*66.162.4(*)(
ck
ck eluc
f
f K
24 ***10 eee cba K
7.16928,18925*3,75 a
5.7345,87425*6,5 b
121325*04,0 c
181.1²94.20*1294.20*5.7347.1692*10 4 e K
422,1705
1002
271.0)*62.7969.165(*)*66.162.4(
*)(
ck
ck eluc
f
f K
Calculation of reduced moment:
301.067.16*²707.0*40.0
002.1
*²* cd w
Ed cu f d b
M
271.0301.0 luccu
Reference reinforcement calculation at SLU:
Calculation of the tension steel section (Section A1 ):
lu
Nm f d b M cd wlu Ed
6
1 10*902.067.16*²707.0*40.0*271.0*²** ■ α
404.0)271.0*21(1*25.1)*21(1*25.1 lulu
■
md z lulu 593.0)404.0*4.01(*707.0)*4.01(*
■
17.55.3*404.0
404.01*
12
cu
u
u su
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■
MPa su su 454*38,95271,432
Mpa MPa su 45463.43700517.0*38,95271,432
■
²76.3463.437*593.0
10*902.0
.
6
11 cm
MPam
Nm
f z
M A
yd lu
Ed
Compressed steel reinforcement reduction (Section As2 ):
9572045070704040707040401000
53 .
.1000
53 0
MPa sc yd sc 52.43500295.0*38.95271.43200217.000295.0
²47.352.435*)045.0707.0(
902.0002.1
)'(
12 cm
d d
M M A
sc
Ed Ed s
²48.364.437
52.435*47.3.22 cm
f A A
yd
sc s
Total area to be implemented:
■
■
Reference reinforcement calculation at SLS:
■ 705
■ 9420.
■
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Calculation of the resistance moment M Rd for detecting the presence of the compressed reinforcement:
md x sce
ce 311.0707.0*40015*94.20
15*94.20*
*
*1
N xb F cwc
6
1 10*933.015311.0*40.0*2
1
***2
1
m x
d z c 603.03
311.0707.0
3
1
Nm z F M ccrb
6610*563.0603.0*10*933.0*
Theoretical section 1 (tensioned reinforcement only)
Nm M M rb
6
1 10*563.0
440.0707.0
311.011
d
x
m x
d z c 603.03
311.0707.0
3
1
²34.23400603.0
563.011 cm
z
M A
sc
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
Nm M M M rbcq ser
66
,2 10*142.010*)591.0705.0(
d
d d ce sc
*
'***
1
1
MPa sc 66.268707.0*440.0
045.0707.0*440.0*15*94.20
²98.766.268*045.0707.0
142.0
*)'('
2
cmd d
M
A sc
²36.5400
66.268*98.7*' cm A A
s
sc s
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Section area:
Reference reinforcement third calculation at SLS:
MPa
A
A s
ELU
ELS s 21.300400*
24.38
70.28*
md x sce
ce 361.0707.0*21.30015*94.20
15*94.20*
*
*1
N xb F cwc
6
1 10*083.115361.0*40.0*2
1***
2
1
m x
d z c 587.03
361.0707.0
3
1
Nm z F M ccrb
66 10*636.0587.0*10*083.1*
Theoretical section 1 (tensioned reinforcement only)
Nm M M rb
6
1 10*636.0
511.0707.0
361.011
d
x
m
x
d z c 587.03
361.0
707.03
1
²09.36400*578.0
36.0
*
11 cm
z
M A
sc
Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)
Nm M M M rbcq ser
66
,2 10*069.010*)636.0705.0(
MPad
d d ce sc 275
707.0*511.0
045.0707.0*511.0*15*94.20
*
'***
1
1
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²79.3
275*045.0707.0
069.0
*)'(' 2 cm
d d
M A
sc
²47.321.300
275*79.3*' cm A A s
sc s
Section area:
Reference solution for minimal reinforcement area
d b
d b f
f
Max A
w
w yk
eff ct
s
**0013.0
***26.0
,
min,
MPa f f ctmeff ct 56.2,
²77.3
²10*68.3707.0*40.0*0013.0
²10*77.3707.0*40.0*500
56.2*26.0
max
4
4
min, cm
m
m A s
Finite elements modeling
■
■
■
ULS and SLS load combinations(kNm)
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Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
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5.20.2.4Reference results
Result name Result description Reference value
5.20.3Calculated results
Result name Result description Value Error
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5.21 EC2 Test28: Verifying the shear resistance for a T concrete beam with inclined transversalreinforcement - Bilinear stress-strain diagram (Class X0)
Passed
5.21.1Description
5.21.2Background
5.21.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
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■
■
■
■
Units
Geometry
■
■
■
■
■ θ ˚
Boundary conditions
■
►
►
■
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Loading:
2
**)( l P x P xV u
u
kN l P
V u Ed 9,292
2
10*57,58
2
*
5.21.2.2Reference results in calculating the maximum design shear resistance
cot
***1max,
tg
b z f vV wucd
Rd
30 90
1v
54.0250
251*6.0
2501*6,01
ck f v
md z u 688.0764.0*9.0*9.0
MN tg
V Rd 590.030cot30
22.0*688.0*67.16*54.0max,
MN V MN V Rd Ed 590.0293.0 max,
Calculation of transverse reinforcement:
ml cmtg
f z
tg V
s
A
ywd u
Ed sw /²66.5
15,1
500*688.0
30*293.0
*
*.
Finite elements modeling
■
■
■
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Advance Design gives the following results for Atz (cm2 /ml)
5.21.2.3Reference results
Result name Result description Reference value
5.21.3Calculated results
Result name Result description Value Error
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5.22 EC2 Test 40: Verifying a square concrete column subjected to a small compression force andsignificant rotation moment to the top - Bilinear stress-strain diagram (Class XC1)
Passed
5.22.1Description
5.22.2Background
5.22.2.1Model description
■
■
■
■
►
►
►
■
►
►
■
■
■
■
■
■■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■ 4511030750
35200
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5.22.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 60.11*20
37.8050.0
60.11*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
EQP M
Ed M
ieee 01
mei 03.0
kN N Eqp 10.177*30.0151
MNmkNme N M Eqp Eqp 181.058.18056.10*10.17* 111
0,t
)(*)(*),( 00 t f t cm RH
72.2830
8.168.16)(
cm
cm f
f
488.0281.0
1
1.0
1
)( 20.020.0
0
0 t t
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213
0
***1.0
1001
1
h
RH
RH
MPa f cm 35944.0
38
3535 7.07.0
1
cm f 984.0
38
3535 2.02.0
2
cm f
72.1984.0*944.0*
250*1.0
100
501
1250500500*2
500*500*2*230
RH mm
u
Ach
28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH
17.1352.0
181.0*28.2*, 0
Ed
EQP
ef M
M t
The second order effects; The buckling calculation:
n
C B A ***20lim
0062.020*²50.0
031.0
*
cd c
Ed
f A
N n
81.0
17.1*2.01
1
*2,01
1
ef
A
1.1*21 B
70.07,1 mr C
42.1580062.0
7.0*1.1*81.0*20lim
42.15837.80 lim
Calculation of the eccentricities and solicitations corrected for ULS:
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■
■
m N
M
e Ed
Ed
46.1103075.0
3525.00
ml
ei 03.0400
6.11
400
0
The first order eccentricity: stresses correction:
MN N Ed 03075.0
meee i 49.1101 MNm N e M Ed Ed 353.003075.0*49.11*1
0* e N M Ed
mmmm
mmmm
mmhmm
e 207.16
20max
30
50020
max
30
20max0
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Reinforcement calculation in the first order situation:
141.020*50.0*50.0
353.0
** 22
cd
Ed
f hb
M
00615.020*5.0*5.0
03075.0
**
cd
Ed
f hb
N
35.0
22
25.4078.434
20*50.0*cm A s
Finite elements modeling
■■
■
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Theoretical reinforcement area(cm2 )
Theoretical value (cm2 )
5.22.2.3Reference results
Result name Result description Reference value
5.22.3Calculated results
Result name Result description Value Error
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5.23 EC2 Test 42: Verifying a square concrete column subjected to a significant rotation momentand small compression force to the top with Nominal Curvature Method - Bilinear stress-straindiagram (Class XC1)
Passed
5.23.1Description
5.23.2Background
Nominal curvature method.
Verifies the
5.23.2.1Model description
■
■
■
■
►
►
►
■
►
►
■ 3,02 ■
■■
■
■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
►
►
■ 4511030750
35200
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5.23.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 60.11*20
37.8050.0
60.11*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
EQP M
Ed M
ieee 01
mei 03.0
me N
M eee i
Eqp
Eqp
i 56.1030.07*30.015
100*30.0150
0
0
01
kN N Eqp 10.177*30.0151
MNmkNme N M Eqp Eqp 181.058.18056.10*10.17* 111
MNm M Ed 3523.01
0,t
)(*)(*),( 00 t f t cm RH
72.2830
8.168.16)(
cm
cm f
f
488.0281.0
1
1.0
1)( 20.020.0
0
0 t t
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213
0
***1.0
1001
1
h
RH
RH
MPa f cm 35 944.038
3535 7.07.0
1
cm f 984.0
38
3535 2.02.0
2
cm f
72.1984.0*944.0*
250*1.0
100
501
1250500500*2
500*500*2*230
RH mm
u
Ach
28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH
The effective creep coefficient calculation:
17.1352.0
181.0*28.2*, 0
Ed
EQP
ef M
M t
The second order effects; The buckling calculation:
n
C B A ***20lim
0062.020*²50.0
031.0
*
cd c
Ed
f A
N n
81.0
17.1*2.01
1
*2,01
1
ef
A
1.1*21 B
70.07,1 mr C
42.1580062.0
7.0*1.1*81.0*20lim
42.15837.80 lim
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Calculation of the eccentricities and solicitations corrected for ULS:
■
■
■
■
m N
M e
Ed
Ed 46.1103075.0
3525.00
ml
ei 03.0400
6.11
400
0
MN N Ed 03075.0
meee i 49.1101
MNm N e M Ed Ed 353.003075.0*49.11*1
0* e N M Ed
207162030
50020
30200
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Reinforcement calculation in the first order situation:
141.020*50.0*50.0
353.0
** 22
cd
Ed
f hb
M
00615.020*5.0*5.0
03075.0**
cd
Ed
f hb N
35.0
22
25.4078.434
20*50.0*cm A s
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The nominal curvature methods (second order effect):
0
1
**
1
r K K r r
1
0
0107.045.0*45,0
1000
78.434
)*45,0()*45,0(
1 md
E
f
d r
s
yd
yd
1
bal u
ur
nn
nn K
00615.020*²50.0
03075.0
*
cd c
Ed
f A
N n
350.020*²50.0
78.434*10*25.40
*
* 4
cd c
yd s
f A
f A
350.1350.011 un
4,0bal n
41.140.0350.1
00615.0350.1
r K
1r K 1r K
K 1*1 ef K
Calculation moment:
20 M M M Ed Ed
Ed M 0
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2 M
22 * e N M Ed
mcl
r e 144.0
10²60.11*0107.0*1
2
02
MNme N M Ed 00443.0144.0*03075.0* 22
MNm M M M Ed Ed 357.000443.0353.020
MN N Ed 03075.0
MNm M Ed 357.0
Reinforcement calculation according the second order effects:
35.0 ²25.40078.434
20*²50.0*35.0***cm
f
f hb A
yd
cd s
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Finite elements modeling
■
■
■
Theoretical reinforcement area(cm2
)
Theoretical value (cm2 )
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5.23.2.3Reference results
Result name Result description Reference value
5.23.3Calculated results
Result name Result description Value Error
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5.24 EC2 Test30: Verifying the shear resistance for a T concrete beam with inclined transversalreinforcement - Bilinear stress-strain diagram (Class XC1)
Passed
5.24.1Description
5.24.2Background
5.24.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
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■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
Units
Geometry
■
■
■ θ ˚
Boundary conditions
■
►
►
■
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Loading:
5.24.2.2Reference results in calculating the longitudinal reinforcement
5.24.2.3Reference results in calculating the transversal reinforcement
ywd u
Ed sw f z
tg V
s
A
*
*.
md z u 469.0521.0*9.0*9.0
225.2cot
90
ml cmtg
f z
tg V
s
A
ywd u
Ed sw /²67.11
15,1
500*469.0
22*589.0
*
*.
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ml cmb f
f b
s
Aw
yk
ck
ww sw /²01.790sin*80.0*
500
30*08,0sin**
*08,0sin**min,
Finite elements modeling
■
■
■
5.24.2.4Reference results
Result name Result description Reference value
5.24.3Calculated results
Result name Result description Value Error
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5.25 EC2 Test 37: Verifying a square concrete column using the simplified method – Professionalrules - Bilinear stress-strain diagram (Class XC1)
Passed
5.25.1Description
5.25.2Background
5.25.2.1Model description
■
■
■
■
►
►
■►
■
■
■
■
■
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Units
Geometry
■■
■
■
Boundary conditions
Loading
■
kN N ED 2280800*5.1800*35.1
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5.25.2.2Reference results in calculating the concrete column
Scope of the method:
25.244.0
12*8.212*
121212
00
2
0
2
4
000 a
L
a
L
a
L
a
a
L
A
I
L
i
L
120
MPa f ck 5020
mh 15.0
Reinforcement calculation:
6025.24
746.0
62
25.241
86.0
621
86.022
93.0hk
1500
500*6.06.1
500*6.06.1 yk
s
f k
²24,1467.16*4.0*4.0746.0*1*93.0
280.2*
78.434
1**
***
1cm f hb
k k
N
f A cd
sh
ed
yd
s
Finite elements modeling
■
■
■
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Theoretical reinforcement area(cm2)
5.25.2.3Reference results
Result name Result description Reference value
5.25.3Calculated results
Result name Result description Value Error
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5.26 Verifying the minimum transverse reinforcement area results for an articulated beam (TTAD#11342)
Passed
5.26.1Description
5.27 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - horizontallevel behavior law
Passed
5.27.1Description
5.28 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - bilinearstress-strain diagram
Passed
5.28.1Description
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5.29 EC2 Test 46 I: Verifying a square concrete beam subjected to a normal force of traction -Inclined stress-strain diagram (Class X0)
Passed
5.29.1Description
5.29.2Background
Tie sizing
Bilinear stress-strain diagram
5.29.2.1Model description
■■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
■
■
Boundary conditions
■
►
►
■
5.29.2.2Reference results in calculating the concrete beam
Calculations according a bilinear stress-strain diagram
²50,11²10*50,11
15.1
400
400.0 4
,
, cmm N
AU s
Ed U s
Calculations according a inclined stress-strain diagram
MPaClassBS U s 373400 ,
²70,10²10*72,10373
400.0 4
,
, cmm N
AU s
Ed U s
Checking the condition of non-fragility:
yk
ctmc s
f
f A A *
MPa f f ck ctm 21,220*30.0*30.0 3/23/2
²0225.015.0*15.0 m Ac
²24.1400
21.2*0225.0* cm
f
f A A
yk
ctmc s
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Finite elements modeling
■
■
■
ULS load combinations (kNm)
5.29.2.3Reference results
Result name Result description Reference value
5.29.3Calculated results
Result name Result description Value Error
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5.30 EC2: Verifying the transverse reinforcement area for a beam subjected to linear loads
Passed
5.30.1Description
5.31 Verifying the reinforced concrete results on a fixed beam (TTAD #11836)
Passed
5.31.1Description
5.32 Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular crosssection
Passed
5.32.1Description
5.33 Verifying the minimum transverse reinforcement area results for articulated beams (TTAD#11342)
Passed
5.33.1Description
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5.34 Verifying Aty and Atz for a fixed concrete beam (TTAD #11812)
Passed
5.34.1Description
5.35 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - inclinedstress strain behavior law
Passed
5.35.1Description
5.36 Verifying the longitudinal reinforcement for linear elements (TTAD #11636)
Passed
5.36.1Description
5.37 Verifying the longitudinal reinforcement for a fixed linear element (TTAD #11700)
Passed
5.37.1Description
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5.38 EC2 : calculation of a square column in traction (TTAD #11892)
Passed
5.38.1Description
5.39 EC2: column design with “Nominal Stiffness method” square section (TTAD #11625)
Passed
5.39.1Description
5.40 Verifying concrete results for linear elements (TTAD #11556)
Passed
5.40.1Description
5.41 Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678)
Passed
5.41.1Description
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5.42 Verifying the reinforced concrete results on a structure with 375 load cases combinations(TTAD #11683)
Passed
5.42.1Description
5.43 Verifying the reinforcement of concrete columns (TTAD #11635)
Passed
5.43.1Description
5.44 Verifying concrete results for planar elements (TTAD #11583)
Passed
5.44.1Description
5.45 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load
Passed
5.45.1Description
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5.46 EC2 Test 7: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram
Passed
5.46.1Description
5.46.2Background
5.46.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
Materials properties
■
■
■
■
■ MPa f
f c
ck cd 67.10
5,1
25
■ MPa f f ck ctm 56.225*30.0*30.0 3/23/2
■ MPa f
f s
yk
yd 83.34715,1
400
Boundary conditions
■
►
►■
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Loading
■
■
kNm M Ed 40.2188
²8*3.27
kNm M Ecq 60.1458
²8*20.18
5.46.2.2Reference results in calculating the concrete beam moment
kNm f h
d hb M cd
f
f eff btu 92867.16*2
12.0482.0*12.0*10.1*
2**
Reference longitudinal reinforcement calculation:
051.067.16*²482.0*10.1
218.0
*²*
1 cd w
Ed cu
F d b
M
066.0051.0*211*25.1)*21(1*25.1 cuu
md z uc 469.0)066.0*40.01(*482.0)*4.01(*
²38.1383.347*469.0
218.0
*cm
f z
M A
yd c
Ed
Finite elements modeling
■
■
■
kNm M kNm M btu Ed 92840.218
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ULS load combinations(kNm)
Theoretical reinforcement area(cm2 )
5.46.2.3Reference results
Result name Result description Reference value
5.46.3Calculated results
Result name Result description Value Error
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5.47 EC2 Test 23: Verifying the shear resistance for a rectangular concrete - Bilinear stress-straindiagram (Class XC1)
Passed
5.47.1Description
5.47.2Background
5.47.2.1Model description
■
■
■
■
■
■
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■
■
■
■
■
■
■
Units
Geometry
■
■
■
■
■
■ ■ ˚
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Boundary conditions
■
►
►
■
Loading
406
76373 51239
82210 5770
0
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5.47.2.2Reference results in calculating the maximum design shear resistance
21max,cot1
cotcot*****
wucd cw Rd b z f V
1cw
250
1*6,01ck f
v
cot
***1max,
tg
b z f vV wucd
Rd
45 90
1v
54.0250
251*6.0
2501*6,01
ck f v
md z u 561.0623.0*9.0*9.0
MN V Rd 757.02
30.0*561.0*67.16*54.0max,
MN V MN V Rd Ed 757.0406.0 max,
Calculation of transversal reinforcement:
ml cmtg
f z
tg V
s
A
ywd u
Ed sw /²65,16
15,1
500*561,0
45*406,0
*
*.
ml cmtg s
A sw /²82,9
15,1
500*561,0
45*23951,0
ml cmtg
s
A sw /²48,8
15,1
500*561,0
45*20674,0
Finite elements modeling
■
■
■
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Advance Design gives the following results for Atz (cm2 /ml)
The reinforcement theoretical value is calculated using the formula:
ml cmb f
f b
s
Aw
yk
ck
ww sw /²4.290sin*30.0*
500
25*08,0sin**
*08,0sin**min,
5.47.2.3Reference results
Result name Result description Reference value
5.47.3Calculated results
Result name Result description Value Error
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5.48 EC2 Test 41: Verifying a square concrete column subjected to a significant compression forceand small rotation moment to the top - Bilinear stress-strain diagram (Class XC1)
Passed
5.48.1Description
5.48.2Background
5.48.2.1Model description
■
■
■
■
►
►
►
■
►
►
■ 3,02 ■
■
■
■■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■ 08703530
0370500
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5.48.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 60.11*20
37.8050.0
60.11*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
EQP M
Ed M
ieee 01
mei 03.0
me N
M eee i
Eqp
Eqp
i 125.030.0100*30.0150
7*30.015
0
0
01
kN N Eqp 180100*30.01501
MNmkNme N M Eqp Eqp 0225.050.22125.0*180* 111
MNm M Ed
041.01
0,t
)(*)(*),( 00 t f t cm RH
72.2830
8.168.16)(
cm
cm f
f
488.0281.0
1
1.0
1
)( 20.020.00
0 t t
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213
0
***1.0
1001
1
h
RH
RH
MPa f cm 35 944.038
3535 7.07.0
1
cm f 984.0
38
3535 2.02.0
2
cm f
72.1984.0*944.0*
250*1.0
100
501
1250500500*2
500*500*2*230
RH mm
u
Ach
28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH
25.1041.0
0225.0*28.2*, 0
Ed
EQP
ef M
M t
The second order effects; The buckling calculation:
n
C B A ***20lim
071.020*²50.0
353.0
*
cd c
Ed
f A
N n
80.0
25.1*2.01
1
*2,01
1
ef
A
1.1*21 B
70.07,1 mr C
24.46071.0
7.0*1.1*80.0*20lim
24.4637.80 lim
Calculation of the eccentricities and solicitations corrected for ULS:
■
■
■
■
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m N
M e
Ed
Ed 087.0353.0
03075.00
ml ei 03.0400
6.11
4000
The first order eccentricity: stresses correction:
MN N Ed 353.0
meee i 117.001
MNm N e M Ed Ed 041.0353.0*117.0*1
0* e N M Ed
mmmm
mmmm
mmhmm
e 207.16
20max
30
50020
max
30
20max0
Reinforcement calculation in the first order situation:
MNmh
d N M M Gua 112.02
50.045.0*353.0041.0)
2(*0
Verification about the partially compressed section:
494,0)45,0
50,0*4,01(*
45,0
50,0*8,0)*4,01(**8,0
d
h
d
h BC
055.020*²45.0*50.0
112.0*²*
cd w
uacu
f d b M
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494.0055.0 BC cu
The calculation for the tensioned steel in pure bending:
055.0cu
071,0)055,0*21(1*25,1 u
md z uc 437,0)071,0*4,01(*45,0)*4,01(*
²89,578,434*437,0
112,0
*cm
f z
M A
yd c
ua
The calculation for the compressed steel in bending:
24 23.278.434
353.010*89.5 cm
F
N A A
yd
²81.078.434
353.0*10.0*10,0min, cm
F
N A
yd
Ed s
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5.48.2.3Calculation of the second order effects:
Estimation of the nominal rigidity:
s s sccd c I E K I E K EI ****
2.1
cmcd
E E
Mpa Mpa f f ck cm 388
Mpa f
E cmcm 57.32836
10
38*22000
10*22000
3.03.0
Mpa E
E cmcd 27364
2.1
32837
2.1
4343
10.208,512
50.0
12
*m
hb I c
Mpa E s 200000
s I
002.050.0*50.0
10.5 4
c
s
A
A
01.0002.0 c
s
A
A
22.120
30
201 ck f
k
071.020*²50.0
353.0
*
cd c
Ed
f A
N n
20.00336.0170
37.80*071.0
170*2
nk
45
242
10.205.02
50.0*
2
10*5*2
2*
2*2 mc
h A I s
s
1 s K 018.0
25.11
0336.0*22.1
1
* 21
ef
c
k k K
²56.610*2*200000*110*208,5*27364*018.0 53 MNm EI
Stress correction:
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1
1*0
Ed
B Ed Ed
N
N M M
MNm M Ed 041.00
MN N Ed 353.0
0
²
c
80 c
234.18
²
MN l
EI N B
48.0²60.11
56.6*²*²2
0
MNm M Ed 182.0
1353.0
48.0
234.11*041.0
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Calculation of the flexural combined reinforcement
MNm M Ed 182.0
MN N Ed 353.0
073.020*50.0*50.0
182.0
** 22
cd
Ed
f hb
M
071.020*5.0*5.0
353.0
**
cd
Ed
f hb
N
11.0
22
65.1278.434
20*50.0*11.0***cm
f
f hb A
yd
cd s
Additional iteration:
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Estimation of the nominal rigidity:
s s sccd c I E K I E K EI ****
2.1cmcd
E E
Mpa Mpa f f ck cm 388
Mpa f
E cmcm 57.32836
10
38*22000
10*22000
3.03.0
Mpa E
E cmcd 27364
2.1
32837
2.1
4343
10.208,512
50.0
12
*
m
hb
I c
Mpa E s 200000
s I
0026.050.0*50.0
10*5.6 4
c
s
A
A
01.0002.0 c
s
A
A
22.120
30
201 ck f
k
071.020*²50.0
353.0
*
cd c
Ed
f A
N n
20.00336.0170
37.80*071.0
170*2
nk
45
242
106.205.0
2
50.0
2
10*5.6*2
2
*
2
*2 mch A
I s s
1 s K 018.0
25.11
0336.0*22.1
1
* 21
ef
c
k k K
²78.710*6.2*200000*110*208,5*27364*018.0 53 MNm EI
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Stress correction:
11*0
Ed
B Ed Ed
N N M M
MNm M Ed 041.00
MN N Ed 353.0
0
²
c
80 c
234.18
²
MN l
EI N B 57.0
²60.11
78.7*²*²
2
0
MNm M Ed 123.0
1
353.0
57.0
234.11*041.0
Calculation of the flexural compound reinforcement
MNm M Ed 123.0
MN N Ed 353.0
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049.020*50.0*50.0
123.0
** 22
cd
Ed
f hb
M
071.020*5.0*5.0
353.0
** cd
Ed
f hb
N
05.0
22
75.578.434
20*50.0*05.0***cm
f
f hb A
yd
cd s
Finite elements modeling
■
■
■
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Theoretical reinforcement area(cm2 )
Theoretical value (cm2 )
5.48.2.4Reference results
Result name Result description Reference value
5.48.3Calculated results
Result name Result description Value Error
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5.49 EC2 Test32: Verifying a square concrete column subjected to compression and rotationmoment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (ClassXC1)
Passed
5.49.1Description
5.49.2Background
5.49.2.1Model description
■
■
■
■
►
►
►
■
►
►
► 3,02 ►
►
►
►
►
►
►
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■ 1003000
03100
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5.49.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 12*20
Calculating the slenderness of the column:
10440.0
12*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef M
M
t ., 0
0,t
EQP M
Ed M
ieee 01
mei 03.0
me N
M eee i
Eqp
Eqp
i 13.030.065.0*30.015
70.0*30.050.1
0
0
01
t N Eqp 95.1670.0*30.050.11
MNmtme N M Eqp Eqp 022.020.213.0*95.16* 111
0,t
)(*)(*),( 00 t f t cm RH
92.2825
8.168.16)(
cm
cm f
f
488.0281.0
11.0
1)(20.020.0
0
0
t
t
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30*1.0
1001
1h
RH
RH
85.1
200*1.0
100
50
11200400400*2400*400*2*2
30
RH mmu Ach
64.2488.0*92.2*85.1)(*)(*),( 00 t f t cm RH
49.1039.0
022.0*64.2*, 0
Ed
EQP
ef M
M t
The necessity of buckling calculation (second order effect):
n
C B A ***20lim
112.067.16*²40.0
300.0
*
cd c
Ed
f A
N n
77.049.1*2.01
1
*2,01
1
ef A
1.1*21 B
70.07,1 mr C
43.35112.0
7.0*1.1*77.0*20lim
43.35104 lim
The second order effects; The buckling calculation:
■
■
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m N
M e
Ed
Ed 10.0300.0
031.00
ml ei 03.0400
12
400
0
The first order eccentricity: stresses correction:
MN N Ed 300.0
meee i 13.001
MNm N e M Ed Ed 039.0300.0*)03.010.0(*1
0* e N M Ed
mmmm
mmmm
mmhmm
e 203.13
20max
30
40020
max
30
20max0
Reinforcement calculation in the first order situation:
MNmh
d N M M Gua 084.02
40.035.0*300.0039.0)
2(*0
496,0)35,0
40,0*4,01(*
35,0
40,0*8,0)*4,01(**8,0
d
h
d
h BC
103.0
67.16*²35.0*40.0
084.0
*²*
cd w
uacu
f d b
M
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BC cu 496.0103.0
Calculations of steel reinforcement in pure bending:
Calculations of steel reinforcement in combined bending:
24 06.178.434
300.010*84.5' cm
f
N A A
yd
2
min, 02.3*002.0²69.078.434
300.0*10.0*10,0cm Acm
f
N A c
yd
Ed s
The second order effects calculation:
Calculation of nominal curvature:
0
1**
1
r K K
r r
1
0
0138.0
35.0*45.0
200000
78.434
*45,0*45,0
1 m
d
E
f
d r
s
yd
yd
r K 1
bal u
ur
nn
nn K
112.067.16*²40.0
300.0
*
cd c
Ed
f A
N n
0512.067.16*²40.0
78.434*10.14,3
*
* 4
cd c
yd s
f A
f A
0512.10512.011 u
n
4,0bal n
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1144.140.00512.1
112.00512.1
r r K K
K 1*1 ef K
218.0150
104
200
2535.015020035,0
ck
f
11675.049.1*218.01*1 K K ef
1
0
0138.01
**1 m
r K K
r r
20 M M M Ed Ed
Ed M 0
2 M
22 * e N M Ed
mc
l
r e 199.0
10
²12*0138.0*
1 2
02
MNme N M Ed 0597.0199.0*300.0* 22 MNm M M M Ed Ed 099.00597.0039.020
MN N Ed 300.0
MNm M Ed 099.0
Reinforcement calculation corresponding to the second order:
093.067.16*²40.0*40.0
099.0** 2
cd
Ed
f hb M
112.067.16*40.0*40.0
300.0
**
cd
Ed
f hb
N v
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150& 20978434
6716400150&
Buckling checking
The curvature evaluation:
r K
1
0
0138.01 m
r
1
bal u
ur
nn
nn K
112.0n
15.067.16*²40.0
78.434*10*20,9
*
* 4
cd c
yd s
f A
f A
15.115.011 un
4,0bal n
11384.140.015.1
112.015.1
r r K K
1 K 1*1 ef K
1
0
0138.01
**1 m
r K K
r r
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Finite elements modeling
■
■
■
Theoretical reinforcement area (cm2
)
Theoretical value (cm2 )
5.49.2.3Reference results
Result name Result description Reference value
5.49.3Calculated results
Result name Result description Value Error
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5.50 EC2 Test 38: Verifying a rectangular concrete column using the simplified method –Professional rules - Bilinear stress-strain diagram (Class XC1)
Passed
5.50.1Description
5.50.2Background
5.50.2.1Model description
■
■
■
■
►
►
■
■
■■
■
Units
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Geometry
■
■
■
■
Boundary conditions
Loading
■
kN N ED 1080800*35.1 5.50.2.2Reference results in calculating the concrete column
Scope of the method:
06.753.0
12*5.612*
121212
00
2
0
2
4
000 a
L
a
L
a
L
a
a
L
A
I
L
i
L
120
MPa f ck 5020
mh 15.0
Reinforcement calculation:
12006.7560
33.0
06.75
3232 3.13.1
93.0)**61(*)*5.075.0( hk h
1500
500*6.06.1
500*6.06.1 yk
s
f k
²43.2367.16*5.0*3.033.0*1*93.0
08.1*
78.434
1**
***
1cm f hb
k k
N
f A cd
sh
ed
yd
s
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Finite elements modeling
■
■
■
Theoretical reinforcement area(cm2
)
5.50.2.3Reference results
Result name Result description Reference value
5.50.3Calculated results
Result name Result description Value Error
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5.51 EC2 Test 18: Verifying a rectangular concrete beam subjected to a uniformly distributed load,with compressed reinforcement - Bilinear stress-strain diagram (Class XD1)
Passed
5.51.1Description
5.51.2Background
Simple Bending Design for
σσ
5.51.2.1Model description
■
■
■
■
■
■
■
■
■
■ σ
■ σ
Units
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Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■ 25■ 500
■ 7037
■ 286
Boundary conditions
■
►
►
■
Loading
■
►
■
►
►
►
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5.51.2.2Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 92.2825
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
7.0
1
35
cm f
2.0
2
35
cm f
121
mm
u
Ach 48.243
800350*2
800*350*220
56.2488.0*92.2*80.1)(*)(*),(80.148.243*1.0
100
501
1 003
t f t cm RH RH
Ecar
Eqp
cm
se
M
M t
E
E
*),(1 0
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MPa f
E ck cm 31476
10
825*22
10
8*22
3.03.0
MPa E s 200000
664.2600
390*56.21*),(1 0
Ecar
Eqp
M
M t
90.16
664.2
31476
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
Mpa f ck bc 1525*6,06,0
Mpa f yk s 320*8,0
Neutral axis position calculation:
0..2
1111
413435
286470377290163522867037901628670379016
.2..1
Calculating the second moment:
443
11
31
0147201472097445349016286453472901670373
413435
..3
Stresses calculation:
Mpa Mpa x I
M
c
ser
c 12143441,0*01472,0
600,0
* 1
Mpa Mpa x
xd sce st 400259
3441,0
3441,072,0*14*90.16**
1
1
Finite elements modeling
■
■
■
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SLS load combinations (kNm) and stresses (MPa)
σ
σ
σ
σ
5.51.2.3Reference results
Result name Result description Reference value
σ σ
σ σ
5.51.3Calculated results
Result name Result description Value Error
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5.52 EC2 Test 19: Verifying the crack openings for a rectangular concrete beam subjected to auniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram(Class XD1)
Passed
5.52.1Description
5.52.2Background
σσ
5.52.2.1Model description
■
■
■
■
■
■
■
■
■ σ
■ σ
■
■
Units
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Geometry
■
■
■
■■
■
Materials properties
■
■
■
■
■ 25
■ 500
■ 4615
Boundary conditions
■
►
►
■
Loading
■
►
►
►
►
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5.52.2.2Reference results in calculating the concrete final value of creep coefficient
)()(),( 00 t f t cm RH
MPa f f cm
cm 92.2825
8.168.16
)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
21
0
3 **
*1.0
1001
1
h
RH
RH
121 Mpa f cm 35
7.0
1
35
cm f
2.0
2
35
cm f
mm
u
Ach 150
600200*2
600*200*220
77.2488.0*92.2*94.1)(*)(*),(94.1150*1.0
100
501
1 003
t f t cm RH RH
Ecar
Eqp
cm
se
M
M t
E
E
*),(1 0
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MPa f
E ck cm 31476
10
825*22
10
8*22
3.03.0
200000
75.2160
101*77.21*),(1 0
Ecar
Eqp
M
M t
75.2
160
101*77.21
31476
200000
*),(1 0
Ecar
Eqp
cm
se
M
M t
E
E
Material characteristics:
Mpa f ck bc 1525*6,06,0
Mpa f yk s 320*8,0
Neutral axis position calculation:
0..2
1111
2.. 2
1
0 sc A
672620
46155347172024615471746154717
.2.. 2
1
Calculating the second moment:
443
11
31
003140267053047171046153
267020
..3
Stresses calculation:
Mpa x I
M ser c 59.8267,0*
00314,0
101,0* 1
Mpa Mpa x
xd sce st 40082.147
267,0
267,053,0*59.8*47.17**
1
1
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Maximum spacing of cracks:
2
, 0222.0111.0*2.0
3.02
6.0
111.03
)267.06.0(175.0)53.06.0(*5.2
min*20.0
2
3
)()(*5.2
min* m
h
xhd h
b A eff c
070.00222.0
10*46.15 4
,
,
eff c
seff p
A
A
mmnn
nneq 22.18
**
**
2211
2
22
2
11
eff p
r
k k ck s
,
213max,
***425.0*
114.251
25*4.3
25*4.3
3/23/2
3
ck
mmk k
ck seff p
r 15207.0
)10*22.18(*5.0*8.0*425.0051.0*114.2
***425.0*
3
,
213max,
Calculation of average strain:
35.631476
200000
cm
se
E
E
44
,
,
,
10*43.4*6,010*33.6
200000
)07.0*35.61(*07.0
56.2*4.082.147).1(**
s
s
s
eff pe
eff p
eff ct
t s
cm sm
E E
f k
Calculation of crack widths:
mm sw cm smr k 096.0)1033.6(*152)(* 4
max,
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Finite elements modeling
■
■
■
SLS load combinations (kNm) and stresses (MPa)
σ
σ
σ
σ
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5.52.2.3Reference results
Result name Result description Reference value
σ σ
σ σ
5.52.3Calculated results
Result name Result description Value Error
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5.53 EC2 Test31: Verifying a square concrete column subjected to compression and rotationmoment to the top - Bilinear stress-strain diagram (Class XC1)
Passed
5.53.1Description
5.53.2Background
5.53.2.1Model description
■
■
■
■
►
►
►
■
►
►
■ 3,02 ■
■
■
■■
■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■ 1003000
03100
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5.53.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 12*20
10440.0
12*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
ieee 01
mei 03.0
me N
M eee i
Eqp
Eqp
i 13.030.065.0*30.015
70.0*30.050.1
0
0
01
t N Eqp 95.1670.0*30.050.11
MNmtme N M Eqp Eqp 022.020.213.0*95.16* 111
)(*)(*),( 00 t f t cm RH
92.2825
8.168.16)(
cm
cm f
f
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
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30*1.0
1001
1h
RH
RH
85.1
200*1.0
100
50
11200400400*2400*400*2*2
30
RH mmu Ach
64.2488.0*92.2*85.1)(*)(*),( 00 t f t cm RH
49.1039.0
022.0*64.2*, 0
Ed
EQP
ef M
M t
The second order effects; The buckling calculation:
n
C B A ***20lim
112.067.16*²40.0
300.0
*
cd c
Ed
f A
N n
77.0
49.1*2.01
1
*2,01
1
ef
A
1.1*21 B
70.07,1 mr C
43.35112.0
7.0*1.1*77.0*20lim
43.35104 lim
The second order effects; The buckling calculation:
■
■
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m N
M e
Ed
Ed 10.0300.0
031.00
ml ei 03.0400
12
400
0
The first order eccentricity: stresses correction:
MN N Ed 300.0
meee i 13.001
MNm N e M Ed Ed 039.0300.0*)03.010.0(*1
0* e N M Ed
Reinforcement calculation in the first order situation:
MNmh
d N M M Gua 084.02
40.035.0*300.0039.0)
2(*0
496,0)35,0
40,0*4,01(*
35,0
40,0*8,0)*4,01(**8,0
d
h
d
h BC
103.0
67.16*²35.0*40.0
084.0
*²*
cd w
uacu
f d b
M
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BC cu 496.0103.0
Calculations of steel reinforcement in pure bending:
103.0cu
136,0)103,0*21(1*25,1 u
md z uc 331,0)136,0*4,01(*35,0)*4,01(*
Calculations of steel reinforcement in combined bending:
24 06.1
78.434
300.010*84.5' cm
f
N A A
yd
2
min, 02.3*002.0²69.078.434
300.0*10.0*10,0cm Acm
f
N A c
yd
Ed s
The second order effects calculation:
Calculation of nominal rigidity:
s s sccd c I E K I E K EI ****
2.1
cmcd
E E
Mpa Mpa f f ck cm 338
Mpa f E cmcm 31476
1033*22000
10*22000
3.03.0
Mpa E
E cmcd 26230
2.1
31476
2.1
410.133,212
40.0
12
* 343
mhb
I c
Mpa E s 200000
s
I
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002.040.040.0
10.14,3 4
c
s
A
A
01.0002.0 c
s
A
A
Mpa f
k ck 12.120
25
201
112.067.16*²40.0
300.0
*
cd c
Ed
f A
N n
20.0069.0170
104*112.0
170*2
nk
410*06,705.02
40.0*
2
10*14,3*2
2*
2*2 6
242
mch A
I s
s
1 s K
031.049.11
069.0*12.1
1
* 21
ef
c
k k K
²15.310*06,7*200000*110*133,2*26230*031.0 63 MNm EI
Corrected stresses:
1
1*0
Ed
B Ed Ed
N
N M M
MNm M Ed 039.00
MN N Ed 300.0
0
²
c
80 c
234.18
²
MN l
EI N B 216.0
²12
15.3*²*²
2
0
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MNm M Ed 133.0
1300.0
216.0
234.11*039.0
s s sccd c I E K I E K EI ****
Mpa E cd 26230 410.133,2 3 m I c
Mpa E s 200000
s I
²8²40,0*005,0*005,0005,0 cm A A c s
²04,8005,0 cm A s
01.0002.0 c
s
A
A
45
242
10*81,105.0
2
40.0*
2
10*04,8*2
2
*
2
*2 mch A
I s s
1 s K
031.049.11
069.0*12.1
1
* 21
ef
c
k k K
²35.510.81,1*200000*110.133,2*26230*031.0 53 MNm EI
11*0
Ed
B Ed Ed
N
N M M
234.1
MN l
EI N B 367.0
²12
35.5*²*²
2
0
MNm M Ed 254.0
1300.0
367.0
234.11*039.0
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Reinforcement calculation for combined bending
238.067.16*²40.0*40.0
254.0
** 2
cd
Ed
f hb
M
112.067.16*40.0*40.0
300.0
**
cd
Ed
f hb
N v
485.0 ²75.2978.434
67.16*²40.0*485.0***cm
f
f hb A
yd
cd s
Buckling checking
The normal rigidity evaluation:
s s sccd c I E K I E K EI ....
s I
03.040.0*40.0
10*25,48 4
c
s
A A
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44
242
10.086,105.02
40.0*
2
10*25,48*2
2*
2*2 mc
h A I s
s
12.11 k
069.02 k
1 s K
031.049.11
069.0*12.1
1
* 21
ef
c
k k K
01.0c
s
A
A
ef
c K *5,01
3,0
²45.2310*086,1*20000010*133,2*26230*031.0 43 MNm EI
Corrected stresses
1
1*0
Ed
B Ed Ed
N
N M M
MNm M Ed 039.00
MN N Ed 300.0
234.1
MN l
EI N B 607.1
²12
45.23*²*²
2
0
MNm M Ed 05.0
1300.0
607.1
234.11*039.0
Reinforcement calculation:
047.067.16*²40.0*40.0
05.0
**05.0
2
cd
Ed Ed
f hb
M MNm M
112.0
67.16*40.0*40.0
300.0
**
300.0 cd
Ed Ed
f hb
N v MN N
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Additional iteration:
Nominal rigidity evaluation:
s s sccd c I E K I E K EI ....
s I
00754.040.0*40.0
10*06,12 4
c
s
A
A
01.0002.0 c
s
A
A
45
242
10.71,205.0240.0*
210*06,12*2
2*
2*2 mch A I s
s
1 s K
031.049.11
069.0*12.1
1
* 21
ef
c
k k K
²15.710*71,2*200000*110*133,2*26230*031.0 53 MNm EI
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Second order loads calculation:
1
1*0
Ed
B Ed Ed
N
N M M
MNm M Ed 039.00
MN N Ed 300.0
234.1
MN l
EI N B 49.0
²12
15.7*²*²
2
0
MNm M Ed 115.0
1300.0
49.0234.11*039.0
Reinforcement calculation:
108.067.16*²40.0*40.0
115.0
** 2
cd
Ed
f hb
M
112.0
67.16*40.0*40.0
300.0
**
cd
Ed
f hb
N v
18.0 ²04.1178.434
67.16*²40.0*18.0***cm
f
f hb A
yd
cd s
Finite elements modeling
■
■
■
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Theoretical reinforcement area(cm2 )
Theoretical value (cm2 )
5.53.2.3Reference results
Result name Result description Reference value
5.53.3Calculated results
Result name Result description Value Error
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5.54 EC2 Test 25: Verifying the shear resistance for a rectangular concrete beam with inclinedtransversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Passed
5.54.1Description
5.54.2Background
5.54.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
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Units
Geometry
■■
■
■
■
■
■ θ ˚
Boundary conditions
■
►
►
■
Loading:
l
a P V u Ed 1
KN V Ed 5,1893
11*25,284
Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to anunreduced shear, that will be used for further calculations.
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5.54.2.2Reference results in calculating the maximum design shear resistance
1cw
250
1*6,01ck f
v
30 45
1v
55.0250
201*6.0
2501*6,01
ck f v
md z u 392.0435.0*9.0*9.0
3940
301
3420039203313540
MN V MN V Rd Ed 394.01895.0 max,
Calculation of transverse reinforcement:
ml cm f z
V
s
A
ywd u
Ed sw /²76.545sin*)45cot30(cot*78.434*392.0
18950.0
sin*)cot(cot**
ml cm s
A sw /²88.245sin*)45cot30(cot*78.434*392.0
09475.0
Finite elements modeling
■
■
■
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5.54.2.3Reference results
Result name Result description Reference value
5.54.3Calculated results
Result name Result description Value Error
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5.55 EC2 Test 4 I: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclinedstress-strain diagram
Passed
5.55.1Description
5.55.2Background
d d d
x
cuud
cuu
cuud
cuu ..x 2
2
2
2
■ 13460.50220
■ 0720.450
■ 0490.50670
MPa AS su su 454.38,95271,432500
MPa BS su su 466.27,72771,432500
MPaC S su su 493.52,89571,432500
■
■■
8/22/2019 AD Validation Guide 2014 En
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5.55.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
8/22/2019 AD Validation Guide 2014 En
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Units
Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■
■ 671651
25
■ 56225300300 3232
■ 78434151
500
Boundary conditions
■
►
►
■
8/22/2019 AD Validation Guide 2014 En
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Loading
■
■
■
kNm M Ed 16.5218
²80.5*94.123
kNm M Ecq 68.3628
²80.5*25.86
5.55.2.2Reference results in calculating the concrete beam reduced moment limit
372.0lu
Reference reinforcement calculation:
■
■
096,067,16*²806.0*50,0
10*52116.0
*²* 2
3
MPamm
Nm
f d b
M
cd w
Ed
cu
■ α
■ 13460..
50220
■ 4540225038952714321500
27.3127.01
127.0
*50.221* cu
cu
sucu
8/22/2019 AD Validation Guide 2014 En
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cd f
md z cuc 765.0)127.0*4.01(*806.0)*4,01(*
²00.15
454*765.0
52116.0
*
cm
f z
M A
yd c
Ed u
Reference solution for minimal reinforcement area
d b
d b f
f
Max A
w
w
yk
eff ct
s
**0013.0
***26.0 ,
min,
MPa f f ctmeff ct 56.2,
²38.5
²10*24.5806.0*50.0*0013.0
²10*38.5806.0*50.0*500
56.2*26.0
max4
4
min, cm
m
m A s
Finite elements modeling
■
■
■
ULS and SLS load combinations (kNm)
8/22/2019 AD Validation Guide 2014 En
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Theoretical reinforcement area (cm2 )
Minimum reinforcement area(cm2 )
5.55.2.3Reference results
Result name Result description Reference value
5.55.3Calculated results
Result name Result description Value Error
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5.56 EC2 Test 1: Verifying a rectangular cross section beam made from concrete C25/30 to resistsimple bending - Bilinear stress-strain diagram
Passed
5.56.1Description
5.56.2Background
5.56.2.1Model description
■
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■ 671651
25
■ 56225300300 3232
■ 78434151
500
■ 3147610
8252200010
822000
3030
Boundary conditions
■
►
►
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 596/1115
Loading
■
■
■
mkN M Ed .59,232
8
²80,5*31,55
mkN M Ecq .94,1628
²80,5*75,38
5.56.2.2Reference results in calculating the concrete beam reduced moment limit
■
■
Reference reinforcement calculation:
■
■
203,067,16*²524,0*25,0
233,0
*²*
cd w
Ed cu
f d b
M
285.0203.0 lucu
■
md z uc 464,0)287,0*4,01(*524,0)4,01(*
■
²55,11²10*55,1178,434*464,0
233,0
*
4 cmm f z
M A
yd c
Ed u
8/22/2019 AD Validation Guide 2014 En
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Reference solution for minimal reinforcement area
00130
260
562
761
10701524025000130
107615240250500
562260
4
4
Finite elements modeling
■■
■
ULS and SLS load combinations(kNm)
Theoretical reinforcement area(cm2 )
8/22/2019 AD Validation Guide 2014 En
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Minimum reinforcement area(cm2 )
5.56.2.3Reference results
Result name Result description Reference value
5.56.3Calculated results
Result name Result description Value Error
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5.57 Modifying the "Design experts" properties for concrete linear elements (TTAD #12498)
Passed
5.57.1Description
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5.58 EC2 Test 4 II: Verifying a rectangular concrete beam subjected to Pivot B efforts – Inclinedstress-strain diagram
Passed
5.58.1Description
5.58.2Background
d d d
x
cuud
cu
ucuud
cuu ..x 2
2
2
2
■ 13460.50220
■ 0720.450
■ 0490.50670
MPa AS su su 454.38,95271,432500
MPa BS su su 466.27,72771,432500
MPaC S su su 493.52,89571,432500
■
■
■
8/22/2019 AD Validation Guide 2014 En
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5.58.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
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Units
Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■
■ 671651
25
■ 56225300300 3232
■ 78434151
500
Boundary conditions
■
►
►
■
8/22/2019 AD Validation Guide 2014 En
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Loading
■
■
■
kNm M Ed 16.5218
²80.5*94.123
kNm M Ecq 68.3628
²80.5*25.86
5.58.2.2Reference results in calculating the concrete beam reduced moment limit
372.0lu
Reference solution for reinforcement from Class B steel ductility
■
■
096,067,16*²806.0*50,0
10*52116.0*²* 2
3
MPamm Nm
f d b M
cd w
Ed cu
■ α
cd f
md z cuc 765.0)127.0*4.01(*806.0)*4,01(*
■
1.245.3*
127,0
127,01*
12cu
u
u su
■
MPa su su 466*27,72771,432
MPa Mpa su 46625.4500241,0*27,72771,432
²13.1525.450*765.0
52116.0
*cm
f z
M A
yd c
Ed u
Finite elements modeling
■
■
■
8/22/2019 AD Validation Guide 2014 En
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ULS and SLS load combinations (kNm)
Theoretical reinforcement area (cm2 )
5.58.2.3Reference results
Result name Result description Reference value
5.58.3Calculated results
Result name Result description Value Error
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5.59 EC2 Test 47: Verifying a rectangular concrete beam subjected to tension load - Bilinear stress-strain diagram (Class XD2)
Passed
5.59.1Description
5.59.2Background
5.59.2.1Model description
■
■
■
■
■
■
■
■
Units
Geometry
■
■
■
■
■
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Materials properties
■
■
■■
■
Boundary conditions
■
►
►
■
Loading
■
4071505113535151351
■
285150135
■
1801503013530
8/22/2019 AD Validation Guide 2014 En
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5.59.2.2Reference results in calculating the longitudinal reinforcement and the crack width
Calculating for ultimate limit state:
²37.9²10*37.9
15.1
500
407.0 4
,
, cmm N
AU s
Ed U s
Calculating for serviceability limit state:
S
ser ser s
N A
,
Mpa f yk s 400500*8,0*8,0
²13.7²10*13.7400
285.0 4
, cmm A ser s
Final reinforcement
MPa MPa A
N s s
ser s
ser s 40080.226
001257.0
285.0
,
Crack opening verification:
Calculating the maximum spacing between the cracks
eff p
r
k k ck s
,
213max,
***425.0*
cmd 4.13
2
26.0520
020.0100.0
2
20.0165.0)134.020.0(*5.2
min*20.0
2
)(*5.2min*,
hd h
b A eff c
²040.0, m A eff c
031.0040.0
10*57.12 4
,
,
eff c
seff p
A
A
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mmnn
nneq 20
**
**
2211
2
22
2
11
eff pr
k k
ck s,
21
3max,
***425.0
*
801
12
99156
2543
2543
3232
3
mmk k
ck s
eff p
r 331
031.0
20*1*8.0*425.056*99.1
***425.0*
,
213max,
Calculation of average deformations
35.631476
200000
cm
se
E
E
Mpa A
N
ser s
qp ser
s 2.143001257.0
180.0
,
44 103041
6010175
200000
031035610310
56240143!.1
! 1
00
mm sw cm smr k 171.0)10*17.5(*331)(* 4
max,
Checking the condition of non-fragility:
yk
ctmc s
f
f A A .
Mpa f f ck ctm 56,225*30.0*30.0 3/23/2
8/22/2019 AD Validation Guide 2014 En
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²04.020.0*20.0 m Ac
²05.2500
56.2*04.0* cm
f
f A A
yk
ctmc s
Finite elements modeling
■
■
■
Longitudinal reinforcement for ULS and SLS load combinations (kNm)
Stresses in the steel reinforcement
Maximum spacing of cracks S r, max (m)
Crack width w k (mm)
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5.59.2.3Reference results
Result name Result description Reference value
5.59.3Calculated results
Result name Result description Value Error
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5.60 EC2 Test 2: Verifying a rectangular concrete beam subjected to a uniformly distributed load,without compressed reinforcement - Bilinear stress-strain diagram
Passed
5.60.1Description
5.60.2Background
5.60.2.1Model description
■
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■ 671651
25
■ 56225300300 3232
■ 78434151
500
■ 3147610
8252200010
822000
3030
Boundary conditions
■
►
►
■
8/22/2019 AD Validation Guide 2014 En
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Loading
■
■
■
mkN M Ed .59,232
8
²80,5*31,55
mkN M Ecq .94,1628
²80,5*75,38
mkN M Eqp .07,1048
²80,5*75,24
8/22/2019 AD Validation Guide 2014 En
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5.60.2.2Reference results in calculating the equivalent coefficient
00
9252825
816816
48802810
1
10
1
2002000
0
φ
213
0
..10
100
1
1
1.. 21 35
70
135
.
20
235
.
1..338 21
8914717610
100
501
1471766002502
60025022
30
7024880925289100
01
.
941628
8057538
071048
8057524
32.17
94.16207.104*70.21
31476
200000
e
8/22/2019 AD Validation Guide 2014 En
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5.60.2.3Reference results in calculating the concrete beam reduced moment limit
MPa f ck 50
207690159701694 .
24 ..10.
8,189*3,75 ck f 108*2,71 ck f
5,874*6,5 ck f 4,847*2,5 ck f
13*04,0 ck f 5,12*03,0 ck f
■ 188810825*2,71 a
■ 4.7174,84725*2,5 b
■ 75.115,1225*03,0 c
■ 079.1²27.17*75.1127.17*4.717188810 4
e K
■ 43,1
94,162
58,232
■ 2504.0)*20.7690.159(*)*70.169.4(
).(
ck
ck eluc
f
f K
Reference reinforcement calculation at SLU:
■
■
207,067,16*²519,0*25,0
233,0
*²*
cd w
Ed
cu f d b
M
■
md z uc 458,0)294,0*4,01(*519,0)*4,01(*
■
8/22/2019 AD Validation Guide 2014 En
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Reference reinforcement calculation at SLS:
■ 3940400153217
153217 11.
1. . 1
rbcq ser M kNm M 163,
■ 394,01
■ md z c 451.0
3
394.01*519.0
3
1* 1
■ ²03.9400*451.0
163.0
*,1 cm
z
M A
sc
Ecq
ser s
Reference solution for minimal reinforcement area
■
00130
260
562
►
731
10701519025000130
107315190250500
562260
4
4
Finite elements modeling
■
■
■
8/22/2019 AD Validation Guide 2014 En
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ULS and SLS load combinations(kNm)
Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
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5.60.2.4Reference results
Result name Result description Reference value
5.60.3Calculated results
Result name Result description Value Error
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5.61 EC2 Test 8: Verifying a rectangular concrete beam without compressed reinforcement –Inclined stress-strain diagram
Passed
5.61.1Description
5.61.2Background
5.61.2.1Model description
■
■■
■
■
■
■
■
■
■
■
■
■
■
■
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Units
Geometry
■
■
■
■
■
■
Materials properties
■
■
■
■
■
■ MPa f
f c
ck cd 20
5,1
30
■ MPa f f ck ctm
90.230*30.0*30.0 3/23/2
■ 3283710
83022000
10
822000
3030
■ 78434151
500
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Boundary conditions
■
►
►
■
Loading
■
■
kNm M Ed 20.4038
²40.6*75.78
kNm M Ecq 60.2818
²40.6*55
kNm M Eqp 16.2208
²40.6*43
8/22/2019 AD Validation Guide 2014 En
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5.61.2.2Reference results in calculating the equivalent coefficient
)(*)(*),( 00 t f t cm RH
φ
213
0
..10
1001
1
1.. 21 35
70
135
.
20
235
.
388
944038
3535 .
7070
1
984038
3535 .
2020
2
7819840719510
100
501
171956502802
65028022
30
372488073278100
4017
60281
162203721
32837
200000
1
.
0
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5.61.2.3Reference results in calculating the concrete beam reduced moment limit
MPa f ck 50
)*62.7969.165(*)*66.162.4(*)(
ck
ck eluc
f
f K
24 ***10 eee cba K
8,189*3,75 ck f 108*2,71 ck f
5,874*6,5 ck f 4,847*2,5 ck f
13*04,0 ck f 5,12*03,0 ck f
22069818925375
570658743025
8111330040
087.1²60.17*80.1160.17*5.7062.206910 4 e K
425,12.307
76.437
272.0)*62.7969.165(*)*66.162.4(
).(
ck
ck eluc
f
f K
225,020*²566.0*50,0
10*403.0
*²* 2
3
MPamm
Nm
f d b
M
cd w
Ed cu
272.0225.0 luccu
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Reference reinforcement calculation at SLS:
■
■
225.0cu
■
md z uc 493,0)323,0*4,01(*566,0)*4,01(*
■
²68,11²10*68,1178,434*458,0
233,0
*
4 cmm f z
M A
yd c
Ed u
■
35.75.3*323,0
323,01*
12
cu
u
u su
MPa MPa su su 45471.43900735.0*38,95271,432*38,95271,432
■
²60.18²10*60.1871.439*493.0
403.0
*
4 cmm f z
M A
yd c
Ed u
Reference solution for minimal reinforcement area
00130
260
902
²39.2
²10*06.2566.0*28.0*0013.0
²10*39.2566.0*28.0*500
90.2*26.0
max4
4
min, cm
m
m A s
Finite elements modeling
■
■
■
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ULS and SLS load combinations (kNm)
Theoretical reinforcement area(cm2 )
Minimum reinforcement area(cm2 )
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5.61.2.4Reference results
Result name Result description Reference value
5.61.3Calculated results
Result name Result description Value Error
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5.62 EC2 Test29: Verifying the shear resistance for a T concrete beam with inclined transversalreinforcement - Inclined stress-strain diagram (Class XC1)
Passed
5.62.1Description
5.62.2Background
5.62.2.1Model description
■
■
■
■
■
■
■
■
■
■
■
■
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■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■ 80.0
■
■
■
Units
Geometry
■
■
■ θ ˚
Boundary conditions
■
►
►
■
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Loading:
5.62.2.2Reference results in calculating the longitudinal reinforcement
5.62.2.3Reference results in calculating the transversal reinforcement
ywd u
Ed sw
f z
tg V
s
A
*
*.
md z u 0125.1125.1*9.0*9.0
74.2975.1cot
90
ml cmtg
f z
tg V
s
A
ywd u
Ed sw /²50.19
15,1
500*0125.1
74.29*502.1
*
*.
8/22/2019 AD Validation Guide 2014 En
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ml cmb f
f b
s
Aw
yk
ck
ww sw /²21.590sin*55.0*
500
35*08,0sin**
*08,0sin**min,
Finite elements modeling
■
■
■
5.62.2.4Reference resultsResult name Result description Reference value
5.62.3Calculated results
Result name Result description Value Error
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5.63 EC2 Test 27: Verifying the shear resistance for a rectangular concrete beam with verticaltransversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Passed
5.63.1Description
5.63.2Background
5.63.2.1Model description
■
■
■
■
■
■
■
■
■
■
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■
■
Units
Geometry
■
■
■
■
■
■
■ θ ˚
Boundary conditions
■
►
►
■
Loading:
2
* l P
V
u
Ed
KN V Ed 2632
5.91*75.5
8/22/2019 AD Validation Guide 2014 En
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5.63.2.2Reference results in calculating the lever arm zc:
ml kN P u /5.91
kNm M Ed 15.3788
²75.5*5.91
167.067.16*²623.0*35.0
378.0
*²*
cd w
Ed cu
f d b
M
230.0167.0*211*25.1*211*25.1 cuu
md z uc 566.0230.0*4.01*623.0*4.01*
Calculation of reduced shear force
cot. z x
2
**)( l P x P xV u
u
m z x 566.045cot*566.0cot*
kN V red Ed 211263566.0*5.91,
Calculation of maximum design shear resistance:
1
.. 1
1cw
250
1*6,01ck f
v
cot
***1max,
tg
b z f vV wucd
Rd
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45 90
1v
54.0250
25
1*6.02501*6,01
ck f
v
md z u 56.0623.0*9.0*9.0
kN MN V Rd 891891.02
35.0*566.0*67.16*54.0max,
kN V kN V Rd Ed 891236 max,
Calculation of transversal reinforcement:
ml cmtg
f z
tg V
s
A
ywd u
Ed sw /²67,8
15,1
500*566,0
45*211,0
*
*.
Calculation of theoretical reinforcement value:
sin**min, ww sw b
s
A
4
min, 1015.7500
20*08,0*08,0 yk
ck
w f
f
ml cmb s
Aww
sw /²43.190sin*2.0*10*15.7sin** 4
min,
Finite elements modeling
■
■
■
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5.63.2.3Reference results
Result name Result description Reference value
5.63.3Calculated results
Result name Result description Value Error
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5.64 EC2 Test34: Verifying a rectangular concrete column subjected to compression on the top –Method based on nominal curvature - Bilinear stress-strain diagram (Class XC1)
Passed
5.64.1Description
5.64.2Background
5.64.2.1Model description
■
■
■
■
►
►
■
►
■ 3,02 ■
■
■
■
■
■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■
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5.64.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 84*2*20
28.6940.0
8*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
EQP M
Ed M
39.0155.1450.0
**
1
1
0
0 ed
eqp
ed
eqp
Ed
Eqp
N N
e N e N
M M
0,t
)(*)(*),( 00 t f t cm RH
MPa f
f cm
cm 73.2830
8.168.16)(
488.0
281.0
1
1.0
1)(
20.020.0
0
0
t
t
213
0
***1.0
1001
1
h
RH
RH
121 MPa f cm 357.0
1
35
cm f
2.0
2
35
cm f
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mm
u
Ach 240
600400*2
600*400*2*20
MPa MPa f cm 3538
944.038
3535 7.07.0
1
cm f 984.0
38
3535 2.02.0
2
cm f
73.1984.0*944.0*240*1.0
100
501
1***1.0
1001
1321
30
h
RH
RH
30.2488.0*73.2*73.1)(*)(*),( 00 t f t cm RH
The effective creep coefficient calculation:
90.039.0*30.2*, 0 Ed
EQP
ef M
M t
The necessity of buckling calculation (second order effect):
n
C B A ***20lim
241.020*60.0*40.0
155.1
*
cd c
Ed
f A
N n
85.0
90.0*2.01
1
*2,01
1
ef
A
1.1*21 B
70.07,1 mr C
66.26112.0
7.0*1.1*85.0*20lim
66.2628.69 lim
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5.64.2.3The eccentricity calculation and the corrected loads on ULS:
Initial eccentricity:
Additional eccentricity:
ml
ei 02.0400
8
400
0
First order eccentricity- stresses correction:
mmmm
mmmm
mmhmm
e 203.13
20max
30
40020
max
30
20max0
Reinforcement calculation in the first order situation:
MNmh
d N M M Gua 196.0
2
40.035.0*155.10231.0)
2
(*0
496,0)35,0
40,0*4,01(*
35,0
40,0*8,0)*4,01(**8,0
d
h
d
h BC
133.020*²35.0*60.0
196.0
*²*
cd w
uacu
f d b
M
BC cu 496.0133.0
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Calculations of steel reinforcement in pure bending:
133.0cu
179,0)133,0*21(1*25,1 u
md z uc 325,0)179,0*4,01(*35,0)*4,01(*
²87.1378,434*325,0
196,0
*cm
f z
M A
yd c
ua
Calculations of steel reinforcement in combined bending:
24 69.1278.434
155.110*87.13' cm
f
N A A
yd
2
min, 80.4*002.0²66.278.434
155.1*10.0*10,0cm Acm
f
N A c
yd
Ed s
The second order effects calculation:
Calculation of nominal rigidity:
s s sccd c I E K I E K EI ****
2.1
cmcd
E E
MPa MPa f f ck cm 388
MPa f
E cmcm 32837
10
38*22000
10*22000
3.03.0
MPa E E cmcd 27364
2.132837
2.1
410.2,312
40.0*60.0
12
* 333
mhb
I c
MPa E s 200000
s I
0026.0
40.0*60.0
10.28,6 4
c
s
A
A
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01.0002.0 c
s
A
A
Mpa f
k ck 22.120
30
201
241.020*60.0*40.0
155.1
.
cd c
Ed
f A
N n
20.0098.0170
28.69*241.0
170*2
nk
410.41,105.02
40.0*
2
10.28,6*2
2*
2*2 5
242
mch A
I s s
1 s K 063.0
90.01
098.0*22.1
1
* 21
ef
c
k k K
²34.810*41,1*200000*110*2,3*27364*063.0 53 MNm EI
Stresses correction:
1
1*0
Ed
B Ed Ed
N
N M M
MNm M Ed 0231.00
0
²
c
80 c
234.18
²
MN l
EI N B 27.1
²8
25.8*²*²
2
0
MN N Ed 155.12
m MN M Ed .31.02
The reinforcement calculations in the second order effect:
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Finite elements modeling
■
■
■
Theoretical reinforcement area (cm2 ) and minimum reinforcement area (cm
2 )
Theoretical value (cm2 )
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5.64.2.4Reference results
Result name Result description Reference value
5.64.3Calculated results
Result name Result description Value Error
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5.65 EC2 Test 26: Verifying the shear resistance for a rectangular concrete beam with verticaltransversal reinforcement - Bilinear stress-strain diagram (Class XC1)
Passed
5.65.1Description
5.65.2Background
5.65.2.1Model description
■
■
■
■
■
■
■
■
■
■
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■
■
Units
Geometry
■
■
■
■
■
■
■ θ ˚
Boundary conditions
■
►
►
■
Loading:
2
* l P
V
u
Ed
KN V Ed 2632
5.91*75.5
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5.65.2.2Reference results in calculating the lever arm zc:
ml kN P u /5.91
kNm M Ed 15.3788
²75.5*5.91
167.067.16*²623.0*35.0
378.0
*²*
cd w
Ed cu
f d b
M
230.0167.0*211*25.1*211*25.1 cuu
md z uc 566.0230.0*4.01*623.0*4.01*
Calculation of reduced shear force
2
**)( l P x P xV u
u
m z x 566.045cot*566.0cot*
kN V red Ed 211263566.0*5.91,
Calculation of maximum design shear resistance:
1
.. 1
1cw
250
1*6,01
ck f v
1
45 90
1v
54.0250
25
1*6.02501*6,01
ck f
v
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md z u 56.0623.0*9.0*9.0
kN MN V Rd 891891.02
35.0*566.0*67.16*54.0max,
kN V kN V Rd Ed
891236max,
Calculation of transversal reinforcement:
ml cmtg
f z
tg V
s
A
ywd u
Ed sw /²67,8
15,1
500*566,0
45*211,0
*
*.
Finite elements modeling
■
■
■
5.65.2.3Reference results
Result name Result description Reference value
5.65.3Calculated results
Result name Result description Value Error
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5.66 EC2 Test33: Verifying a square concrete column subjected to compression by nominal rigiditymethod- Bilinear stress-strain diagram (Class XC1)
Passed
5.66.1Description
5.66.2Background
5.66.2.1Model description
■
■
■
■
►
►
■
■
■
■
■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■
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5.66.2.2Reference results in calculating the concrete column
Load calculation:
cm M
M e
u
u 07.1
0
0
cmcmcmcm L
cmei 2805.0;2max400
322;2max
400;2max 0
meee i 02.001
The necessity of buckling calculation (second order effect):
n
C B A ***20lim
944.020*²30.0
7.1
*
cd c
Ed
f A
N n
7.0 A ef ef
A *2,01
1
1.1 B ω cd c
yd s
f A
f A B
*
**21*21
27.120*²3.0
78.434*10*56.12*21
4
B
70.0C mr C 7,1
81.12944.0
7.0*27.1*7.0*20***20lim
n
C B A
34.383.0
12*32.312*0 h
L
81.1234.38 lim
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Effective creep coefficient calculation:
Ed
EQP
ef M
M t ., 0
0,t
EQP M
Ed M
74.0
7.1
26.1
*
*
1
1
0
0 ed
eqp
ed
eqp
Ed
Eqp
N
N
e N
e N
M
M
),( 0t
)(*)(*),( 00 t f t cm RH
213
0
***1.0
1001
1
h
RH
RH
121 MPa f cm 35
7.0
1
35
cm f
2.0
2
35
cm f
mm
u
Ach 150
300300*2
300*300*2*20
MPa MPa f cm 3538944.0
38
35 7.0
1
984.038
35 2.0
2
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86.1984.0*944.0*150*1.0
100
501
1***1.0
1001
1321
30
h
RH
RH
48.2488.0*73.2*86.1)(*)(*),( 00 t f t cm RH
835.174.0*48.2*),(0
0
0 Ed
Eqp
ef M
M t
835.2835.111 ef
Calculation of nominal rigidity:
s s sccd c I E K I E K EI ****
e
c
k k K
1
* 21
835.21 ef
22.120
30
201 ck f
k
170
*2
nk
944.020*²3.0
7.1
*
cd c
Ed
f A
N n
34.38
20.0213.0170
34.38*944.0
170
*22 k
nk
086.0835.2
20.022.1
1
21
e
c
k k K
CE
cmcd
E E
MPa f
E ck cm 32837
10
825*22
10
8*22
3.03.0
MPa E
E CE
cmcd 27364
2.1
32837
433
000675.012
3.0*3.0
12
*m
hb I c
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45
242
10*256.12
05.025.0*
2
10*56.12*2
2
'*
2*2 m
d d A I theo
s
²1.410*256.1*200000*1000675.0*27364*086.0 5 MNm EI
MN L
EI N
f
B 67.3²32.3
10.4*²
²
*²
234.18
²²
0
c
MNm
N
N M M
Ed
B Ed Ed 070.0
17.1
67.3
234.11*034.0
1
1*0
MN N Ed 7.1 MNm M Ed 070.0
Finite elements modeling
■
■
■
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Theoretical reinforcement area(cm2 )
Theoretical value (cm2 )
5.66.2.3Reference results
Result name Result description Reference value
5.66.3Calculated results
Result name Result description Value Error
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5.67 EC2 Test 39: Verifying a circular concrete column using the simplified method – Professionalrules - Bilinear stress-strain diagram (Class XC1)
Passed
5.67.1Description
5.67.2Background
5.67.2.1Model description
■
■
■
■
►
►
■
■
■
■
■
Units
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Geometry
■
■
■
Boundary conditions
Loading
■
kN N ED 67505000*35.1
5.67.2.2Reference results in calculating the concrete column
Scope of the method:
■
■■
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Reinforcement calculation:
6025
722.0
62
251
84.0
62
1
84.022
)**81(*)*5.070.0( Dk h
ρ δ
95.0)**81(
045.195.0*)8.0*5.070.0()**61(*)*5.070.0( Dk h
95.0500
500*65.06.1
500*65.06.1 yk
s
f k
MPa f
f ck
cd
67.165.1
25
5.1
MPa f
f yk
yd 33.3335.1
500
5.1
²14.3167.16*4.0*722.0*95.0*045.1
750.6*
33.333
1**
***
1 22 cm f r k k
N
f A cd
sh
ed
yd
s
Finite elements modeling
■
■
■
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Theoretical reinforcement area(cm2 )
5.67.2.3Reference results
Result name Result description Reference value
5.67.3Calculated results
Result name Result description Value Error
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5.68 Verifying the capacity design results according to Eurocode EC2 and EC8 French standards.(DEV2013 #8.3)
Passed
5.68.1Description
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5.69 EC2 Test 47 I: Verifying a rectangular concrete beam subjected to a tension distributed load -Bilinear stress-strain diagram (Class XD2)
Passed
5.69.1Description
5.69.2Background
5.69.2.1Model description
■
■
■
■
■
■
■
Units
Geometry
■
■
■
■
■
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Materials properties
■
■
■■
■
Boundary conditions
■
►
►
■
Loading
■
285150135
■
1801503013530
5.69.2.2Reference results in calculating the longitudinal and the minimum reinforcement
Calculating the longitudinal reinforcement for serviceability limit state:
S
ser ser s
N A
,
Mpa f yk s 400500*8,0*8,0
²13.7²10*13.7400
285.0 4
, cmm A ser s
Calculating the minimum reinforcement:
²04.020.0*20.0 m Ac ²05.2
500
56.2*04.0* cm
f
f A A
yk
ctmc s
Finite elements modeling
■
■
■
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Longitudinal reinforcement for SLS load combinations
Minimum longitudinal reinforcement
5.69.2.3Reference results
Result name Result description Reference value
5.69.3Calculated results
Result name Result description Value Error
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5.70 EC2 Test35: Verifying a rectangular concrete column subjected to compression to top – Basedon nominal rigidity method - Bilinear stress-strain diagram (Class XC1)
Passed
5.70.1Description
5.70.2Background
5.70.2.1Model description
■
■
■
■
►
►
►
■
►
►
■ 3,02 ■
■
■
■■
■
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Units
Geometry
■
■
■
■
Boundary conditions
Loading
■
■ me 166.0
3575.1
225.00
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5.70.2.2Reference results in calculating the concrete column
Geometric characteristics of the column:
ml l 950.4*2*20
28.6945.0
9*32*32 0 a
l
Effective creep coefficient calculation:
Ed
EQP
ef
M
M t ., 0
0,t
EQP M
Ed M
MNm M Eqp 118.006.0*3.010.00
MNm M Ed 225.00
524.0225.0
118.0
0
0 Ed
Eqp
M
M
0,t
)(*)(*),( 00 t f t cm RH
MPa f f cm
cm 73.2830
8.168.16)(
488.0281.0
1
1.0
1)(
20.020.0
0
0
t
t
213
0
***1.0
1001
1
h
RH
RH
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121 MPa f cm 357.0
1
35
cm f
2.0
2
35
cm f
mm
u
Ach 274
700450*2
700*450*2*20
MPa MPa f cm 3538
944.038
3535 7.07.0
1
cm f 984.0
38
3535 2.02.0
2
cm f
70.1984.0*944.0*274*1.0
100
501
1***1.0
1001
1321
30
h
RH
RH
26.2488.0*73.2*70.1)(*)(*),( 00 t f t cm RH
18.1524.0*26.2*, 0 Ed
EQP
ef M
M t
18.218.111 ef
The necessity of buckling calculation (second order effect):
n
C B A ***20lim
215.020*70.0*45.0
3575.1
*
cd c
Ed
f A
N n
81.0
18.1*2.01
1
*2,01
1
ef
A
90.26215.0
7.0*1.1*85.0*20lim
90.2628.69 lim
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5.70.2.3The eccentricity calculation and the corrected loads on ULS:
Initial eccentricity:
me 166.03575.1
225.00
Additional eccentricity:
First order eccentricity- stresses correction:
MN N Ed 3575.1
cmmeee i 85.181885.001
MNm M Ed 256.0
Reinforcement calculation in the first order situation:
MNmh
d N M M Gua 494.02
45.040.0*3575.1256.0)
2(*0
495,0)40,0
45,0*4,01(*
40,0
45,0*8,0)*4,01(**8,0
d
h
d
h BC
220.020*²40.0*70.0
495.0
*²*
cd w
uacu
f d b
M
BC cu 495.0220.0
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Calculations of steel reinforcement in pure bending:
220.0cu
315,0)220,0*21(1*25,1 u
md z uc 350,0)315,0*4,01(*40,0)*4,01(*
²46.3278,434*350,0
495,0
*cm
f z
M A
yd c
ua
Calculations of steel reinforcement in combined bending:
24 24.178.434
3575.110*46.32' cm
f
N A A
yd
2
min, 3.6*002.0²12.378.434
3575.1*10.0*10,0cm Acm
f
N A c
yd
Ed s
The second order effects calculation:
Calculation of nominal rigidity:
s s sccd c I E K I E K EI ****
2.1
cmcd
E E
MPa MPa f f ck cm 388
MPa f
E cmcm 32837
10
38*22000
10*22000
3.03.0
MPa E
E cmcd 273642.1
32837
2.1
410*316.512
45.0*70.0
12
* 333
mhb
I c
MPa E s 200000
s I
002.040.0*70.0
10*28,6 4
c
s
A
A
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01.0002.0 c
s
A
A
Mpa f
k ck 22.120
30
201
215.020*70.0*45.0
3575.1
.
cd c
Ed
f A
N n
20.0088.0170
28.69*215.0
170*2
nk
45
242
10.92,105.02
45.0*
2
10.28,6*2
2*
2*2 mc
h A I s
s
1 s K 049.0
18.11
088.0*22.1
1
* 21
ef
c
k k K
²97.1010*92.1*200000*110*316.5*27364*049.0 53 MNm EI
Stresses correction:
1
1*0
Ed
B Ed Ed
N
N M M
MNm M Ed 0256.00
MN N Ed 3575.1
0
²
c
80 c
234.18
²
MN l
EI N B 32.1²982.10*²*²
2
0
MNm M Ed 18.11
13575.1
32.1
234.11*256.02
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Additional iteration:
0029.045.0*70.0
10*05,9 4
MN N Ed 3575.12
MNm M Ed 48.22
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008.0
MN N Ed 3575.12
MNm M Ed 563.02
0095.0
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MN N Ed 3575.12
MNm M Ed 499.02
Finite elements modeling
■
■
■
Theoretical reinforcement area(cm2 )
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Theoretical value (cm2 )
5.70.2.4Reference results
Result name Result description Reference value
5.70.3Calculated results
Result name Result description Value Error
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5.71 EC2 Test 46 II: Verifying a square concrete beam subjected to a normal force of traction -Bilinear stress-strain diagram (Class X0)
Passed
5.71.1Description
5.71.2Background
5.71.2.1Model description
■■
■
■
■
■
Units
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Geometry
■
■
■
■
■
■
■
■
Boundary conditions
■
►
►
■
5.71.2.2Reference results in calculating the concrete beam
Calculations according a bilinear stress-strain diagram
²50,11²10*50,11
15.1
400
400.0 4
,
, cmm N
AU s
Ed U s
Calculations according a inclined stress-strain diagram
MPaClassBS U s 373400 ,
²70,10²10*72,10373
400.0 4
,
, cmm N
AU s
Ed U s
Checking the condition of non-fragility:
yk
ctmc s
f
f A A *
MPa f f ck ctm 21,220*30.0*30.0 3/23/2
²0225.015.0*15.0 m Ac
²24.1400
21.2*0225.0* cm
f
f A A
yk
ctmc s
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Finite elements modeling
■
■
■
ULS load combinations (kNm)
5.71.2.3Reference results
Result name Result description Reference value
5.71.3Calculated results
Result name Result description Value Error
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6 General application
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6.1 Verifying element creation using commas for coordinates (TTAD #11141)
Passed
6.1.1 Description
6.2 Verifying the precision of linear and planar concrete covers (TTAD #12525)
Passed
6.2.1 Description
6.3 Verifying mesh, CAD and climatic forces - LPM meeting
Passed
6.3.1 Description
6.4 Creating system trees using the copy/paste commands (DEV2012 #1.5)
Passed
6.4.1 Description
6.5 Creating and updating model views and post-processing views (TTAD #11552)
Passed
6.5.1 Description
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6.6 Defining the reinforced concrete design assumptions (TTAD #12354)
Passed
6.6.1 Description
6.7 Verifying the synthetic table by type of connection (TTAD #11422)
Passed
6.7.1 Description
6.8 Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724)
Passed
6.8.1 Description
6.9 Verifying material properties for C25/30 (TTAD #11617)
Passed
6.9.1 Description
6.10 Creating system trees using the copy/paste commands (DEV2012 #1.5)
Passed
6.10.1Description
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6.11 Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238)
Passed
6.11.1Description
6.12 Verifying the appearance of the local x orientation legend (TTAD #11737)
Passed
6.12.1Description
6.13 Verifying the objects rename function (TTAD #12162)
Passed
6.13.1Description
6.14 Verifying geometry properties of elements with compound cross sections (TTAD #11601)
Passed
6.14.1Description
6.15 Launching the verification of a model containing steel connections (TTAD #12100)
Passed
6.15.1Description
6.16 Importing a cross section from the Advance Steel profiles library (TTAD #11487)
Passed
6.16.1Description
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6.17 Creating a new Advance Design file using the "New" command from the "Standard" toolbar(TTAD #12102)
Passed
6.17.1Description
6.18 Changing the default material (TTAD #11870)
Passed
6.18.1Description
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7 Import / Export
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7.1 Importing IFC files containing continuous foundations (TTAD #12410)
Passed
7.1.1 Description
7.2 Exporting a meshed model to GTC (TTAD #12550)
Passed
7.2.1 Description
7.3 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3)
Passed
7.3.1 Description
7.4 Exporting an Advance Design model to DO4 format (DEV2012 #1.10)
Passed
7.4.1 Description
7.5 Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952)
Passed
7.5.1 Description
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7.6 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS(TTAD #12197)
Passed
7.6.1 Description
7.7 Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9)
Passed
7.7.1 Description
7.8 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3)
Passed
7.8.1 Description
7.9 System stability when importing AE files with invalid geometry (TTAD #12232)
Passed
7.9.1 Description
7.10 Exporting linear elements to IFC format (TTAD #10561)
Passed
7.10.1Description
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7.11 Verifying the releases option of the planar elements edges after the model was exported andimported via GTC format (TTAD #12137)
Passed
7.11.1Description
7.12 Importing GTC files containing "PH.RDC" system (TTAD #12055)
Passed
7.12.1Description
7.13 Verifying the load case properties from models imported as GTC files (TTAD #12306)
Passed
7.13.1Description
7.14 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS(TTAD #12197)
Passed
7.14.1Description
7.15 Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172)
Passed
7.15.1Description
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8 Joint Design
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8.1 Creating connections groups (TTAD #11797)
Passed
8.1.1 Description
8.2 Deleting a welded tube connection - 1 gusset bar (TTAD #12630)
Passed
8.2.1 Description
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9 Mesh
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9.1 Verifying the mesh for a model with generalized buckling (TTAD #11519)
Passed
9.1.1 Description
9.2 Verifying mesh points (TTAD #11748)
Passed
9.2.1 Description
9.3 Creating triangular mesh for planar elements (TTAD #11727)
Passed
9.3.1 Description
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10 Old Climatic Generator
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10.1 NV2009: Verifying wind and snow reports for a protruding roof (TTAD #11318)
Passed
10.1.1Description
10.2 NV2009: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604)
Passed
10.2.1Description
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11 Reports Generator
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11.1 Verifying the steel shape sheet display (TTAD #12657)
Passed
11.1.1Description
11.2 Verifying the modal analysis report (TTAD #12718)
Passed
11.2.1Description
11.3 Verifying the global envelope of linear elements displacements (on start and end of superelement) (TTAD #12230)
Passed
11.3.1Description
11.4 Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD#12230)
Passed
11.4.1Description
11.5 Verifying the global envelope of linear elements stresses (on the end point of super element)(TTAD #12230, TTAD #12261)
Passed
11.5.1Description
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11.6 Verifying the global envelope of linear elements forces result (on the start point of superelement) (TTAD #12230)
Passed
11.6.1Description
11.7 Verifying the global envelope of linear elements displacements (on the end point of superelement) (TTAD #12230, TTAD #12261)
Passed
11.7.1Description
11.8 Verifying the global envelope of linear elements stresses (on start and end of super element)(TTAD #12230)
Passed
11.8.1Description
11.9 Verifying the global envelope of linear elements forces result (on all quarters of super element)(TTAD #12230)
Passed
11.9.1Description
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11.10 Creating the steel materials description report (TTAD #11954)
Passed
11.10.1Description
11.11 Verifying the global envelope of linear elements displacements (on the start point of superelement) (TTAD #12230)
Passed
11.11.1Description
11.12 Verifying the model geometry report (TTAD #12201)
Passed
11.12.1Description
11.13 Verifying the global envelope of linear elements forces result (on start and end of superelement) (TTAD #12230)
Passed
11.13.1Description
11.14 Verifying the global envelope of linear elements displacements (on all quarters of superelement) (TTAD #12230)
Passed
11.14.1Description
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11.15 Verifying the global envelope of linear elements displacements (on end points and middle ofsuper element) (TTAD #12230)
Passed
11.15.1Description
11.16 Verifying the Min/Max values from the user reports (TTAD# 12231)
Passed
11.16.1Description
11.17 Verifying the shape sheet strings display (TTAD #12622)
Passed
11.17.1Description
11.18 Verifying the shape sheet for a steel beam (TTAD #12455)
Passed
11.18.1Description
11.19 Verifying the global envelope of linear elements forces result (on the end point of superelement) (TTAD #12230, #12261)
Passed
11.19.1Description
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11.20 Creating the rules table (TTAD #11802)
Passed
11.20.1Description
11.21 System stability when the column releases interfere with support restraints (TTAD #10557)
Passed
11.21.1Description
11.22 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element)(TTAD #12230)
Passed
11.22.1Description
11.23 Verifying the global envelope of linear elements stresses (on all quarters of super element)(TTAD #12230)
Passed
11.23.1Description
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11.24 Generating a report with modal analysis results (TTAD #10849)
Passed
11.24.1Description
11.25 Generating the critical magnification factors report (TTAD #11379)
Passed
11.25.1Description
11.26 Verifying the Max row on the user table report (TTAD #12512)
Passed
11.26.1Description
11.27 Verifying the EC2 calculation assumptions report (TTAD #11838)
Passed
11.27.1Description
11.28 Verifying the global envelope of linear elements forces result (on end points and middle of
super element) (TTAD #12230)
Passed
11.28.1Description
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11.29 Modal analysis: eigen modes results for a structure with one level
Passed
11.29.1Description
11.30 Verifying the global envelope of linear elements stresses (on end points and middle of superelement) (TTAD #12230)
Passed
11.30.1Description
11.31 Verifying the global envelope of linear elements stresses (on the start point of super element)(TTAD #12230)
Passed
11.31.1Description
11.32 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element)(TTAD #12230)
Passed
11.32.1Description
11.33 Verifying the shape sheet report (TTAD #12353)
Passed
11.33.1Description
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12 Seismic analysis
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12.1 Verifying the displacements results of a linear element according to Marocco seismicstandards (RPS 2011) (DEV2013 #3.6)
Passed
12.1.1Description
12.2 EC8 French Annex: verifying torsors on grouped walls from a multi-storey concrete structure
Passed
12.2.1Description
12.3 EC8 French Annex: verifying torsors on walls
Passed
12.3.1Description
12.4 EC8: verifying the sum of actions on supports and nodes restraints (TTAD #12706)
Passed
12.4.1Description
12.5 Verifying the displacements results of a linear element according to Algerian seismic
standards (RPA 99/2003) (DEV2013 #3.5)
Passed
12.5.1Description
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12.6 Verifying the displacements results of a linear element for an envelope spectrum according toEurocode EC8 standard (DEV2013 #8.2)
Passed
12.6.1Description
12.7 Verifying seismic efforts on planar elements with Q4 and T3-Q4 mesh type (TTAD #14244)
Passed
12.7.1Description
12.8 Verifying the displacements results of a linear element for an envelope spectrum according toRomanian EC8 appendix (DEV2013 #8.2)
Passed
12.8.1Description
12.9 EC8 French Annex: verifying torsors on walls, elastic linear supports and user-definedsection cuts (TTAD #14460)
Passed
12.9.1Description
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12.10 Verifying the spectrum results for EC8 seism (TTAD #11478)
Passed
12.10.1Description
12.11 Verifying the spectrum results for EC8 seism (TTAD #12472)
Passed
12.11.1Description
12.12 Verifying the combinations description report (TTAD #11632)
Passed
12.12.1Description
12.13 Verifying the damping correction influence over the efforts in supports (TTAD #13011).
Passed
12.13.1Description
12.14 EC8 French Annex: verifying seismic results when a design spectrum is used (TTAD #13778)
Passed
12.14.1Description
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12.15 Verifying the displacements results of a linear element for an envelope spectrum according toFrench PS92/100 standard (DEV2013 #8.2)
Passed
12.15.1Description
12.16 Seismic norm PS92: verifying efforts and torsors on planar elements (TTAD #12974)
Passed
12.16.1Description
12.17 Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517)
Passed
12.17.1Description
12.18 EC8 : Verifying the displacements results of a linear element according to Czech seismicstandards (CSN EN 1998-1) (DEV2012 #3.18)
Passed
12.18.1Description
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12.19 EC8 Fr Annex: Generating forces results per modes on linear and planar elements (TTAD#13797)
Passed
12.19.1Description
12.20 Verifying the displacements results of a linear element for an envelope spectrum according toMarocco seismic standards (RPS 2011) (DEV2013 #8.2)
Passed
12.20.1Description
12.21 EC8 Romanian Annex: verifying action results and torsors per modes on point, linear andplanar supports (TTAD #14840)
Passed
12.21.1Description
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13 Steel design
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13.1 EC3 Test 15: Verifying a rectangular hollow section column subjected to bending and axialefforts
Passed
13.1.1Description
13.1.2Background
13.1.2.1Model description
■
■
■
■
►
►
■
■
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Units
Geometry
■■
■
■
■
■
■
■ 0.10 M
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
►
■
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13.1.2.2Reference results for calculating the column subjected to bending and axial force
Cross section class
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1510
10215220022
mm
mmmmmm
t
t r b
t
c
0.1235
y f
333315 t
c
5.0832.063.26
132
1
MPa
MPa
2510
10215230022
mm
mmmmmm
t
t r h
t
c
0.1235
y f
34.40113
39625
t
c
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Verifying the design resistance for uniform compression
N MPamm f A
N M
y
Rd c 22301500.1
2359490 2
0
,
%42.221002230150
500000100100
,,
N
N
N
F
N
N
Rd c
z
Rd c
Ed
Verifying the design plastic moment resistance
Nmm
MPamm f W
M M M
y y pl
Rd pl Rd c 2246600000.1
235956000 3
0
,
,,
%82.27100224660000
2
50005000/5
1002100,,
Nmm
mmmmmm N
M
L Lq
M
M
Rd c Rd c
Ed
Finite elements modeling
■
■
■
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Finite elements results
Work ratio of the design resistance for uniform compression
Work ratio of the design resistance for bending
13.1.2.3Reference results
Result name Result description Reference value
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13.1.3Calculated results
Result name Result description Value Error
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13.2 EC3 Test 45: Comparing the shear resistance of a welded built-up beam made from differentsteel materials
Passed
13.2.1Description
13.2.2Background
13.2.2.1Model description
■
■
■
■
►
■
■
Units
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Geometry
■
■
■
■
■
■
■ 0.10 M
Materials properties
■
■
Boundary conditions
■
►
►
►
Loading
■
►
■
13.2.2.2Reference results for calculating the design plastic shear resistance of the cross section
0.1,
Rd pl
Ed
V
V
■
N mmml N Lq
V Ed 500002
5000/20000
2
■
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Shear area of the cross section made of 500 MPa yield strength user-defined material
206.1978)1.76.278(0.1 mmmmmmt h A wwv
Shear area of the cross section made of S275 steel
267.2373)1.76.278(2.1 mmmmmmt h A wwv
Design plastic shear resistance of the cross section made of 500 MPa yield strength user-defined material
N
MPamm
f A
V M
y
v
Rd pl 7.5710160.1
3
50006.1978
3
2
0
,
0.10876.07.571016
50000
,
N
N
V
V
Rd pl
Ed
%76.81007.571016
50000100,
N
N
V
V
Rd pl
Ed
Design plastic shear resistance of the cross section made of S275 steel
N
MPamm
f A
V M
y
v
Rd pl 7.3768700.1
3
27567.2373
3
2
0
,
0.1133.07.376870
50000
,
N
N
V
V
Rd pl
Ed
%27.131007.376870
50000100
,
N
N
V
V
Rd pl
Ed
Finite elements modeling
■
■■
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Finite elements results
Work ratio of the design shear resistance (beam made of 500 MPa yield strength user-defined material)
Work ratio of the design shear resistance (beam made of S275 steel)
13.2.2.3Reference results
Result name Result description Reference value
13.2.3Calculated results
Result name Result description Value Error
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13.3 EC3 Test 28: Verifying an user defined I section class 1, column fixed on base and without anyother restraint
Passed
13.3.1Description
13.3.2Background
13.3.2.1Model description
■
■
■
■
■
■
Units
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Geometrical properties
■
■ 2
06.4904 mm A ■ mmb 150
■ mmt f 70.10■ mmr 0
■ mmt w 10.7
■ mmhw 260
■ 3
, 63.445717 mmW yel
■ 318.501177 mmW y
■ 3
, 89.80344 mmW z el
■ 3, 96.123381 mmW z pl
■ 464.57943291 mm I y
■ 446.6025866 mm I z
■ 497.149294 mm I t
■ 688.19351706542 mm I w
Materials properties
■
■
■
Boundary conditions
■
►
►
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Loading
■
13.3.2.2Cross section Class
1253.006.179
30.45
sup
inf
Mpa
Mpa
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73.190
36.224
06.1796.238
175.4836.224
3.456.238
36.224
6.238
06.17930.4506.17930.45 y
x y x y x
5.080.06.238
73.1906.238
x
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924.0275
235235
y f
93.3818.013
924.0396
113
3966.33
924.0
61.331.7
7.102260
t
cmm
mmmm
t
c
316.8924.0968.6
924.0
68.67.10
2
1.7150
t
c
t
c
13.3.2.3Buckling verification
a) over the strong axis of the section, y-y:
α
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34.0
y y
11
22
y y y
y
y
ycr
y
y N
f A
,
*
206.4904 mm A 464.57943291 mm I y
kN N
mm
mmmm N
l
I E N
fy
y
ycr 33.380295.3802327²5620
64.57943291/210000²
²
² 42
,
5956.095.3802327
/27506.4904 22
,
N
mm N mm
N
f A
ycr
y y
7446.05956.02.05956.034.015.0²)2.0(15.0 2 y y y
839.0
1
839.05956.07446.07446.0
11
2222
y
y
y y y
y
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b) over the weak axis of the section, z-z
α
49.0
z z
1122
z z z
z
z
z cr
y z
N
f A
,
*
446.6025866 mm I z
206.4904 mm A
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kN N
mm
mmmm N
l
I E N
fz
z z cr 43.39563.395426
²5620
46.6025866/210000²
²
² 42
,
847.1
63.395426
/27506.4904 22
,
N
mm N mm
N
f A
z cr
y z
609.2847.12.0847.149.015.0²)2.0(15.0 2 z z z
225.0
1
225.0847.1609.2609.2
112222
z
z
z z z
z
13.3.2.4Lateral torsional buckling verification
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
77.1050
01
,
, C M
M
top y
botom y
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464.57943291 mm I y
446.6025866 mm I z
497.149294 mm I t
688.19351706542 mm I w
4
2
f z
w
t h I I
mmh 260
f t mmt f 7.10
6
24
093627638294
7.1026046.6025866mm
mmmmmm I w
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mmmmmm N
I E I G L
I I
L I E C M
z
t
z
w z cr
150181.150184702
58.21463.39542677.146.6025866/210000
97.149294/808005620
46.6025866
09362763829
562046.6025866/21000077.1
²²
²²
422
422
4
6
2
422
1
958.01.150184702
/27518.501177 23,
Nmm
mm N mm
M
f W
cr
y y pl LT
LT LT
1²²
1
LT LT LT
LT
²2.015.0 LT LT LT LT
2733.1150
260
mm
mm
b
h
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α
49.0
145.1²958.02.0958.049.015.0²2.015.0 LT LT LT LT
1564.0²958.0²145.1145.1
1
²²
1
LT LT LT
LT
13.3.2.5Internal factor, yyk , calculation
yyk
yy
ycr
Ed
y
mLT my yyC
N
N C C k
1̀
1,
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ycr
Ed y
ycr
Ed
y
N
N
N
N
,
,
1
1
839.0 y
kN N Ed 328
kN N l
I E N
fy
y
ycr 33.380295.3802327²
²,
985.0
95.3802327
328000839.01
95.3802327
3280001
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
myC
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0
0
,0
cr
y y pl
M
f W
318.501177 mmW y
0cr M 11 C 02 C
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
85.8427.84850646
58.21463.395426146.6025866/210000
97.149294/808005620
46.6025866
88.19351706542
5620
46.6025866/2100001
²
²
²
²
422
422
4
6
2
422
1
274.127.84850646
/27518.501177 23,
0
Nmm
mm N mm
M
f W
cr
y y pl
4
,,
1 1120.0
TF cr
Ed
z cr
Ed
N
N
N
N C
T cr TF cr N N ,,
²1
,
2
0
,
T cr
wt T cr
L I E I G
I N
0 I
4442
0 1.6396915846.602586664.57943291 mmmmmm I I z A I I I z y g z y
497.149294 mm I t
688.19351706542 mm I w
T cr L ,
m L T cr 62.5,
mm
mmmm N mmmm N
mm
mm N T cr
607.1395246
²5620
88.19351706542/21000097.149294/80800
1.63969158
06.4904 62242
4
2
,
N N Ed 328000
N N N T cr TF cr 607.1395246,,
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kN N
mm
mmmm N
l
I E N
fz
z z cr 43.39563.395426
²5620
46.6025866/210000²
²
² 42
,
120.0
607.1395246
3280001
63.395426
3280001120.01120.0 44
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
1
11
11
120.01120.0274.1
120.01120.0
274.1
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C
a
a
C C C
N
N
N
N C
N
N
N
N C
myC
LT y
LT y
mymymya
aC C C
1
1 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
3
,, 63.445717 mmW W yeff yel
677.163.445717
06.4904
328000
10503
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
1997.064.57943291
97.14929411
4
4
mm
mm
I
I a
y
t LT
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0mC
0
ycr
Ed
ycr
Ed my
N
N
N
N C
,,
0, 33.036.079.033.036.021.079.0
kN N l
I E N
fy
y
ycr 33.380295.3802327²
²,
N N Ed 328000
780.095.3802327
32800033.036.079.00,
N
N C my
904.01677.11
1677.1780.01780.0
11 0,0,
LT y
LT y
mymymya
aC C C
mLT C
256.2
1
256.2
607.1395246
3280001
63.395426
3280001
997.0904.0
11
2
,,
2
mLT
mLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
N
N
N
N
aC C
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yyC
y pl
yel
LT pl my
y
my
y
y yyW
W bnC
wC
wwC
,
,2
max2
max2 6.16.1
2)1(1
1534.0
/27596.123381
10000000
/27518.501177564.0
50000000274.1997.05.0
5.05.0
2323
2
,
,
,
,
2
0
,,
,
,,
,
2
0
mm N mm
Nmm
mm N mm
Nmm
f W
M
f W
M a
M
M
M
M ab
y z pl
Ed z
y y pl LT
Ed y LT
Rd z pl
Ed z
Rd y pl LT
Ed y LT LT
5.1124.163.445717
18.5011773
3
,
, mm
mm
W
W w
yel
y pl
y
243.0
1
/27506.4904
32800022
1
mm N mm
N
N
N n
M
Rk
Ed pl
847.1847.1;5956.0max;maxmax
z y
857.01534.0243.0²847.1²904.0124.1
6.1847.1²904.0
124.1
6.12)1124.1(1
yyC
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889.0889.018.501177
63.445717
857.0
,
,
3
3
,
,
yy
y pl
yel
yy
y pl
yel
yy
C
W
W C
mm
mm
W
W
C
kN N l
I E N
fy
y
ycr 33.380295.3802327²
²,
yyk
47.2889.0
1
95.3802327
3280001
985.0256.2904.0
1̀
1,
N
N C
N
N C C k
yy
ycr
Ed
y
mLT my yy
13.3.2.6Internal factor, yz k , calculation
y
z
yz
z cr
Ed
y
mz yz w
w
C
N
N C k
6.0
1
1,
691.063.395426
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
985.0
95.3802327
328000839.01
95.3802327
328000
1
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
kN N l
I E N
fz
z z cr 43.39563.395426
²
²,
LT pl
z
mz z yz cn
w
C wC
5
2
max2
142)1(1
Rd y pl lt my
Ed y
z
LT LT M C
M ac
,,
,
4
2
0
5
10
997.064.57943291
97.14929411
4
4
mm
mm
I
I a
y
t LT
274.127.84850646
/27518.501177 23,
0
Nmm
mm N mm
M
f W
cr
y y pl
847.163.395426
/27506.4904
22
, N
mm N mm N
f A z cr
y z
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Nm M Ed y 50000,
904.01
1 0,0,
LT y
LT y
mymymya
aC C C
1564.0²958.0²145.1145.1
1²²
1
LT LT LT
LT
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
664.05.137823724564.0904.0
50000000
847.15
²274.1997.010
5
10
4
,,
,
4
2
0
Nmm
Nmm
M C
M ac
Rd y pl lt my
Ed y
z
LT LT
LT pl
z
mz z yz cn
w
C wC
5
2
max2
142)1(1
5.1
5.1
5.1536.189.80344
96.1233813
3
,
,
z
z
z el
z pl
z w
w
mm
mm
W
W w
kN N l
I E N
fz
z z cr 43.39563.395426
²
²,
691.063.395426
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
847.1847.1;5956.0max;maxmax
z y
243.0
1
M
Rk
Ed pl N
N n
532.0691.0243.05.1
847.1691.0142)15.1(1
142)1(1
5
22
5
2
max
2
LT pl
z
mz z yz cn
wC wC
20.5124.1
5.16.0
532.0
1
63.395426
3280001
985.0691.06.0
1
1,
N
N w
w
C
N
N C k
y
z
yz
z cr
Ed
y
mz yz
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13.3.2.7Internal factor, zyk , calculation
y
z
zy
ycr
Ed
z mLT my zy
w
w
C
N
N C C k
6.0
1
1,
210.0
63.395426
328000225.01
63.395426
3280001
1
1
,
,
N
N N
N
N
N
N
N
z cr
Ed z
z cr
Ed
z
225.0 z
y pl
yel
z
y
LT pl
y
my
y zyW
W
w
wd n
w
C wC
,
,
5
2
max2
6.0142)1(1
Nmmmm N mm f W M z z Rd z pl 33930039/27596.123381 23
,,
771.0
33930039691.0
10000000
5.137823724564.0904.0
50000000
847.11.0
274.1997.02
1.0
2
4
,,
,
,,
,
4
0
Nmm
Nmm
Nmm
Nmm
M C
M
M C
M ad
Rd z pl mz
Ed z
Rd y pl LT my
Ed y
z
LT LT
310.0771.0243.0124.1
847.1904.0142)1124.1(1
142)1(1
5
22
5
2
max
2
LT pl
y
my y zy d n
wC wC
462.018.501177
63.445717
5.1
124.16.06.0
3
3
,
, mm
mm
W
W
w
w
y pl
yel
z
y
462.0
462.06.0
310.0
6.0
,
,
,
,
zy
y pl
yel
z
y
zy
y pl
yel
z
y
zy
C
W
W
w
w
C
W
W
w
wC
256.2mLT C
529.05.1
124.16.0
462.0
1
95.3802327
3280001
210.0256.2904.06.0
1
1,
N
N w
w
C
N
N C C k
z
y
zy
ycr
Ed
z mLT my zy
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13.3.2.8Internal factor, zz k , calculation
zz
z cr
Ed
z mz zz
C
N
N C k
1
1,
z pl
z el
LT pl mz
z
mz
z
z zz W
W enC
wC
wwC
,
,max
2max
2 6.16.12)1(1
131.0/27518.501177564.0904.0
50000000
847.11.0
274.1997.07.1
1.0
7.1
234
,,
,
4
0
mmmm
Nmm
M C
M ae
Rd y pl lt my
Ed y
z
LT LT
926.0131.0243.0847.1691.05.1
6.1847.1691.0
5.1
6.12)15.1(1
6.16.12)1(1
222
2
max2
max2
LT pl mz
z
mz
z
z zz enC wC wwC
919.0926.0
1
63.395426
3280001
210.0691.0
1
1,
N
N C
N
N C k
zz
z cr
Ed
z mz zz
13.3.2.9Bending and axial compression verification
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rk z
Rd z Ed z
zz
M
Rk y
LT
Rd y Ed y
zy
M
Rk z
Ed
M
Rk z
Rd z Ed z
yz
M
Rk y
LT
Rd y Ed y
yy
M
Rk y
Ed
M
M M k
M
M M k
N
N
M
M M k
M
M M k
N
N
i y Rk A f N
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69.127.034.008.1
1
/27596.123381
1010919.0
1
/27518.501177564.0
1050529.0
1
/27506.4904225.0
328000
41.353.159.129.0
1/27596.123381
101020.5
1
/27518.501177564.0
1050479.2
1
/27506.4904839.0
328000
23
6
23
6
22
23
6
23
6
22
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
Finite elements modeling
■
■
■
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y coefficient corresponding to non-dimensional slenderness y
y
z z
z
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Internal factor, yyk
yyk
Internal factor, yz k
yz k
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Internal factor, zyk
zyk
Internal factor, zz k
zz k
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LT
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13.3.2.1 Reference results
Result name Result description Reference value
y y y
z z z
yyk yyk
yz k yz k
zyk zyk
zz k zyk
LT
13.3.3Calculated results
Result name Result description Value Error
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13.4 EC3 Test 40: Verifying CHS508x8H class 3, simply supported beam, loaded with centriccompression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle
Passed
13.4.1Description
13.5 EC3 Test 43: Determining lateral torsional buckling parameters for a I-shaped laminated beamconsidering the load applied on the lower flange
Passed
13.5.1Description
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13.6 EC3 Test 6: Class section classification and combined biaxial bending verification of anIPE300 beam
Passed
13.6.1Description
13.6.2Background
13.6.2.1Model description
■
■
■
■
■
Units
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Materials properties
Boundary conditions
■
►►
■
13.6.2.2Reference results for calculating the cross section class
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13.6.2.3Reference results for calculating the combined biaxial bending
1,
,
,
,
Ed Nz
Ed z
Rd Ny
Ed Y
M
M
M
M
α β
2
)1;max(n
00
,,
Rd pl Rd pl
Ed
N N
N n 1
Bending around Y:
a
n M M
Rd y pl
Rd Ny*5.01
)1(*,,
,
Rd ply Rd Ny M M ,,
00
,,
Rd pl Rd pl
Ed
N N
N n
5.0403.010*81.53
)0107.0*15.0*210*81.53()**2(4
4
A
t b Aa
f
8.0)403.0*5.01(
,,,,
,,
Rd y pl Rd y pl
Rd y N
M M M
Rd y pl Rd y N Rd y pl Rd y N M M M M ,,,,,,,,*8.0 Rd ply Rd Ny M M ,,
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Bending around Y:
a
n M M
Rd z pl
Rd Nz *5.01
)1(*,,
,
Rd plz Rd Nz M M ,,
Rd z pl Rd z N Rd z pl Rd z N M M M M ,,,,,,,,*8.0 Rd plz Rd Nz M M ,,
MNm f w
M M M
y z pl
Rd z pl Rd z N 030.01
235*10*20.125* 6
0
,
,,,,
11086.1029375.0
03125.0
148.0
03125.0 12
,
,
,
,
Ed Nz
Ed z
Rd Ny
Ed Y
M
M
M
M
Finite elements modeling
■
■
■
Finite elements results
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13.6.2.4Reference results
Result name Result description Reference value
13.6.3Calculated results
Result name Result description Value Error
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13.7 EC3 test 10: Verifying the classification and the bending resistance of a welded built-up beam
Passed
13.7.1Description
13.7.2Background
13.7.2.1Model description
■
■
■
■
►
■
■
Units
Geometry
■
■
■
■
■
■
■ 0.10 M
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Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►■
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13.7.2.2Cross-section classification
67.1318
2/)8500(
mm
mmmm
t
c
8136.0235
y f
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39.111467.13 t
c
25.748
218630
mm
mmmm
t
c
8136.0235
y f
89.10012425.74 t
c
13.7.2.3Reference results for calculating the design resistance for bending
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The buckling factor (k ) and the stress ratio( ) - for flanges
43.00.11
2
k
The plate modified slenderness ( p )
902.0
43.08136.04.28
18/2/8500
4.28
/
mmmmmm
k
t c p
The reduction factor ( )
0.1188.0
2
p
p
mm
mmmmcb f eff 89.215
2
85008776.0,
Effective section modulus
►
►
►
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3
,sup, 91.5204392 mmW yeff
3
,inf , 4.5736064 mmW yeff
3
,inf ,,sup,min,, 91.5204392,min mmW W W yeff yeff yeff
Design resistance for bending
Nmm MPamm f W
M M
yeff
Rd c 18475594830.1
35591.5204392 3
0
min,
,
Work ratio
%54.2100
1847559483
8/5000/15100
8/100
2
,
2
,
Nmm
mmmm N
M
Lq
M
M
Rd c Rd c
Finite elements modeling
■
■
■
Work ratio of the design resistance for bending
13.7.2.4Reference results
Result name Result description Reference value
13.7.3Calculated results
Result name Result description Value Error
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13.8 EC3 test 9: Verifying the classification and the compression resistance of a welded built-upcolumn
Passed
13.8.1Description
13.8.2Background
13.8.2.1Model description
■
■
■
■
►
■
■
Units
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Geometry
■
■
■
■
■
■
■ 0.10 M
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
■
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13.8.2.2Cross-section classification
67.1318
2/)8500(
mm
mmmm
t
c
81.0235
y f
34.111467.13 t c
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25.748
218630
mm
mmmm
t
c
81.0235
y f
02.344225.74 t
c
13.8.2.3Reference results for calculating the compression resistance of the cross-section
The buckling factor (k ) and the stress ratio( ) - for flanges
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43.00.11
2
k
The buckling factor (k ) and the stress ratio( ) - for web
0.40.11
2
k
The plate modified slenderness ( p ) – for flanges
906.0
43.081.04.28
18/2/8500
4.28
/
mmmmmm
k
t c p
The plate modified slenderness ( p) – for web
614.1
0.481.04.28
8/182630
4.28
/
mmmmmm
k
t b p
The reduction factor ( ) – for flanges
0.1188.0
2
p
p
mm
mmmmcb f eff 25.215
2
8500875.0,
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The reduction factor ( ) – for web
0.1
3055.02
p
p
mmmmmmbb weff 8.317182630535.0,
Effective area
weff ww f eff f eff f eff bt t bbt A ,,, )(2
Compression resistance of the cross section
N MPamm f A
N M
yeff
Rd c 65065820.1
3554.18328 2
0
,
Work ratio
%54.11006506582
100000100
,
N
N
N
N
Rd c
Finite elements modeling
■
■
■
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Work ratio of the design resistance for uniform compression
13.8.2.4Reference results
Result name Result description Reference value
13.8.3Calculated results
Result name Result description Value Error
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13.9 EC3 Test 21: Verifying the buckling resistance of a CHS219.1x6.3H column
Passed
13.9.1Description
13.9.2Background
13.9.2.1Model description
■■
■
■
■
■
Units
Geometrical properties
■
■
■
■
■
■
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Materials properties
■
■
Boundary conditions
■
►
►
■
■
Loading
■
■
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13.9.2.2Buckling in the strong inertia of the profile (along Y-Y)
%100100,
Rd b
Ed
N N
778.343.6
1.219
mm
mm
t
d
814.0355
235235
y f
381.46814.05080778.34 2
t
d
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1
,
M
y
Rd b
f A N
Coefficient corresponding to non-dimensional slenderness after the Y-Y axis
11
22
y y y
y
1
*
i
L
N
f Acr
cr
y
y
41.76355
2100001
y f
E
mmmm
mm
A
I i
y
283.754210
238600002
4
043.141.76283.75
6000
1
mm
mm
i
Lcr y
132.1²043.1)2.0043.1(21.015.0)2.0(15.0 2 y y y
1636.0²043.1²132.1132.1
1
y
1 M
11 M
N mm N mm f A
N M
y y
Rd b 8.9505331
/3554210636.0 22
1
,
N N Ed 100000
%520.101008.950533
100000100
,
N
N
N
N
Rd b
Ed
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Finite elements modeling
■
■
■
Finite elements results
The appropriate non-dimensional slenderness
LT
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Ratio of the design normal force to design buckling resistance in the strong inertia of the profile
13.9.2.3Reference results
Result name Result description Reference value
y y y
ySN
13.9.3Calculated results
Result name Result description Value Error
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13.10 EC3 Test 17: Verifying a simply supported rectangular hollow section beam subjected totorsional efforts
Passed
13.10.1Description
13.11 EC3 Test 18: Verifying a simply supported circular hollow section element subjected totorsional efforts
Passed
13.11.1Description
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13.12 EC3 Test 24: Verifying an user defined I section class 4 column fixed on the bottom andwithout any other restraint
Passed
13.12.1Description
13.12.2Background
13.12.2.1Model description
■
■
■
■
■
■
Units
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Geometrical properties
■
■ 210850mm A
■ mmb 220
■ mmt f 15■ mmr 0
■ mmt w 5
■ mmhw 880
■ 33
, 1066.3387 mmW yel
■ 331062.3757 mmW y
■ 33
, 1008.242 mmW z el
■ 33, 1031.368 mmW z pl
■
■
■
■
Materials properties
■
■
■
Boundary conditions
■
►
►
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Loading
■
13.12.2.2Cross section Class
1873.030.406
84.354
sup
inf
Mpa
Mpa
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27.396
14.761
84.354850
73.45314.761
3.406850
14.761
850
84.3543.40684.3543.406 y
x y x y x
5.05338.0850
73.453
850
x
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924.0275
235235
y f
61.101)873.0(33.067.0
924.042
33.067.0
42170
1
1705
152880
t
cmm
mmmm
t
c
9961.7
1
61.715
5.107
t
ct
c
13.12.2.3Effective cross-sections of Class4 cross-sections
1705
152880
mm
mmmm
t
c
41 k
k t
bw
p
4.28
wb mmbw 850
9244.0275
235235
y f
261.349244.04.28
5
850
mm
mm
p
286.0
261.3
4055.0261.33055.0
043
673.0261.322
p
p p
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mmmmb
mmmmb
mmmmb
bb
bb
bb
e
e
eff
eff e
eff e
weff
55.1211.2435.0
55.1211.2435.0
1.243850286.0
5.0
5.0
2
1
2
1
2
21, 5.121555.121555.1215 mmmmmmmmmmbt bt A ewewwebeff 2
, 330022015 mmmmmmbt A f f flangeeff
222
,, 2.7815330025.12152 mmmmmm A A A flangeeff webeff eff
13.12.2.4Effective elastic section modulus of Class4 cross-sections
mmmm
bb t c 4252
8501
sup
inf
9.231 k
1705
152880
mm
mmmm
t
c
k
t
bw
p
4.28
wb mmbw 850
9244.0275
235235
y f
325.19.239244.04.28
5
850
mm
mm
p
692.0325.1
2055.0325.13055.0
023
673.0325.122
p
p p
mmmmb
mmmmb
mmmm
b
bb
bb
bbb
e
e
eff
eff e
eff e
wceff
46.1761.2946.0
64.1171.2944.0
1.29411
850692.0
6.0
4.0
1
2
1
2
1
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mm
yG
53.155.10195
095.158330
15220546.601564.11715220
27.124546.6015.4321522018.366564.1175.43215220
4
23
23
23
23
14482344295.6993800726.748854356
97.4161522012
2201574.107546.601
12
546.601
71.381564.11712
564.11703.44815220
12
22015
mm
I y
423
23
23
23
63.2662749001522012
152200546.601
12
46.6015
0564.11712
64.1175015220
12
15220
mm
I z
34
max,
533.317922953.455
1448234429mm
mm
mm
z
I W
y
yel
34
max
, 10.242068110
63.26627490mm
mm
mm
y
I W z
z el
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13.12.2.5Buckling verification
a) over the strong axis of the section, y-y:
α
34.0
y y
11
22
y y y
y
y
ycr
yeff
y N
f A
,
*
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22.7815 mm Aeff
kN N
mm
mmmm N
l
I E N
fy
y
ycr 37.9503544.95035371
²5620
1448234429/210000²
²
² 42
,
15.044.95035371
/2752.7815 22
,
N
mm N mm
N
f A
ycr
yeff y
503.015.02.015.034.015.0²)2.0(15.0 2 y y y
1
1
017.115.0503.0503.0
11
2222
y
y
y y y
y
b) over the weak axis of the section, z-z
α
49.0
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z z
11
22
z z z
z
z
z cr
yeff z
N
f A
,
*
22.7815 mm Aeff
kN N mm
mmmm N l
I E N fz
z z cr 34.1747905.1747336
²562063.26627490/210000²
²²
42
,
109.1905.1747336
/2752.7815 22
,
N
mm N mm
N
f A
z cr
yeff z
338.1109.12.0109.149.015.0²)2.0(15.0 2 z z z
479.0
1
479.0109.1338.1338.1
11
2222
z
z
z z z
z
13.12.2.6Lateral torsional buckling verification
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
01274
0
kNm
kNm
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77.101274
01 C
kNm
kNm
41448234429mm I y
463.26627490 mm I z
4
2
f z
w
t h I I
f t mmt f 15
611
24
10808.494
15880mm326627490.6mm
mmmm I w
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
163.14201420163158
185.459905.174733677.163.26627490/210000
514614/808005620
63.26627490
10808.49
5620
63.26627490/21000077.1
²
²
²
²
422
422
4
611
2
422
1
34
max
, 533.317922953.455
1448234429mm
mm
mm
z
I W
y
yel
785.01420163158
/275533.3179229 23,
Nmm
mm N mm
M
f W
cr
y yeff LT
LT LT
1²²
1
LT LT LT
LT
²2.015.0 LT LT LT LT
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24220
880
mm
mm
b
h
α
76.0
030.1²785.02.0785.076.015.0²2.015.0 LT LT LT LT
1589.0²785.0²030.1030.1
1
²²
1
LT LT LT
LT
13.12.2.7Internal factor, yyk , calculation
yyk
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ycr
Ed
y
mLT my yy
N
N C C k
,
1
ycr
Ed y
ycr
Ed
y
N
N N
N
,
,
1
1
1 y
kN N Ed 328
kN N l
I E N
fy
y
ycr 37.9503544.95035371²
²,
1
44.95035371
32800011
44.950353713280001
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
myC
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0
0
,0
cr
y yeff
M
f W
33
, 1066.3387 mmW yel
kNm Nmm M cr 163.14201420163158
810.01420163158
/2753387660 23
0
,0
Nmm
mm N mm
M
f W
cr
y yeff
4
,,
1 1120.0
TF cr
Ed
z cr
Ed
N
N
N
N C
T cr TF cr N N ,,
²,
2
0
,
T cr
wt T cr
L
I E I G
I
A N
0 I
2
0 g z y z A I I I 467.1490580416 mm I
y
417.26628854 mm I z
210850mm A
0 g z
4442
0 151720927017.2662885467.1490580416 mmmmmm I I z A I I I z y g z y
T cr L ,
m L T cr 62.5, 475.514614 mm I t 61110808.49 mm I w
mm
mmmm N mmmm N
mm
mm N T cr
14.2634739
²5620
10808.49/21000075.514614/80800
1517209270
10850 6112242
4
2
,
N N Ed 328000
N N N T cr TF cr 14.2634739,,
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N l
I E N
fz
z z cr 905.1747336
²
²,
244.0
14.2634739
3280001
905.1747336
328000177.120.01120.0 44
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
1
11
11
244.01120.0810.0
244.01120.0
810.0
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
T cr
Ed
z cr
Ed
LT
mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C
a
aC C C
N
N
N
N C
N
N
N
N C
myC
LT y
LT y
mymymya
aC C C
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
3
,, 533.3179229 mmW W yeff yel
55.9533.3179229
2.7815
328000
1012743
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
19996.01448234429
75.51461411
4
4
mm
mm
I
I a
y
t LT
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0mC
0
ycr
Ed my
N
N C
,
0, 33.036.021.079.0
kN N l
I E N
fy
y
ycr 37.9503544.95035371²
²,
79.044.95035371
32800033.0036.079.00,
N
N C my
949.0155.91
155.979.0179.0
11 0,0,
LT y
LT y
mymymya
a
C C C
1
11,,
2
T cr
Ed
z cr
Ed
LT mymLT
N
N
N
N
aC C
068.1
14.2634739
3280001
905.1747336
3280001
1949.0 2
N
N
N
N C mLT
0161.1
44.95035371
3280001
1068.1949.0
1,
N
N
N
N C C k
ycr
Ed
y
mLT my yy
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13.12.2.8Internal factor, yz k , calculation
z cr
Ed
y
mz yz
N
N C k
,1
7677.0905.1747336
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
945.0
905.1747336
3280001
17677.0
1,
N
N
N
N C k
z cr
Ed
y
mz yz
13.12.2.9Internal factor, zyk , calculation
ycr
Ed
z mLT my zy
N
N C C k
,
1
893.0
905.1747336
328000479.01
905.1747336
3280001
1
1
,
,
N
N N
N
N
N
N
N
z cr
Ed z
z cr
Ed
z
908.0
44.950353713280001
893.0068.1949.0
1,
N N
N N
C C k
ycr
Ed
z mLT my zy
13.12.2.10Internal factor, zz k , calculation
844.0
905.1747336
3280001
893.07677.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
13.12.2.11Bending and axial compression verification
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rk z
Rd z Ed z
zz
M
Rk y
LT
Rd y Ed y
zy
M
Rk z
Ed
M
Rk z
Rd z Ed z
yz
M
Rk y
LT
Rd y Ed y
yy
M
Rk y
Ed
M
M M k
M
M M k
N
N
M
M M k
M
M M k
N
N
i y Rk A f N
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14.461.124.232.0
1
/27510.242068
104.127844.0
1
/275533.3179229589.0
101274908.0
1
/2752.7815479.0
328000
47.4808.151.215.0
1
/27510.242068
104.127945.0
1
/275533.3179229589.0
1012740161.1
1
/2752.78151
328000
23
6
23
6
22
23
6
23
6
22
mm N mm
Nmm
mm N mm Nmm
mm N mm N
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
Finite elements modeling
■
■
■
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y coefficient corresponding to non-dimensional slenderness y
y
z z
z
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Internal factor, yyk
yyk
Internal factor, yz k
yz k
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Internal factor, zyk
zyk
Internal factor, zz k
zz k
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13.12.2.12Reference results
Result name Result description Reference value
y y y
z z z
yyk yyk
yz k yz k
zyk zyk
zyk zyk
13.12.3Calculated results
Result name Result description Value Error
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13.13 EC3 Test 26: Verifying an user defined I section class 3 column fixed on the bottom
Passed
13.13.1Description
13.13.2Background
13.13.2.1Model description
■
■
■
■
■
■
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Units
Geometrical properties
■
■ 2
72.9108 mm A ■ mmb 190
■ mmt f 6.14
■ mmr 0
■ mmt w 4.9
■ mmhw 408
■ 3
, 06.1261435 mmW yel
■ 378.1428491 mmW y
■ 3
, 65.175962 mmW z el
■ 69.271897, z pl W
■
■
■
■
Materials properties
■
■
■
Boundary conditions
■
►
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Loading
■
13.13.2.2Cross section Class
118.083.268
76.48
sup
inf
Mpa
Mpa
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909.320
59.317
83.26876.378
152.5859.317
76.4876.378
59.317
76.378
83.26876.4883.26876.48 y
x y x y x
5.0153.076.378
152.5876.378
x
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924.0275
235235
y f
56.63)18.0(33.067.0
924.042
33.067.0
42170
924.0
30.404.9
6.142408
t
cmm
mmmm
t
c
316.8924.0918.6
924.0
18.66.14
30.90
t
ct
c
13.13.2.3Buckling verification
a) over the strong axis of the section, y-y:
α
34.0
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y y
11
22
y y y
y
y
ycr
y
y N
f A
,
*
272.9108 mm A
kN N mm
mmmm N
l
I E N
fy
y ycr 05.1333387.133338053²2000
257332751/210000²
²
² 42
,
137.07.133338053
/27572.9108 22
,
N
mm N mm
N
f A
ycr
y y
499.0137.02.0137.034.015.0²)2.0(15.0 2 y y y
1
1
022.1137.0499.0499.0
11
2222
y
y
y y y
y
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b) over the weak axis of the section, z-z
α
49.0
z z
1122
z z z
z
z
z cr
y z
N
f A
,
*
ml fz 00.2
272.9108 mm A
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kN N
mm
mmmm N
l
I E N
fz
z z cr 70.866138.8661700
²2000
10.16716452/210000²
²
² 42
,
538.0
38.8661700
/27572.9108 22
,
N
mm N mm
N
f A
z cr
y z
728.0538.02.0538.049.015.0²)2.0(15.0 2 z z z
821.0
1
821.0538.0728.0728.0
11
2222
z
z
z z z
z
13.13.2.4Lateral torsional buckling verification
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
11 C
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4
2
f z
w
t h I I
f t mmt f 6.14
611
24
1046774.64
6.144080mm16716452.1mm
mmmm I w
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
089.18021802088994
052.208384.8661700110.16716452/210000
13.492581/808002000
10.16716452
1046774.6
2000
10.16716452/2100001
²
²
²
²
422
422
4
611
2
422
1
34
max
, 172.1261434204
257332751mm
mm
mm
z
I W
y
yel
439.01802088994
/275172.1261434 23,
Nmm
mm N mm
M
f W
cr
y yeff LT
LT LT
1
²²
1
LT LT LT
LT
²2.015.0 LT LT LT LT
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2147.2190
408
mm
mm
b
h
α
76.0
687.0²439.02.0439.076.015.0²2.015.0 LT LT LT LT
1813.0²439.0²687.0687.0
1
²²
1
LT LT LT
LT
13.13.2.5Internal factor, yyk , calculation
yyk
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ycr
Ed
y
mLT my yy
N
N C C k
,
1
ycr
Ed y
ycr
Ed
y
N
N N
N
,
,
1
1
022.1 y
kN N Ed 1000
kN N l
I E N
fy
y
ycr 05.1333387.133338053²
²,
1
7.133338053
100000011
7.13333805310000001
1
1
,
,
N
N N
N
N
N
N N
ycr
Ed
y
ycr
Ed
y
myC
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0
0
,0
cr
y yeff
M
f W
34
max
, 172.1261434204
257332751mm
mm
mm
z
I W
y
yel
0cr M 11 C 02 C
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
089.18021802088994
052.208384.8661700110.16716452/210000
13.492581/808002000
10.16716452
1046774.6
2000
10.16716452/2100001
²
²
²
²
422
422
4
611
2
422
1
439.01802088994
/275172.1261434 23,
0
Nmm
mm N mm
M
f W
cr
y yeff
4
,,
1 1120.0
TF cr
Ed
z cr
Ed
N
N
N
N C
T cr TF cr N N ,,
²
1
,
2
2
0
,
T cr
wt T cr
L
I E I G
i N
0 I
0301.02
0
2
0
222
0 z yiii z y
T cr
L,
m L T cr 00.2,
mm
mmmm N mmmm N
mm
mm N T cr
13
62242
4
2
,
10244.1
²2000
33.746457599819/21000013.492581/80800
0301.0
1
N N Ed 1000000
N N N T cr TF cr
13
,, 10244.1
N
mmmmmm N
l I E N
fz
z z cr 38.8661700
²200010.16716452/210000²
²²
42
,
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172.010244.1
10000001
38.8661700
10000001120.01120.0 4
134
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
1
11
11
172.01120.0469.0
172.01120.0
469.0
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C
a
aC C C
N
N
N
N C
N
N
N
N C
myC
LT y
LT y
mymymya
aC C C
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
3
,, 06.1261435 mmW W yeff yel
444.106.1261435
72.9108
1000000
102003
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
1998.0257332751
492581.1311
4
4
mm
mm
I
I a
y
t LT
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0mC
1
ycr
Ed my
N
N C
,
0, 33.036.021.079.0
N
mm
mmmm N
l
I E N
fy
y
ycr 7.133338053²2000
257332751/210000²
²
² 42
,
N N Ed 1000000
002.17.133338053
100000033.0136.0121.079.00, N
N C my
001.11444.11
1444.1002.11002.1
11 0,0,
LT y
LT y
mymymya
aC C C
mLT C
065.1
1
065.1
10244.1
10000001
38.8661700
10000001
1001.1
11 13
2
,,
2
mLT
mLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
N
N
N
N
aC C
yyk
074.1
7.133338053
10000001
1065.1001.1
1,
N
N
N
N C C k
ycr
Ed
y
mLT my yy
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13.13.2.6Internal factor, yz k , calculation
z cr
Ed
y
mz yz
N
N C k
,
1
mz C
1
200
200
,sup
,inf
kNm
kNm
M
M
Ed
Ed
kN N
mm
mmmm N
l
I E N
fz
z z cr 70.866138.8661700
²2000
10.16716452/210000²
²
² 42
,
776.038.8661700
100000033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
878.0
38.8661700
10000001
1776.0
1,
N
N
N
N C k
z cr
Ed
y
mz yz
13.13.2.7Internal factor, zyk , calculation
ycr
Ed
z mLT my zy
N
N C C k
,
1
799.0
38.8661700
1000000821.01
38.8661700
10000001
1
1
,
,
N
N N
N
N
N
N
N
z cr
Ed
z
z cr
Ed
z
821.0 z
050.1
7.133338053
10000001
977.0065.1001.1
1,
N
N
N
N C C k
ycr
Ed
z mLT my zy
13.13.2.8Internal factor, zz k , calculation
857.0
38.8661700
10000001
977.0776.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
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13.13.2.9Bending and axial compression verification
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rk z
Rd z Ed z
zz
M
Rk y
LT
Rd y Ed y
zy
M
Rk z
Ed
M
Rk z
Rd z Ed z
yz
M
Rk y
LT
Rd y Ed y
yy
M
Rk
y
Ed
M
M M k
M
M M k
N
N
M
M M k
M
M M k
N
N
i y Rk A f N
41.118.074.049.0
1
/2755.175962
1010857.0
1
/27506.1261435813.0
10200050.1
1
/27572.9108821.0
1000000
34.118.076.040.0
1
/2755.175962
1010878.0
1
/27506.1261435813.0
10200074.1
1
/27572.91081
1000000
23
6
23
6
222
23
6
23
6
22
mm N mm
Nmm
mm N mm
Nmm
mm N mmmm
N
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
Finite elements modeling
■
■■
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y coefficient corresponding to non-dimensional slenderness y
y
z z
z
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Internal factor, yyk
yyk
Internal factor, yz k
yz k
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Internal factor, zyk
zyk
Internal factor, zz k
zz k
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13.13.2.10Reference results
Result name Result description Reference value
y y y
z z z
yyk yyk
yz k yz k
zyk zyk
zyk zyk
13.13.3Calculated results
Result name Result description Value Error
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13.14 Verifying the cross section optimization according to EC3 (TTAD #11516)
Passed
13.14.1Description
13.15 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method
Passed
13.15.1Description
13.16 EC3: Verifying the buckling length results (TTAD #11550)
Passed
13.16.1Description
13.17 Verifying the buckling coefficient Xy on a class 2 section
Passed
13.17.1Description
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13.18 EC3 fire verification: Verifying the work ratios after performing an optimization for steelprofiles (TTAD #11975)
Passed
13.18.1Description
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13.19 EC3 Test 13: Verifying the resistance of a rectangular hollow section column subjected tobending and shear efforts
Passed
13.19.1Description
13.19.2Background
13.19.2.1Model description
■
■
■
■
►
■
■
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Units
Geometry
■■
■
■
■
■
■
■ 0.10 M
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
■
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13.19.2.2Reference results for calculating the design plastic shear resistance of the cross-section
Shear area of the cross section
22
5694200300
3009490mm
mmmm
mmmm
hb
h A Av
Design plastic shear resistance of the cross section
N
MPa
mm
f
AV
M
y
v
Rd pl 6.7725460.1
3
235
56943
2
0
,
Work ratio
%89.251006.772546
200000100
,
Rd pl V
V
13.19.2.3Reference results for calculating the design resistance for bending
Cross section class
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1510
10215220022
mm
mmmmmm
t
t r b
t
c
0.1235
y f
333315 t
c
2510
10215230022
mm
mmmmmm
t
t r h
t
c
0.1235
y f
727225 t
c
Design resistance for bending
Nmm MPamm f W
M M
y y pl
Rd c 2246600000.1
235956000 3
0
,
,
Work ratio
%56.222100224660000
2
5000200000
1002100,,
Nmm
mm N
M
LV
M
M
Rd c Rd c
Ed
Finite elements modeling
■
■
■
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Finite elements results
Work ratio of the design shear resistance
Work ratio of the design resistance for bending
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13.19.2.4Reference results
Result name Result description Reference value
13.19.3Calculated results
Result name Result description Value Error
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13.20 EC3 Test 19: Verifying the buckling resistance for a IPE300 column
Passed
13.20.1Description
13.20.2Background
13.20.2.1Model description
■■
■
■
■
■
Units
Geometrical properties
■
■
■
Materials properties
■
■
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Boundary conditions
■
►
►■
■
Loading
■
■
13.20.2.2Buckling in the strong inertia of the profile (along Y-Y)
%100100,
Rd b
Ed
N
N
1
,
M
y
Rd b
f A N
Coefficient corresponding to non-dimensional slenderness for Y-Y axis
11
22
cr
y
N
f A*
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N
mm
mm MPa
L
I E N
fy
z cr 70.1731878
10000
108356210000²
²
²2
44
854.070.1731878
/2355380 22
N
mm N mm
N
f A
cr
y
²)2.0(15.0
933.0854.02.0854.021.015.0²)2.0(15.0 2
1764.0854.0933.0933.0
112222
1 M 11 M
N
mm N mm
N Rd b 089.9660511
/2355380764.0 22
,
N N Ed 200000
%703.20100089.966051
200000100
,
N
N
N
N
Rd b
Ed
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13.20.2.3Buckling in the weak inertia of the profile (along Z-Z)
%100100,
Rd b
Ed
N N
1
,
M
y
Rd b
f A N
Coefficient corresponding to non-dimensional slenderness for Z-Z axis
11
22
cr
y
N
f A*
N
mm
mm MPa
L
I E N
fy
z cr 610.125144
10000
1080.603210000²
²
²2
44
178.3610.125144
/2355380 22
N
mm N mm
N
f A
cr
y
²)2.0(15.0
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056.6178.32.0178.334.015.0²)2.0(15.0 2
1089.0178.3056.6056.6
112222
1 M 11 M
N mm N mm
N Rd b 78.1127711
/2355380089.0 22
,
N N Ed 200000
%349.17710078.112771
200000100
,
N
N
N
N
Rd b
Ed
Finite elements modeling
■
■
■
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Finite elements results
Coefficient corresponding to non-dimensional slenderness after Y-Y axis
Ratio of the design normal force to design buckling resistance (strong inertia)
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13.20.2.4Reference results
Result name Result description Reference value
y
z
13.20.3Calculated results
Result name Result description Value Error
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13.21 EC3 test 8: Verifying the classification and the resistance of a column subjected to bendingand axial load
Passed
13.21.1Description
13.21.2Background
13.21.2.1Model description
■
■
■
■
►
►
■
■
Units
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Geometry
■
■
■
■
■
■ 3
, 3512000 mmW y pl
■ 0.10 M
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►►
■
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13.21.2.2Reference results for calculating the column subjected to bending and axial force
Cross section class
Verifying the design resistance for uniform compression
N MPamm f A
N M
y
Rd c 36660000.1
23515600 2
0
,
%100%3.270.1273.03666000
1000000
,,
N
N
N
N
N
N
Rd c Rd c
Ed
Verifying the column subjected to bending and axial force
N N N N N Rd pl Ed 916500366600025.0100000025.0 ,
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■
► Nmm MPamm f W
M M
y y pl
Rd y pl 8253200000.1
2353512000 3
0
,
,,
■
► 273.036660001000000
, N
N N
N n Rd pl
■
► 464.015600
1922021560022
2
mm
mmmmmm
A
t b Aa
f
■
►a
n M M Rd y pl Rd y N
5.01
1,,,, Rd y pl Rd y N M M ,,,,
► Nmm Nmm Nmm M Rd y N 825320000781259948464.05.01
273.01
825320000,,
■
► 0.1,,
,
,,
,
Rd z N
Ed z
Rd y N
Ed y
M
M
M
M
►
0.12/
,,
2
,,
,
Rd y N Rd y N
Ed y
M
Lq
M
M
►
%100%9.790.1799.0781259948
2/5000/502/ 2
,,
2
Nmm
mmmm N
M
Lq
Rd y N
Finite elements modeling
■
■
■
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Work ratio of the design resistance for uniform compression
Work ratio of the design resistance for oblique bending
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13.21.2.3Reference results
Result name Result description Reference value
13.21.3Calculated results
Result name Result description Value Error
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13.22 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow sectionbeam
Passed
13.22.1Description
13.22.2Background
13.22.2.1Model description
■
■
■
■
►
■
■
Units
Geometry
■
■
■
■
■
■■ 0.10 M
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Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►■
13.22.2.2Reference results for calculating the design plastic shear resistance of the cross-section
Shear area of the cross section
22
5694200300
3009490mm
mmmm
mmmm
hb
h A Av
Design plastic shear resistance of the cross section
N
MPamm
f A
V M
y
v
Rd pl 9040440.1
3
2755694
3
2
0
,
Work ratio
%83.13100904044
125000100
904044
2
5000/50
1002100,,
N
N
N
mmmm N
V
Lq
V
V
Rd pl Rd pl
Ed
Finite elements modeling
■
■
■
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Finite elements results
Work ratio of the design shear resistance
13.22.2.3Reference results
Result name Result description Reference value
13.22.3Calculated results
Result name Result description Value Error
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13.23 Verifying the shape sheet results for a column (TTAD #11550)
Passed
13.23.1Description
13.24 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770)
Passed
13.24.1Description
13.25 Changing the steel design template for a linear element (TTAD #12491)
Passed
13.25.1Description
13.26 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533)
Passed
13.26.1Description
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13.27 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method
Passed
13.27.1Description
13.28 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545)
Passed
13.28.1Description
13.29 Verifying the steel shape optimization when using sections from Advance Steel Profilesdatabase (TTAD #11873)
Passed
13.29.1Description
13.30 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method
Passed
13.30.1Description
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13.31 EC3 Test 3: Class section classification and share and bending moment verification of anIPE300 column
Passed
13.31.1Description
13.31.2Background
13.31.2.1Model description
■
■
■
■
■
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Units
Materials properties
Boundary conditions
■
►
►
■
13.31.2.2Reference results for calculating the cross section class
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13.31.2.3Reference results in calculating the shear resistance V pl,Rd
0
,
3*
M
y
v
Rd pl
f A
V
f w f v t r t t b A A *)*2(**2
268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmt r t t b A A f w f v
0 M 10 M
kN MN
f A
V M
y
v
Rd pl 42.3483484.01
3
235*10*68.25
3* 4
0
,
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13.31.2.4Reference results in calculating the bending moment resistance
%50%4.5742.348
200
,
Rd pl
Ed
V
V
0223.01348.0
200.0*212 22
,
Rd pl
Ed
V V
f w f v t r t t b A A *)*2(**2
268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmt r t t b A A f w f v
0 M 10 M
MNm
f t
Aw
M M
y
w
v pl
Rd V y 146.01
235*0071.0*4
)²10*68.25(*0223.010*40.628*
4
²* 46
0
,,
MNmkNmkN m M Ed 5.0500200*5.2
%342146.0
500.0
,,
Rd V y
Ed
M
M
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Finite elements modeling
■
■
■
Finite elements results
Shear z direction work ratio
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Combined oblique bending
13.31.2.5Reference results
Result name Result description Reference value
13.31.3Calculated results
Result name Result description Value Error
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13.32 EC3 Test 34: Verifying C310x30.8 class 3beam, loaded with centric compression, uniformlinear horizontal efforts by Y and a vertical punctual load in the middle
Passed
13.32.1Description
13.33 EC3 Test 30: Verifying IPE300 beam, simply supported, loaded with centric compression anduniform linear efforts by Y and Z axis
Passed
13.33.1Description
13.34 EC3 Test 32: Verifying IPE600 simple supported beam, loaded with centric compression and
uniform linear efforts by Y and Z axis
Passed
13.34.1Description
13.35 EC3 Test 39: Verifying CHS323.9x6.3H class 2 beam, loaded with centric compression, uniformlinear horizontal efforts by Y and a vertical punctual load in the middle
Passed
13.35.1Description
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13.36 EC3 Test 14: Verifying the bending resistance of a rectangular hollow section column made ofS235 steel
Passed
13.36.1Description
13.36.2Background
13.36.2.1Model description
■
■
■
■
■
■
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Units
Geometry
■■
■
■
■
■
■ 0.10 M
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
■
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13.36.2.2Reference results for calculating the design resistance for bending
Cross section class
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1510
10215220022
mm
mmmmmm
t
t r b
t
c
0.1235
y f
333315 t
c
2510
10215230022
mm
mmmmmm
t
t r h
t
c
0.1235 y f
727225 t
c
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Design resistance for bending
Nmm MPamm f W
M M
y y pl
Rd c 224660000
0.1
235956000 3
0
,
,
Work ratio
%64.55100224660000
2
500050000
1002100,,
Nmm
mm N
M
LV
M
M
Rd c Rd c
Ed
Finite elements modeling
■■
■
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Finite elements results
Work ratio of the design resistance for bending
13.36.2.3Reference results
Result name Result description Reference value
13.36.3Calculated results
Result name Result description Value Error
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13.37 EC3 Test 25: Verifying an user defined I section class 4 column fixed on the bottom and with adisplacement restraint at 2.81m from the bottom
Passed
13.37.1Description
13.37.2Background
13.37.2.1Model description
■
■
■
■
■
■
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Units
Geometrical properties
■
■ 2
10850mm A ■ mmb 220
■ mmt f 15
■ mmr 0
■ mmt w 5
■ mmhw 880
■ 33
, 1066.3387 mmW yel
■ 331062.3757 mmW y
■ 33
, 1008.242 mmW z el
■ 33
, 1031.368 mmW z pl
■
■
■
■
Materials properties
■
■
■
Boundary conditions
■
►
►
■
►
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Loading
■
13.37.2.2Cross section Class
1873.030.406
84.354
sup
inf
Mpa
Mpa
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27.396
14.761
84.354850
73.45314.761
3.406850
14.761
850
84.3543.40684.3543.406 y
x y x y x
5.05338.0850
73.453
850
x
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924.0275
235235
y f
61.101)873.0(33.067.0
924.042
33.067.0
42170
924
1705
152880
t
cmm
mmmm
t
c
316.8924.0961.7
924
61.715
5.107
t
ct
c
13.37.2.3Effective cross-sections of Class4 cross-sections
1705
152880
mm
mmmm
t
c
41 k
k
t
bw
p
4.28
wb mmbw 850
9244.0275
235235
y f
261.349244.04.28
5
850
mm
mm
p
286.0
261.3
4055.0261.33055.0
043
673.0261.322
p
p p
mmmmb
mmmmb
mmmmb
bb
bb
bb
e
e
eff
eff e
eff e
weff
55.1211.2435.0
55.1211.2435.0
1.243850286.0
5.0
5.0
2
1
2
1
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2
21, 5.121555.121555.1215 mmmmmmmmmmbt bt A ewewwebeff
2
, 330022015 mmmmmmbt A f f flangeeff
222
,, 2.7815330025.12152 mmmmmm A A A flangeeff webeff eff
13.37.2.4Effective elastic section modulus of Class4 cross-sections
mmmm
bb t c 4252
8501
sup
inf
9.231 k
170
5
152880
mm
mmmm
t
c
k
t
bw
p
4.28
wb mmbw 850
9244.0275
235235
y f
325.19.239244.04.28
5
850
mm
mm
p
692.0
325.1
2055.0325.13055.0
023
673.0325.122
p
p p
mmmmb
mmmmb
mmmm
b
bb
bb
bbb
e
e
eff
eff e
eff e
wceff
46.1761.2946.0
64.1171.2944.0
1.294
11
850692.0
6.0
4.01
2
1
2
1
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mm
yG
53.155.10195
095.158330
15220546.601564.11715220
27.124546.6015.4321522018.366564.1175.43215220
4
23
23
23
23
14482344295.6993800726.748854356
97.4161522012
2201574.107546.601
12
546.601
71.381564.11712
564.11703.44815220
12
22015
mm
I y
423
23
23
23
63.2662749001522012
152200546.601
12
46.6015
0564.11712
64.1175015220
12
15220
mm
I z
34
max,
533.317922953.455
1448234429mm
mm
mm
z
I W
y
yel
34
max
, 10.242068110
63.26627490mm
mm
mm
y
I W z
z el
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13.37.2.5Buckling verification
a) over the strong axis of the section, y-y:
α
34.0
y y
11
22
y y y
y
y
ycr
yeff
y N
f A
,
*
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22.7815 mm Aeff
kN N
mm
mmmm N
l
I E N
fy
y
ycr 37.9503544.95035371
²5620
1448234429/210000²
²
² 42
,
15.044.95035371
/2752.7815 22
,
N
mm N mm
N
f A
ycr
yeff y
503.015.02.015.034.015.0²)2.0(15.0 2 y y y
1
1
017.115.0503.0503.0
11
2222
y
y
y y y
y
b) over the weak axis of the section, z-z
α
49.0
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z z
11
22
z z z
z
z
z cr
yeff z
N
f A
,
*
22.7815 mm Aeff
ml fz
81.2
kN N
mm
mmmm N
l
I E N
fz
z z cr 35.698962.6989347
²2810
63.26627490/210000²
²
² 42
,
555.062.6989347
/2752.7815 22
,
N
mm N mm
N
f A
z cr
yeff z
741.0555.02.0555.049.015.0²)2.0(15.0 2 z z z
812.0
1
812.0555.0741.0741.0
11
2222
z
z
z z z
z
13.37.2.6Lateral torsional buckling verification
a) for the top part of the column:
z
t
z
w z cr
I E
I G L
I
I
L
I E
C M
²
²
²
²1
²252.0423.0325.0
11
C
50.01274
637
kNm
kNm
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31.15.01274
6371 C
kNm
kNm
41448234429mm I y
463.26627490 mm I z
4
2
f z
w
t h I I
f t mmt f 15
611
24
10808.49
4
15880mm326627490.6mm
mmmm I w
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
43.40224022433856
32.439626.698934731.163.26627490/210000
514614/808002810
63.26627490
10808.49
2810
63.26627490/21000031.1
²
²
²
²
422
422
4
611
2
422
1
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34
max
, 533.317922953.455
1448234429mm
mm
mm
z
I W
y
yel
466.04022433856
/275533.3179229 23,
Nmm
mm N mm
M
f W
cr
y yeff LT
LT LT
1²²
1
LT LT LT
LT
²2.015.0 LT LT LT LT
24220
880
mm
mm
b
h
α
76.0
710.0²466.02.0466.076.015.0²2.015.0 LT LT LT LT
1803.0²466.0²710.0710.0
1
²²
1
LT LT LT
LT
b) for the bottom part of the column:
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
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0637
0
kNm
77.10 1 C
41448234429mm I y
463.26627490 mm I z
4
2
f z
w
t h I I
f t mmt f 15 611
24
10808.494
15880mm326627490.6mm
mmmm I w
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kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
89.54345434891269
32.439626.698934777.163.26627490/210000
514614/808002810
63.26627490
10808.49
2810
63.26627490/21000077.1
²
²
²
²
422
422
4
611
2
422
1
34
max
, 533.317922953.455
1448234429mm
mm
mm
z
I W
y
yel
401.05434891269
/275533.3179229 23,
Nmm
mm N mm
M
f W
cr
y yeff LT
LT LT
1
²²
1
LT LT LT
LT
²2.015.0 LT LT LT LT
24220
880
mm
mm
b
h
α
76.0
657.0²401.02.0401.076.015.0²2.015.0 LT LT LT LT
1849.0²401.0²657.0657.0
1
²²
1
LT LT LT
LT
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13.37.2.7Internal factor, yyk , calculation
yyk
a) for the top part of the column:
ycr
Ed
ymLT my yy
N
N C C k
,
1
ycr
Ed y
ycr
Ed
y
N
N
N
N
,
,
1
1
1 y
kN N Ed 328
kN N l
I E N
fy
y
ycr 37.9503544.95035371²
²,
1
44.95035371
32800011
44.95035371
3280001
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
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myC
0
0
,0
cr
y yeff
M
f W
34
max
, 533.317922953.455
1448234429mm
mm
mm
z
I W
y
yel
0cr M 11 C 02 C
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
56.30703070560199
32.439626.6989347163.26627490/210000
514614/808002810
63.26627490
10808.49
2810
63.26627490/2100001
²
²
²
²
422
422
4
611
2
422
10
534.03070560199
/275533.3179229 23
0
,0
Nmm
mm N mm
M
f W
cr
y yeff
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4
,,
1 1120.0
TF cr
Ed
z cr
Ed
N
N
N
N C
T cr TF cr N N ,,
²,
2
0
,
T cr
wt T cr
L
I E I G
I
A N
0 I
2
0 g z y z A I I I 467.1490580416 mm I y
417.26628854 mm I z
2
10850mm A 0 g z
4442
0 151720927017.2662885467.1490580416 mmmmmm I I z A I I I z y g z y
T cr L ,
m L T cr 81.2, 475.514614 mm I t
61110808.49 mm I w
mm
mmmm N mmmm N
mm
mm N T cr
24.9646886
²2810
10808.49/21000075.514614/80800
1517209270
10850 6112242
4
2
,
N N Ed 328000
N N N T cr TF cr 24.9646886,,
kN N l
I E N fz
z z cr 35.698962.6989347
²
²,
224.0
24.9646886
3280001
62.6989347
328000131.120.01120.0 44
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
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1
11
1
1
224.01120.0534.0
224.01120.0
534.0
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C a
aC C C
N
N
N
N C
N
N
N
N C
myC
LT y
LT y
mymymy a
a
C C C
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
3
,, 533.3179229 mmW W yeff yel
55.9533.3179229
2.7815
328000
1012743
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
19996.01448234429
75.51461411 4
4
mmmm
I I a
y
t LT
0mC
0
ycr
Ed my
N
N C
,
0, 33.036.021.079.0
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kN N l
I E N
fy
y
ycr 37.9503544.95035371²
²,
79.044.95035371
328000
33.0036.079.00, N
N
C my
949.0155.91
155.979.0179.0
11 0,0,
LT y
LT y
mymymya
aC C C
mLT C
mLT C
1
11,,
2
T cr
Ed
z cr
Ed
LT mymLT
N
N
N
N
aC C
myC mLT C
myC 5.0
LT y
LT y
mymymya
aC C C
11 0,0,
895.044.95035371
32800033.05.036.05.021.079.033.036.021.079.0
,
0, N
N
N
N C
ycr
Ed my
55.9533.3179229
2.7815328000
1012743
26
,
, mm
mm N Nmm
W A
N M
yeff
eff
Ed
Ed y
y
974.0155.91
155.9895.01895.0
11 0,0,
LT y
LT y
mymymya
aC C C
1
1
989.0
644.9
328.01
9897.6
328.01
1974.0
11
2
,,
2
mLT
mLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
aC C
yyk
952.0
44.95035371
3280001
11949.0
1,
N
N
N
N C C k
ycr
Ed
y
mLT my yy
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b) for the bottom part of the column:
ycr
Ed
y
mLT my yy
N
N C C k
,
1
534.03070560199
/275533.3179229 23
0
,0
Nmm
mm N mm
M
f W
cr
y yeff
260.024.9646886
3280001
62.6989347
328000177.120.01120.0 44
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
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1
11
1
1
260.01120.0534.0
260.01120.0
534.0
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C a
aC C C
N
N
N
N C
N
N
N
N C
myC
LT y
LT y
mymymy a
a
C C C
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
3
,, 533.3179229 mmW W yeff yel
55.9533.3179229
2.7815
328000
1012743
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
19996.01448234429
75.51461411 4
4
mmmm
I I a
y
t LT
0mC
01274
0
,sup
,inf kNm M
M
Ed
Ed
ycr
Ed my
N N C
,
0, 33.036.021.079.0
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kN N l
I E N
fy
y
ycr 37.9503544.95035371²
²,
79.044.95035371
328000
33.0036.079.00, N
N
C my
949.0155.91
155.979.0179.0
11 0,0,
LT y
LT y
mymymya
aC C C
mLT C
mLT C
1
11 ,,
2
T cr
Ed
z cr
Ed
LT mymLT
N
N
N
N
aC C
myC mLT C
myC 5.0
LT y
LT y
mymymya
aC C C
11 0,0,
895.0
44.95035371
32800033.05.036.05.021.079.033.036.021.079.0
,
0,
N
N
N
N C
ycr
Ed my
77.4533.3179229
2.7815
328000
106373
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
967.0177.41
177.4895.01895.0
11 0,0,
LT y
LT y
mymymya
aC C C
1
1
974.0
24.9646886
3280001
62.6989347
3280001
1967.0
11
2
,,
2
mLT
mLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
N
N
N
N
aC C
952.0
44.95035371
3280001
11949.0
1,
N
N
N
N C C k
ycr
Ed
y
mLT my yy
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13.37.2.8Internal factor, yz k , calculation
a) for the top part of the column:
z cr
Ed
y
mz yz
N
N C k
,
1
mz C
0
1274
0
,sup
,inf
kNm M
M
Ed
Ed
784.062.6989347
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
823.0
62.6989347
3280001
1784.0
1,
N
N
N
N C k
z cr
Ed
y
mz yz
b) for the bottom part of the column:
823.0
62.6989347
3280001
1784.0
1,
N
N
N
N C k
z cr
Ed
y
mz yz
13.37.2.9Internal factor, zyk , calculation
a) for the top part of the column:
ycr
Ed
z mLT my zy
N
N C C k
,
1
991.0
62.6989347
328000812.01
62.6989347
3280001
1
1
,
,
N
N N
N
N
N
N
N
z cr
Ed z
z cr
Ed
z
944.0
44.95035371
3280001
991.01949.0
1,
N
N
N
N C C k
ycr
Ed
z mLT my zy
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b) for the bottom part of the column:
944.0
44.95035371
3280001
991.01949.0
1,
N
N
N
N C C k
ycr
Ed
z mLT my zy
13.37.2.10Internal factor, zz k , calculation
a) for the top part of the column:
815.0
62.6989347
3280001
991.0784.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
b) for the bottom part of the column:
815.0
62.6989347
3280001
991.0784.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
13.37.2.11Bending and axial compression verification
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rk z
Rd z Ed z
zz
M
Rk y
LT
Rd y Ed y
zy
M
Rk z
Ed
M
Rk z
Rd z Ed z yz
M
Rk y
LT
Rd y Ed y yy
M
Rk y
Ed
M
M M k
M
M M k
N
N
M
M M k M
M M k N N
i y Rk A f N
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a) for the top part of the column:
56.356.171.119.0
1
/27510.242068
104.127815.0
1
/275533.3179229803.0
101274944.0
1
/2752.7815812.0
328000
46.358.173.115.0
1
/27510.242068104.127823.0
1
/275533.3179229803.0
101274952.0
1
/2752.78151
328000
23
6
23
6
22
23
6
23
6
22
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
mm N mm Nmm
mm N mm
Nmm
mm N mm
N
b) for the bottom part of the column:
56.356.171.119.0
1
/27510.242068
104.127815.0
1
/275533.3179229803.0
101274944.0
1
/2752.7815812.0
328000
46.358.173.119.0
1
/27510.242068
104.127823.0
1
/275533.3179229803.0
101274952.0
1
/2752.7815812.0
328000
23
6
23
6
22
23
6
23
6
22
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
Finite elements modeling
■■
■
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y coefficient corresponding to non-dimensional slenderness y
y
z z
z
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Internal factor, yyk
yyk
Internal factor, yz k
yz k
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Internal factor, zyk
zyk
Internal factor, zz k
zz k
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13.37.2.12Reference results
Result name Result description Reference value
y y y
z z z
yyk yyk
yz k yz k
zyk zyk
zyk zyk
13.37.3Calculated results
Result name Result description Value Error
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13.38 EC3 Test 42: Determining lateral torsional buckling parameters for a I-shaped welded built-upbeam considering the load applied on the upper flange
Passed
13.38.1Description
13.39 EC3 Test 38: Verifying RHS350x150x5H class 4 column, loaded with centric compression,punctual horizontal force by Y and a bending moment, all applied to the top
Passed
13.39.1Description
13.40 EC3 Test 33: Verifying UPN300 simple supported beam, loaded with centric compression,uniform linear horizontal efforts by Y and punctual vertical force by Z axis
Passed
13.40.1Description
13.41 EC3 Test 35: Verifying C310x30.8 class 4 cantilever, loaded with centric compression, uniformlinear horizontal efforts by Y and a vertical punctual load applied on the free end
Passed
13.41.1Description
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13.42 EC3 Test 29: Verifying an user defined I section class 1, column hinged on base and restrainedon top for the X, Y translation and Z rotation
Passed
13.42.1Description
13.42.2Background
13.42.2.1Model description
■
■
■
■
■
■
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Units
Geometrical properties
■
■ 2
06.4904 mm A ■ mmb 150
■ mmt f 70.10
■ mmr 0
■ mmt w 10.7
■ mmhw 260
■ 3
, 63.445717 mmW yel
■ 318.501177 mmW y
■ 3
, 89.80344 mmW z el
■ 3
, 96.123381 mmW z pl
■ 464.57943291 mm I y
■ 446.6025866 mm I z
■ 497.149294 mm I t
■ 688.19351706542 mm I w
Materials properties
■
■
■
Boundary conditions
■
►
►
►
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Loading
■
13.42.2.2Cross section Class
1253.006.179
30.45
sup
inf
Mpa
Mpa
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73.190
36.224
06.1796.238
175.4836.224
3.456.238
36.224
6.238
06.17930.4506.17930.45 y
x y x y x
5.080.06.238
73.1906.238
x
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924.0275
235235
y f
93.3818.013
924.0396
113
3966.33
924.0
61.331.7
7.102260
t
cmm
mmmm
t
c
316.8924.0968.6
924.0
68.67.10
2
1.7150
t
c
t
c
13.42.2.3Buckling verification
a) over the strong axis of the section, y-y:
α
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34.0
y y
11
22
y y y
y
y
ycr
y
y N
f A
,
*
206.4904 mm A
4
64.57943291 mm I y
kN N
mm
mmmm N
l
I E N
fy
y
ycr 33.380295.3802327²5620
64.57943291/210000²
²
² 42
,
5956.095.3802327
/27506.4904 22
,
N
mm N mm
N
f A
ycr
y y
7446.05956.02.05956.034.015.0²)2.0(15.0 2 y y y
839.0
1
839.05956.07446.07446.0
11
2222
y
y
y y y
y
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b) over the weak axis of the section, z-z
α
49.0
z z
1122
z z z
z
z
z cr
y z
N
f A
,
*
446.6025866 mm I z
206.4904 mm A
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kN N
mm
mmmm N
l
I E N
fz
z z cr 71.3158151.1581706
²2810
46.6025866/210000²
²
² 42
,
923.0
51.1581706
/27506.4904 22
,
N
mm N mm
N
f A
z cr
y z
103.1923.02.0923.049.015.0²)2.0(15.0 2 z z z
586.0
1
586.0923.0103.1103.1
112222
z
z
z z z
z
13.42.2.4Lateral torsional buckling verification
a) for the top part of the column:
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
31.15.025
251
,
, C kN
kN
M
M
top y
botom y
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464.57943291 mm I y
446.6025866 mm I z
4
97.149294 mm I t 6
88.19351706542 mm I w
4
2
f z
w
t h I
I
mmh 260
f t mmt f 7.10
6
24
093627638294
7.1026046.6025866mm
mmmmmm I w
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mmmmmm N
I E I G L
I I
L I E C M
z
t
z
w z cr
36.31574.315363380
20.15251.158170631.146.6025866/210000
97.149294/808002810
46.6025866
09362763829
281046.6025866/21000031.1
²²
²²
422
422
4
6
2
422
1
661.074.315363380
/27518.501177 23,
Nmm
mm N mm
M
f W
cr
y y pl LT
LT LT
1²²
1
LT LT LT
LT
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²2.015.0 LT LT LT LT
2733.1150
260
mm
mm
b
h
α
49.0 LT
831.0²661.02.0661.049.015.0²2.015.0 LT LT LT LT
1749.0²661.0²831.0831.0
1
²²
1
LT LT LT
LT
b) for the bottom part of the column:
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
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0637
0
kNm
77.10 1 C
464.57943291 mm I y
446.6025866 mm I z
497.149294 mm I t
688.19351706542 mm I w
4
2
f z
w
t h I I
mmh 260
f t mmt f 7.10
6
24
093627638294
7.1026046.6025866mm
mmmmmm I w
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kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
10.4266.426102243
20.15251.158170677.146.6025866/210000
97.149294/808002810
46.6025866
09362763829
2810
46.6025866/21000077.1
²
²
²
²
422
422
4
6
2
422
1
569.06.426102243
/27518.501177 23,
Nmm
mm N mm
M
f W
cr
y y pl LT
LT LT
1²²
1
LT LT LT
LT
²2.015.0 LT LT
LT LT
2733.1150
260
mm
mm
b
h
α
49.0 LT
752.0²569.02.0569.049.015.0²2.015.0 LT LT LT LT
1804.0²569.0²752.0752.0
1
²²
1
LT LT LT
LT
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13.42.2.5Internal factor, yyk , calculation
yyk
a) for the top part of the column:
yy
ycr
Ed
y
mLT my yy C
N
N C C k
1̀
1,
ycr
Ed y
ycr
Ed
y
N
N
N
N
,
,
1
1
839.0 y
kN N Ed 328
kN N l
I E N
fy
y
ycr 33.380295.3802327²
²,
985.0
95.3802327
328000839.01
95.3802327
3280001
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
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myC
0
0
,0
cr
y y pl
M
f W
318.501177 mmW y
0cr M 11 C 02 C
kNm Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
73.2408.240735730
20.15251.1581706146.6025866/210000
97.149294/808002810
46.6025866
88.19351706542
2810
46.6025866/2100001
²
²
²
²
422
422
4
6
2
422
1
757.08.240735730
/27518.501177 23,
0
Nmm
mm N mm
M
f W
cr
y y pl
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4
,,
1 1120.0
TF cr
Ed
z cr
Ed
N
N
N
N C
T cr TF cr N N ,,
²
1
,
2
0
,
T cr
wt T cr
L
I E I G
I N
0 I
4442
0 1.6396915846.602586664.57943291 mmmmmm I I z A I I I z y g z y
497.149294 mm I t
6
88.19351706542 mm I w T cr L ,
m L T cr 81.2,
kN
mm
mmmm N mmmm N
mm
mm N T cr
63.280668.2806625
²2810
88.19351706542/21000097.149294/80800
1.63969158
06.4904 62242
4
2
,
N N Ed 328000
N N N T cr TF cr 68.2806625,,
kN N l
I E N
fz
z z cr 71.3158151.1581706
²
²,
183.0
68.2806625
3280001
51.1581706
3280001120.01120.0 44
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
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1
11
11
183.01120.0757.0
183.01120.0
757.0
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C
a
aC C C
N
N
N
N C
N
N
N
N C
myC
LT y
LT ymymymy
aaC C C
1
1 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
3
,, 63.445717 mmW W yeff yel
677.163.445717
06.4904
328000
10503
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
1997.064.57943291
97.14929411
4
4
mm
mm
I
I a
y
t LT
0mC
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0
ycr
Ed
ycr
Ed my
N
N
N
N C
,,
0, 33.036.079.033.036.021.079.0
kN N l
I E N
fy
y
ycr 33.380295.3802327²
²,
N N Ed 328000
780.095.3802327
32800033.036.079.00,
N
N C my
904.01677.11
1677.1780.01780.0
11 0,0,
LT y
LT y
mymymya
aC C C
mLT C
mLT C
mLT C
1
11,,
2
T cr
Ed
z cr
Ed
LT mymLT
N
N
N
N
aC C
myC mLT C
myC 5.0
LT y
LT y
mymymya
aC C C
11 0,0,
900.095.3802327
23800033.05.036.05.021.079.033.036.021.079.0
,
0, N
N
N
N C
ycr
Ed my
677.1
63.445717
06.4904
328000
10503
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
956.01677.11
1677.1900.01900.0
11 0,0,
LT y
LT y
mymymya
aC C C
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089.1
1
089.1
68.2806625
3280001
51.1581706
3280001
997.0
956.0
11
2
,,
2
mLT
mLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
N
N
N
N
aC C
yyC
y pl
yel
LT pl my
y
my
y
y yyW
W bnC
wC
wwC
,
,2
max2
max2 6.16.1
2)1(1
Rd z pl
Ed z
Rd y pl LT
Ed y
LT LT M
M M
M ab,,
,
,,
,2
05.0
LT b 0
LT
757.0661.031.110
LT C
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041.0
/27596.123381
10000000
/27518.501177749.0
50000000757.0997.05.0
5.05.0
2323
2
,
,
,
,2
0
,,
,
,,
,2
0
mm N mm
Nmm
mm N mm
Nmm
f W
M
f W
M a
M
M
M
M ab
y z pl
Ed z
y y pl LT
Ed y
LT
Rd z pl
Ed z
Rd y pl LT
Ed y
LT LT
5.1124.163.445717
18.5011773
3
,
, mm
mm
W
W w
yel
y pl
y
243.0
1
/27506.4904
32800022
1
mm N mm
N
N
N n
M
Rk
Ed pl
923.0923.0;5956.0max;maxmax
z y
993.0041.0243.0²923.0²904.0124.1
6.1923.0²904.0
124.1
6.12)1124.1(1
yyC
993.0889.018.501177
63.445717
993.0
,
,
3
3
,
,
yy
y pl
yel
yy
y pl
yel
yy
C
W
W C
mm
mm
W
W
C
kN N l
I E N
fy
y
ycr 33.380295.3802327²
²,
yyk
069.1993.0
1
95.3802327
3280001
985.0089.1904.0
1̀
1,
N
N C
N
N C C k
yy
ycr
Ed
y
mLT my yy
b) for the bottom part of the column:
yy
ycr
Ed
y
mLT my yyC
N
N C C k
1̀
1,
8/22/2019 AD Validation Guide 2014 En
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-the terms: mLT C LT b yyC
1
11
,,
2
T cr
Ed
z cr
Ed
LT mymLT
N
N
N
N
aC C
myC )0
780.095.3802327
23800033.036.079.033.036.021.079.0
,
0, N
N
N
N C
ycr
Ed my
839.063.445717
06.4904
328000
10253
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
885.01839.01
1839.0780.01780.0
11 0,0,
LT y
LT y
mymymya
aC C C
1
1
933.0
68.2806625
3280001
51.1581706
3280001
997.0885.0
11
2
,,
2
mLT
mLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
N
N
N
N
aC C
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y pl
yel
LT pl my
y
my
y
y yyW
W bnC
wC
wwC
,
,2
max2
max2 6.16.1
2)1(1
Rd z pl
Ed z
Rd y pl LT
Ed y
LT LT
M
M
M
M ab
,,
,
,,
,2
05.0
LT b 0
LT
757.08.240735730
/27518.501177 23,
0
Nmm
mm N mm
M
f W
cr
y y pl
1804.0²569.0²752.0752.0
1
²²
1
LT LT LT
LT
0095.0
/27596.123381
5000000
/27518.501177804.0
25000000757.0997.05.0
5.05.0
2323
2
,
,
,
,
2
0
,,
,
,,
,
2
0
mm N mm
Nmm
mm N mm
Nmm
f W M
f W
M a
M M
M
M ab
y z pl
Ed z
y y pl LT
Ed y LT
Rd z pl
Ed z
Rd y pl LT
Ed y LT LT
5.1124.163.445717
18.5011773
3
,
, mm
mm
W
W w
yel
y pl
y
243.0
1
/27506.4904328000 22
1
mm N mm N
N N n
M
Rk
Ed pl
923.0923.0;5956.0max;maxmax
z y
997.00095.0243.0²923.0²904.0124.1
6.1923.0²904.0
124.1
6.12)1124.1(1
yyC
997.0889.018.501177
63.445717
997.0
,
,
3
3
,
,
yy
y pl
yel
yy
y pl
yel
yy
C
W
W C
mm
mm
W
W
C
kN N l
I E N
fy
y
ycr 33.380295.3802327²
²,
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yyk
977.0997.0
1
95.3802327
3280001
985.01904.0
1̀
1,
N
N C
N
N C C k
yy
ycr
Ed
y
mLT my yy
13.42.2.6Internal factor, yz k , calculation
y
z
yz
z cr
Ed
y
mz yz w
w
C
N
N C k
6.0
1
1,
mz C )0
765.051.1581706
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
985.0
95.3802327
328000839.01
95.3802327
3280001
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
N l
I E N
fz
z z cr 51.1581706
²
²,
LT pl
z
mz z yz cn
w
C wC
5
2
max2
142)1(1
Rd y pl lt my
Ed y
z
LT LT M C
M ac
,,
,
4
2
0
5
10
997.064.57943291
97.14929411
4
4
mm
mm
I
I a
y
t LT
757.0
8.240735730
/27518.501177 23,
0
Nmm
mm N mm
M
f W
cr
y y pl
923.051.1581706
/27506.4904 22
,
N
mm N mm
N
f A
z cr
y z
Nm M Ed y 50000,
5.1
5.1
5.1536.189.80344
96.1233813
3
,
,
z
z
z el
z pl
z w
w
mm
mm
W
W w
5.1124.163.44571718.501177
3
3
,
, mmmm
W W w
yel
y pl y
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myC
956.0
1677.11
1677.1900.01900.0
1
1 0,0,
LT y
LT y
mymymy
a
aC C C
1749.0²661.0²831.0831.0
1
²²
1
LT LT LT
LT
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
506.05.137823724749.0956.0
50000000
923.05
²757.0
997.010
5
10
4
,,
,
4
2
0
Nmm
Nmm
M C
M ac
Rd y pl lt my
Ed y
z
LT LT
N l
I E N
fz
z z cr 51.1581706
²
²,
765.051.1581706
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
923.0923.0;5956.0max;maxmax
z y
243.0
1
M
Rk
Ed
pl N
N
n
878.0506.0243.05.1
923.0765.014215.11
142)1(1
5
22
5
2
max2
LT pl
z
mz z yz cn
w
C wC
750.0124.1
5.16.0
878.0
1
51.1581706
3280001
985.0765.06.0
1
1,
N
N w
w
C
N
N C k
y
z
yz
z cr
Ed
y
mz yz
885.01839.01
1839.0780.01780.0
11 0,0,
LT y
LT y
mymymya
aC C C
1804.0²569.0²752.0752.0
1
²²
1
LT
LT LT
LT
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Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
254.05.137823724804.0885.0
25000000
923.05
²757.0997.010
5
10
4
,,
,
4
2
0
Nmm
Nmm
M C
M ac
Rd y pl lt my
Ed y
z
LT LT
N l
I E N
fz
z z cr 51.1581706
²
²,
765.051.1581706
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
923.0923.0;5956.0max;maxmax
z y
243.0
1
M
Rk
Ed pl N
N n
0043.1254.0243.0
5.1
923.0765.014215.11
142)1(1
5
22
5
2
max2
LT pl
z
mz z yz cn
w
C wC
656.0124.1
5.16.0
0043.1
1
51.1581706
3280001
985.0765.06.0
1
1,
N
N w
w
C
N
N C k
y
z
yz
z cr
Ed
y
mz yz
13.42.2.7Internal factor, zyk , calculation
a) for the top part of the column:
y
z
zy
ycr
Ed
z mLT my zy
ww
C
N
N C C k
6.011
,
902.0
51.1581706
328000586.01
51.1581706
3280001
1
1
,
,
N
N N
N
N
N
N
N
z cr
Ed z
z cr
Ed
z
586.0 z
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y pl
yel
z
y
LT pl
y
my
y zyW
W
w
wd n
w
C wC
,
,
5
2
max2
6.0142)1(1
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177
23
,, Nmmmm N mm f W M z z Rd z pl 33930039/27596.123381 23
,,
LT d myC mz C
mz C 5.0
908.0
51.1581706
32800033.05.036.05.021.079.0
33.05.036.05.021.079.0,
0,
N
N
N
N C C
z cr
Ed mz mz
301.0
33930039908.0
10000000
5.137823724749.0956.0
50000000
923.01.0
757.0997.02
1.0
2
4
,,
,
,,
,
4
0
Nmm
Nmm
Nmm
Nmm
M C
M
M C
M ad
Rd z pl mz
Ed z
Rd y pl LT my
Ed y
z
LT LT
859.0
6.0
616.018.501177
63.445717
124.1
5.16.06.0
859.0301.0243.0124.1
847.1904.0142)1124.1(1
142)1(1
,
,
3
3
,
,
5
22
5
2
max2
zy
y pl
yel
z
y
zy
y pl
yel
z
y
LT pl y
my
y zy
C
W
W
w
wC
mm
mm
W
W
w
w
d nw
C
wC
zyC myC LT d
588.05.1
124.16.0
859.0
1
95.3802327
3280001
902.0089.1904.06.0
1
1,
N
N w
w
C
N
N C C k
z
y
zy
ycr
Ed
z mLT my zy
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b) for the bottom part of the column:
y
z
zy
ycr
Ed
z mLT my zy
w
w
C
N
N C C k
6.0
1
1,
902.0
51.1581706
328000586.01
51.1581706
3280001
1
1
,
,
N
N N
N
N
N
N N
z cr
Ed z
z cr
Ed
z
586.0 z
y pl
yel
z
y
LT pl
y
my
y zyW
W
w
wd n
w
C wC
,
,
5
2
max2
6.0142)1(1
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
Nmmmm N mm f W M z z Rd z pl 33930039/27596.123381 23
,,
LT d myC mz C
mz C 0
765.051.1581706
32800033.036.079.033.036.079.0
,
0, N
N
N
N C C
z cr
Ed mz mz
090.0
33930039765.0
5000000
5.137823724804.0885.0
25000000
923.01.0
757.0997.02
1.0
2
4
,,
,
,,
,
4
0
Nmm
Nmm
Nmm
Nmm
M C
M
M C
M ad
Rd z pl mz
Ed z
Rd y pl LT my
Ed y
z
LT LT
923.0923.0;5956.0max;maxmax
z y
892.0
6.0
462.018.501177
63.445717
5.1
124.16.06.0
892.0090.0243.0124.1
923.0885.0142)1124.1(1
142)1(1
,
,
3
3
,
,
5
22
5
2
max2
zy
y pl
yel
z
y
zy
y pl
yel
z
y
LT pl
y
my
y zy
C
W
W
w
wC
mm
mm
W
W
w
w
d nw
C wC
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zyC myC LT d
566.05.1
124.16.0
892.0
1
95.3802327
3280001
902.0089.1904.06.0
1
1,
N
N w
w
C
N
N C C k
z
y
zy
ycr
Ed
z mLT my zy
13.42.2.8Internal factor, zz k , calculation
a) for the top part of the column:
zz
z cr
Ed
z mz zz
C
N
N C k
1
1,
z pl
z el
LT pl mz z
mz z
z zz W
W enC
wC
wwC
,
,max
2max
2 6.16.12)1(1
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
LT e myC
787.0/27518.501177749.0956.0
50000000
923.01.0
757.0997.07.1
1.0
7.1
234
,,
,
4
0
mm N mm
Nmm
M C
M ae
Rd y pl lt my
Ed y
z
LT LT
zz C mz C LT e
013.1
651.096.123381
89.80344
013.1243.0787.0847.1765.05.1
6.1847.1765.0
5.1
6.12)15.1(1
6.16.12)1(1
,
,
3
3
,
,
222
2
max2
max2
zz
z pl
z el zz
z pl
z el
pl LT mz
z
mz
z
z zz
C
W W C
mm
mm
W
W
neC w
C w
wC
860.0013.1
1
51.1581706
3280001
902.0765.0
1
1,
N
N C
N
N C k
zz
z cr
Ed
z mz zz
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b) for the bottom part of the column:
zz
z cr
Ed
z mz zz
C
N
N C k
1
1,
z pl
z el LT pl mz
z
mz
z
z zz W W enC
wC
wwC
,
,max
2max
2 6.16.12)1(1
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
LT e myC
396.0/27518.501177804.0885.0
25000000
923.01.0
757.0
997.07.1
1.0
7.1
234
,,
,
4
0
mmmm
Nmm
M C
M ae
Rd y pl lt my
Ed y
z
LT LT
zz C mz C LT e
923.0923.0;5956.0max;maxmax
z y
060.1651.0
96.123381
89.80344
060.1243.0396.0923.0765.05.1
6.1923.0765.0
5.1
6.12)15.1(1
6.16.12)1(1
,
,
3
3
,
,
222
2
max2
max2
zz
z pl
z el
zz
z pl
z el
pl LT mz
z
mz
z
z zz
C
W
W C
mm
mm
W
W
neC w
C w
wC
821.0060.1
1
51.1581706
3280001
902.0765.0
1
1,
N
N C
N
N C k
zz
z cr
Ed
z mz zz
8/22/2019 AD Validation Guide 2014 En
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13.42.2.9Bending and axial compression verification
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rk z
Rd z Ed z
zz
M
Rk y
LT
Rd y Ed y
zy
M
Rk z
Ed
M
Rk z
Rd z Ed z
yz
M
Rk y
LT
Rd y Ed y
yy
M
Rk
y
Ed
M
M M k
M
M M k
N
N
M
M M k
M
M M k
N
N
i y Rk A f N
a) for the top part of the column:
93.025.027.041.0
1
/27596.123381
1010860.0
1
/27518.501177804.0
1050588.0
1
/27506.4904586.0
328000
02.121.052.029.0
1
/27596.123381
1010750.0
1
/27518.501177749.0
1050069.1
1
/27506.4904839.0
328000
23
6
23
6
22
23
6
23
6
22
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 930/1115
b) for the bottom part of the column:
67.012.013.042.0
1
/27596.123381
105821.0
1
/27518.501177804.0
1025566.0
1
/27506.4904586.0
328000
30.197.022.029.0
1
/27596.123381
105656.0
1
/27518.501177804.0
1025977.0
1
/27506.4904839.0
328000
23
6
23
6
22
23
6
23
6
22
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
Finite elements modeling
■
■
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 931/1115
y coefficient corresponding to non-dimensional slenderness y
y
z z
z
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 932/1115
Internal factor, yyk
yyk
Internal factor, yz k
yz k
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 933/1115
Internal factor, zyk
zyk
Internal factor, zz k
zz k
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 934/1115
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 935/1115
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http://slidepdf.com/reader/full/ad-validation-guide-2014-en 936/1115
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http://slidepdf.com/reader/full/ad-validation-guide-2014-en 937/1115
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 938/1115
LT
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 939/1115
13.42.2.10Reference results
a) for the top part of the column:
Result name Result description Reference value
y y y
z z z
yyk yyk
yz k yz k
zyk zyk
zz k zyk
LT
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 940/1115
b) for the bottom part of the column:
Result name Result description Reference value
y y y
z z z
yyk yyk
yz k yz k
zyk zyk
zz k zyk
LT
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 941/1115
13.42.3Calculated results
Result name Result description Value Error
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 942/1115
13.43 EC3 Test 44: Determining lateral torsional buckling parameters for a I-shaped welded built-upbeam considering the load applied on the upper flange
Passed
13.43.1Description
13.43.2Background
13.43.2.1Model description
■
■
■
■
►
►
►
■
■
8/22/2019 AD Validation Guide 2014 En
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Units
Geometry
■■
■
■
■
■
■
■
■
■
■ 0.10 M
Materials properties
■
■
■
Boundary conditions
■
►
►
■
Loading
■
►
►
►
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 944/1115
13.43.2.2Reference results for calculating the elastic critical moment of the cross section
■
8.0142000
113600
Nm
Nm
■
1.1
101428
5000/50
8 6
22
Nmm
mmmm N
M
Lq
■ 2.411.148.014
■ 84.81.182.48
22
■ 738.19126223.06960364.01413364.015.0 2 a
■ 4765.14281556.19240091.01603341.015.0 2 b
■ 0602.09352904.05940757.00900633.01801266.0 2 c
■ 5625.22 cba A
■
2143.42
2 ba B
■ 036.2152.01 d
■ 3.01 e
■
■
0.14773.08
15.0
■ 0.1002.21411 m
■
5363.02
42
110
A
A B Br C
■
065.2675.0425.02 d
■ 37.035.065.02 e
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 945/1115
■
■ 074.1101 C mC
■235.0398.0 1022 C r C
■mm z g 130
■
Nmm z C z C I E
I G L
I
I
L
I E C M g g
z
t
z
w z cr
6
2
2
22
2
2
2
1 1071772.91
13.43.2.3Reference results for calculating the reduction factor for lateral torsional buckling
LT
■ LT
133.11071772.91
/23518.5011776
23,
Nmm
mm N mm
M
f W
cr
y y pl
LT
■
49.0 LT
■
370.1133.12.0133.149.015.02.015.0 22 LT LT LT LT
■
0.1467.0133.137.137.1
11
2222
LT LT LT
LT
Finite elements modeling
■
■
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 946/1115
Finite elements results
C1 parameter
C2 parameter
Elastic critical moment
Reduction factor for lateral torsional buckling
13.43.2.4Reference results
Result name Result description Reference value
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 947/1115
13.43.3Calculated results
Result name Result description Value Error
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 948/1115
13.44 EC3 Test 2: Class section classification and share verification of an IPE300 beam subjected tolinear uniform loading
Passed
13.44.1Description
13.44.2Background
13.44.2.1Model description
■
■
■
■
■
Units
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 949/1115
Materials properties
Boundary conditions
■
►
►
■
13.44.2.2Reference results for calculating the cross section class
■
■
■
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http://slidepdf.com/reader/full/ad-validation-guide-2014-en 950/1115
8/22/2019 AD Validation Guide 2014 En
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276.57.10
45.56
mm
mm
t
c
92.0
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28.892.0*9*9276.57.10
45.56
mm
mm
t
c
13.44.2.3Reference results in calculating the shear resistance Vpl,Rd
0
,
3*
M
y
v
Rd pl
f A
V
■ f w f v t r t t b A A *)*2(**2
■
■
■
■
■
■
■
■
268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmt r t t b A A f w f v
■
■
■ 0 M 10 M
kN MN
f A
V M
y
v
Rd pl 7.4074077.01
3
275*10*68.25
3*
4
0
,
8/22/2019 AD Validation Guide 2014 En
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*72
w
w
t
h
20.11
1*20.1*20.1
0
1 M
M
92.0275
235235
y f
91.9392.0
20.1*72*7201.35
1.7
6.248
w
w
t
h
Finite elements modeling
■
■
■
Finite elements results
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http://slidepdf.com/reader/full/ad-validation-guide-2014-en 954/1115
13.44.2.4Reference results
Result name Result description Reference value
13.44.3Calculated results
Result name Result description Value Error
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 955/1115
13.45 EC3 Test 1: Class section classification and compression verification of an IPE300 column
Passed
13.45.1Description
13.45.2Background
13.45.2.1Model description
■
■
■
■
■
Units
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 956/1115
Materials properties
Boundary conditions
■
►►
■
13.45.2.2Reference results for calculating the cross section class
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 957/1115
1
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 958/1115
276.57.10
45.56
mm
mm
t
c
1
9*9276.57.10
45.56
mm
mm
t
c
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 959/1115
13.45.2.3Reference results in calculating the compressive resistance Nc,Rd
0
,
*
M
y
Rd c
f A N
■
■
■ 0 M 10 M
kN MN f A
N M
y
Rd c 54.1264264535.11
235*10*81.53* 4
0
,
Finite elements modeling■
■
■
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http://slidepdf.com/reader/full/ad-validation-guide-2014-en 960/1115
Finite elements results
13.45.2.4Reference results
Result name Result description Reference value
13.45.3Calculated results
Result name Result description Value Error
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 961/1115
13.46 EC3 Test 7: Class section classification and compression resistance for an IPE600 column
Passed
13.46.1Description
13.46.2Background
13.46.2.1Model description
■
■
■
■
►
■
■
Units
Materials properties
■
■
Boundary conditions
■
►
►
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 962/1115
Loading
■
■
13.46.2.2Reference results for calculating the cross section class
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 963/1115
83.4212
224219600
mm
mmmmmm
t
c
0.1235
y
f
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 964/1115
424283.42 t
c
21.419
2/)22412220(
mm
mmmmmm
t
c
0.1235
y f
9921.4 t
c
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 965/1115
13.46.2.3Reference results for calculating the compression resistance of the cross-section
The buckling factor (k ) and the stress ratio( )
0.11
2
0.4 k
The plate modified slenderness ( p)
754.0
0.40.14.28
12/242192600
4.28
/
mmmmmmmm
k
t b p
The reduction factor ( )
0.1
3055.02
p
p
Effective area
275.1522312242192600939.01156001 mmt b A A weff
Compression resistance of the cross section
N MPamm f A
N M
yeff
Rd c 35775810.1
23575.15223 2
0
,
Work ratio
%3%8.21003577581
100000100
,
N
N
N
N
Rd c
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 966/1115
Finite elements modeling
■
■
■
Finite elements results
Work ratio of the design resistance for uniform compression
13.46.2.4Reference results
Result name Result description Reference value
13.46.3Calculated results
Result name Result description Value Error
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 967/1115
13.47 Verifying shape sheet on S275 beam (TTAD #11731)
Passed
13.47.1Description
13.48 Verifying the calculation results for steel cables (TTAD #11623)
Passed
13.48.1Description
13.49 Generating the shape sheet by system (TTAD #11471)
Passed
13.49.1Description
13.50 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522)
Passed
13.50.1Description
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 968/1115
13.51 EC3 Test 22: Verifying the lateral torsional buckling of a IPE300 beam
Passed
13.51.1Description
13.51.2Background
13.51.2.1Model description
■■
■
■
■
■
Units
Geometrical properties
■
■
■
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 969/1115
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
■
13.51.2.2Buckling in the strong inertia of the profile (along Y-Y)
%100100,
Rd b
Ed
M
M
■
7272014.35
1
014.351.7
6.248
t
cmm
mm
t
c
■
99276.5
1
276.57.10
45.56
t
cmm
mm
t
c
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 970/1115
22150
300
mm
mm
b
h
1
,
M
y y LT
Rd b
f W M
LT
11
22
LT LT LT
LT
2
)2.0(15.0 LT LT LT LT
LT 21.0 LT
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 971/1115
LT
cr
y y LT
M
f W
g g
Z
t z
z
w
w
z
z
z cr z C z C
I E
I G Lk
I
I
k
k
Lk
I E C M 2
2
22
2
2
2
1
■
■
■
■
■
■
■
■
■
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 972/1115
kNm Nmmmm
mmmm N
mmmm N mm
mm
mm
mm
mmmm N M cr
61.1308.13060942480.23044.50057813.1
1080.603/210000²
1012.20/80770)²5000(
1080.603
1059.12
)²5000(
1080.603/210000²13.1
442
442
44
610
442
063.18.130609424
/235104.628 233
Nmm
mm N mm
M
f W
cr
y y LT
156.1063.1)2.0063.1(21.015.0)2.0(15.0 22
LT LT LT LT
1621.0²063.1²156.1156.1
1
²²
1
LT LT LT
LT
Finite elements modeling
■
■
■
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http://slidepdf.com/reader/full/ad-validation-guide-2014-en 973/1115
Finite elements results
Lateral torsional buckling coefficient
Elastic critical moment for lateral-torsional buckling
13.51.2.3Reference results
Result name Result description Reference value
LT
13.51.3Calculated results
Result name Result description Value Error
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 974/1115
13.52 EC3 Test 23: Verifying a IPE400 column for compression, shear, bending moment, buckling,lateral torsional buckling and bending and axial compression
Passed
13.52.1Description
13.52.2Background
13.52.2.1Model description
■
■
■
■
■
■
Units
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 975/1115
Geometrical properties
■
■ 2
8446mm A ■ mmb 180
■ mmt f 5.13■ mmr 21
■ mmt w 6.8
■ mmhw 400
■ 33
, 101156 mmW
yel
■ 33101307 mmW y
■ 33
, 1040.146 mmW z el
■ 33, 10229 mmW z pl
■
■
■
■
Materials properties
■
■
■
Boundary conditions
■
►
►
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 976/1115
Loading
■
■
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 977/1115
CTICM model
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 978/1115
13.52.2.2Cross section Class
727249.38
1
49.386.8
331
t
cmmmm
t c
9950.4
1
50.45.13
47.67
t
cmm
mm
t
c
Class 1
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 979/1115
13.52.2.3Compression verification
%100100,
Rd c
Ed
N
N
0
,
M
y
Rd c
f A N
28446mm A
y f 2/275 mm N f y
0 M 10 M
kN N mm N mm f A
N M
y
Rd c 65.232223226501
/2758446 22
0
,
N N Ed 125000
%100%38.51002322650
125000
,
N
N
N
N
Rd c
Ed
kN N Rd c 2324,
13.52.2.4Shear verification
%100100,
Rd c
Ed
V
V
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 980/1115
0
,
3
M
y
V
Rd c
f A
V
ww f w f V t ht r t t b A A 22
28446mm A
mmb 180
mmt f 5.13
mmr 21mmt w 6.8
mmhw 400
1
22 1.42695.132126.85.1318028446
22
mmmmmmmmmmmmmm
t r t t b A A f w f V
234404006.81 mmmmmmt h ww
22 34401.4269 mmmm AV
y f 2/275 mm N f y
0 M 10 M
N
mm N mm
f A
V M
y
V
Rd c 66.6778101
3
/2751.4269
3
22
0
,
N V Ed 28330
%100%180.410004179.010066.677810
28330100
,
N
N
V
V
Rd c
Ed
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 981/1115
kN V Rd z pl
8.677,,
13.52.2.5Bending moment verification
%100100,
Rd pl
Ed
M
M
0
,
M
y pl
Rd pl
f w M
31307000mmw pl
y f 2/275 mm N f y
0 M 1
0
M
Nmmmm N mm f w
M M
y pl
Rd pl 359425000
1
/2751307000 23
0
,
Nmm M Ed
254970000
%100%938.7010070938.0100359425000
254970000100
,
Nmm
Nmm
M
M
Rd pl
Ed
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 982/1115
kNm M Rd y pl
7.359,,
13.52.2.6Buckling verification
a) over the strong axis of the section, y-y:
α
21.0
Coefficient corresponding to non-dimensional slenderness after Y-Y axis
y y
1122
y y y
y
y
ycr
y
y N
f A
,
*
N
mmmm MPa
L I E N
fz
y
ycr 77.59184729000
1023130210000²²
²2
44
,
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 983/1115
62645.077.5918472
/2758446 22
,
N
mm N mm
N
f A
ycr
y y
740997.062645.02.062645.021.015.0²)2.0(15.0 2 y y y
187968.062645.0740997.0740997.0
112222
y y y
y
y 8796.0 y
b) over the strong axis of the section, z-z
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α
34.0
z z
11
22
z z z
z
z
z cr
y z
N
f A
,
*
15.9cr
N
mm
mmmm N
l
I E N
fy
z z cr 153.1142396
²4890
101318/210000²
²
² 442
,
42588.1153.1142396
/2758446 22
,
mm N mm
N
f A
z cr
y z
72497.142588.12.042588.134.015.0²)2.0(15.0 2 z z z
137096.042588.172497.172497.1
11
2222
z z z
z
z 3711.0 z
8/22/2019 AD Validation Guide 2014 En
http://slidepdf.com/reader/full/ad-validation-guide-2014-en 985/1115
13.52.2.7Lateral torsional buckling verification
a) for the 3m part of the column:
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
66667.097.254
98.169
kNm
kNm
17932.1²66667.0252.066667.0423.0325.0
1
²252.0423.0325.0
11
C
■
■
■
■
■
■
■
Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
2.806585210
33396.22534.303523217932.1101318/210000²
1008.51/808003000
101318
1049
3000
101318/210000²17932.1
²
²
²
²
442
4422
44
610
2
442
1
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LT
cr
y y LT
M
f W
33101307 mmW y
66754.02.806585210
/275101307 233
Nmm
mm N mm
M
f W
cr
y y LT
LT LT
1²²
1
LT LT LT
LT
²2.015.0 LT LT LT LT
α
34.0
80228.0²66754.02.066754.034.015.0 LT
180173.0²66754.0²80228.080228.0
1
²²
1
LT LT LT
LT
LT 7877.0 LT
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b) for the 6m part of the column:
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
²252.0423.0325.0
11
C
0
77.11 C
■■
■
■
■
■
■
Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
406423987
604.302085.75880877.1101318/210000²
1008.51/80800²6000
101318
1049
²6000
101318/210000²77.1
²
²
²
²
442
4422
44
610
2
442
1
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LT
cr
y y LT
M
f W
33101307 mmW y
94040.0406423987
/275101307 233
Nmm
mm N mm
M
f W
cr
y y LT
LT LT
1²²
1
LT LT LT
LT
²2.015.0 LT LT LT LT
α
34.0
06804.1²94040.02.094040.034.015.0 LT
163518.0²94040.0²06804.106804.1
1
²²
1
LT LT LT
LT
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LT 694.0 LT
13.52.2.8Bending and axial compression verification
1
1
,
,,
1
,
,,
1
M
Rk z
Ed z Ed z
yz
M
Rd y
LT
Rd y Ed y
yy
M
Rk y
Ed
M
M M k
M
M M k
N
N
1
1
,
,,
1
,
,,
1
M
Rk z
Ed z Ed z
zz
M
Rd y
LT
Rd y Ed y
zy
M
Rk z
Ed
M
M M k
M
M M k
N
N
0, Ed z M
0, Rd y M 0, Rd z M
62.600.1
61.600.1
1
,
,
1
1
,
,
1
M
Rk y
LT
Ed y
zy
M
Rk z
Ed
M
Rk y
LT
Ed y
yy
M
Rk y
Ed
M
M k
N
N
M
M k
N
N
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13.52.2.9Internal factor, yyk , calculation:
yyk
yy
ycr
Ed
y
mLT my yyC
N
N C C k
1
1,
ycr
Ed y
ycr
Ed
y
N N
N
N
,
,
1
1
87968.0 y
N
mm
mmmm N
l
I E N
fy
y
ycr 773.5918472²9000
41023130/210000²
²
² 42
,
99741.0
773.5918472
12500087968.01
773.5918472
1250001
N N
N
y
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myC
0
LT
C 10
²252.0423.0325.0
11
C
0
77.11 C
66754.02.806585210
/275101307 233
Nmm
mm N mm M
f W cr
y y LT
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88811.066754.077.1110 LT LT C C
4
,,
1 1120.0
TF cr
Ed
z cr
Ed
N
N
N
N C
T cr TF cr N N ,,
²,
2
0
,
T cr
wt T cr
L
I E I G
I
A N
0 I
2
0 g z y z A I I I
28446mm A
0 g z
4444442
0 10244481013181023130 mmmmmm I I z A I I I z y g z y
T cr L ,
m L T cr 6,
N
mm
mmmm N mmmm N
mm
mm N T cr
788.2400423
²6000
1049/2100001008.51/80800
1024448
8446 61022442
44
2
,
N N N T cr TF cr 788.2400423,,
N N z cr 153.1142396,
25505.0
788.2400423
1250001
153.1142396
125000177.120.01120.0 44
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
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1
11
11
25505.01120.088811.0
,,
2
0,
0,0,
4
,,
10
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C
a
aC C C
N N
N N C
myC
LT y
LT y
mymymya
aC C C
11 0,0,
yel Ed
Ed y
yW
A
N
M
,
,
33
, 101156 mmW yel
90119.14101156
8446
125000
1094.25433
26
,
,
mm
mm
N
Nmm
W
A
N
M
yel Ed
Ed y
y
99779.01023130
1008.511144
44
mm
mm
Iy
It a LT
0mC
0
ycr
Ed
my N
N
C ,0, 33.036.021.079.0
N N ycr 773.5918472,
78749.0773.5918472
12500033.0036.079.00,
N
N C my
95619.099779.090119.141
99779.090119.1478749.0178749.0
11 0,0,
LT y
LT y
mymymya
aC C C
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mLT C mLT C
for each of the column parts.
1
11,,
2
T cr
Ed
z cr
Ed
LT mymLT
N
N
N
N
aC C
a) for the 3m part of the column:
LT y
LT y
mymymmya
aC C C
11 0,0,3.
90119.14 y
99779.0 LT a
0mC
66667.097.254
98.169
kNm
kNm
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ycr
Ed my
N
N C
,
0, 33.036.021.079.0
N N ycr 773.5918472,
93256.0773.5918472
12500033.066667.036.066667.021.079.00,
N C my
98609.099779.090119.141
99779.090119.1493256.0193256.0
11 0,0,3,
LT y
LT y
mymymmya
aC C C
N N z cr 153.1142396,
N N T cr 788.2400423,
99779.0 LT
a
05596.1
1
05596.1
788.2400423
1250001
153.1142396
1250001
99779.0²98609.0
11
3,
3,
,,
2
3,3,
mmLT
mmLT
T cr
Ed
z cr
Ed
LT mmymmLT
C
C
N
N
N
N
N
N
N
N
aC C
a) for the 6m part of the column:
LT y
LT y
mymymmya
aC C C
11 0,0,6.
yel Ed
Ed y
yW
A
N
M
,
,
33
, 101156 mmW yel
9353.9101156
8446125000
1098.169 33
26
,
, mmmm
N Nmm
W A
N
M
yel Ed
Ed y y
99779.01023130
1008.5111
44
44
mm
mm
Iy
It a LT
0mC
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0
ycr
Ed my
N
N C
,
0, 33.036.021.079.0
N N ycr 773.5918472,
78749.0773.5918472
12500033.0036.0021.079.00,
N
N C my
94873.099779.09353.91
99779.09353.9
78749.0178749.011 0,0,6,
LT y
LT y
mymymmya
a
C C C
N N z cr 153.1142396,
N N T cr 788.2400423,
1
1
97746.0
788.2400423
1250001
153.1142396
1250001
99779.0
²94873.0
11
6,
6,
,,
2
6,6,
mmLT
mmLT
T cr
Ed
z cr
Ed
LT mmymmLT
C
C
N
N
N
N
N
N
N
N
aC C
1
05596.1
6,
3,
mmLT
mmLT
mLT C
C C
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yyC
y pl
yel
LT pl my
y
my
y
y yyW
W bnC
wC
wwC
,
,2
max
2
max
2 6.16.12)1(1
■ 005.05.0,,
,2
0
,
,
,,
,2
0
Rd y pl LT
Ed y
LT
Rd pl
Ed z
Rd y pl LT
Ed y
LT LT M
M a
M
M
M
M ab
■ 5.1,
, yel
y pl
yW
W w
■ 33
, 101156 mmW yel
■ 33
, 101307 mmW y pl
■ 5.113062.1101156101307 33
33
,
, mmmm
W
W w
yel
y pl y
■
1 M
Rk
Ed pl N
N n
■ N mm N mm f A
N N M
y
Rd c Rk 23226501
/2758446 22
0
,
■ 05382.0
1
2322650
125000
1
N
N
N
N n
M
Rk
Ed pl
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■ z y ;maxmax
■ 62645.0 y
■ 42504.1 z
■ 42504.142504.1;62645.0max;maxmax z y
■ 95619.0myC
98511.0
005382.0²42504.1²95619.013062.1
6.142504.1²95619.0
13062.1
6.12)113062.1(1
yyC
88447.0101307
10115698511.0
33
33
,
,
mm
mm
W
W C
y pl
yel
yy
98902.098511.0
1
773.5918472
1250001
99741.0195619.0
1
1
04409.198511.0
1
773.5918472
1250001
99741.005569.195619.0
1
1
,
6,6,
,
3,3`,
N
N C
N
N C C k
N
N C
N
N C C k
k
yy
ycr
Ed
y
mmLT mym yy
yy
ycr
Ed
y
mmLT mym yy
yy
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13.52.2.10Internal factor, zyk , calculation:
zyk
z
y
zy
ycr
Ed
z mLT my zy
w
w
C
N
N C C k
6.0
1
1,
z cr
Ed z
z cr
Ed
z
N N
N
N
,
,
1
1
37096.0 z
N
mm
mmmm N
l
I E N
fy
z z cr 153.1142396
²4890
101318/210000²
²
² 442
,
92826.0
153.114239612500037096.01
1533.1142396
1250001
N N
N
N
z
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myC
1
11
11
208.01120.088811.0
,,
2
0,
0,0,
4
,,
10
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C
a
aC C C
N
N
N
N C
90119.14101156
8446
125000
1094.25433
26
,
,
mm
mm
N
Nmm
W
A
N
M
yel Ed
Ed y
y
99779.01023130
1008.5111
44
44
mm
mm
Iy
It a LT
0mC
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0
78749.033.036.021.079.0,
0, ycr
Ed my
N
N C
95619.01
1 0,0,
LT y
LT y
mymymya
aC C C
1
05596.1
116,
3,
,,
2
mmLT
mmLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
aC C
zyC
y pl
yel
z
y
LT pl
y
my
y zyW
W
w
wd n
w
C wC
,
,
5
2
max
2
6.0142)1(1
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001.0
21.0
2,,
,
4
0
,,
,
,,
,
4
0
Rd y pl LT my
Ed y
z
LT
Rd z pl mz
Ed z
Rd y pl LT my
Ed y
z
LT LT M C
M a
M C
M
M C
M ad
5.113062.1101156101307
33
33
,
,
mmmm
W W w
yel
y pl
y
5.1,
, z el
z pl
z W
W w
33
, 1040.146 mmW z el
33
, 10229 mmW z pl
5.15.1
564.1
1040.146
1022933
33
,
,
z
z
z el
z pl
z
ww
mm
mm
W
W w
1 M
Rk
Ed pl N
N n
05382.0
1
2322650
125000
1
N
N
N
N n
M
Rk
Ed pl
42504.142504.1;62645.0max;maxmax z y
90887.0005382.013062.1
42504.198609.0142)113062.1(1
5
22
zyC
46073.0101307
101156
5.1
13062.16.06.0
33
33
,
,
mm
mm
W
W
w
w
y pl
yel
z
y
46073.06.090887.0142)1(1,
,
5
2
max
2
y pl
yel
z
y
LT pl
y
my
y zyW
W
w
wd n
w
C wC
56307.0
5.1
13062.16.0
90887.0
1
773.5918472
1250001
923830.005569.198609.06.0
1
1,
N
N w
w
C
N
N C C k
z
y
zy
ycr
Ed
z mLT my zy
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50752.05.1
13062.16.0
98511.0
1
773.5918472
1250001
99741.0195619.0
6.01
1
52306.0
5.1
13062.16.0
90887.0
1
773.59184721250001
99741.005569.195619.0
6.01
1
,
6,6,
,
3,3`,
N
N
w
w
C
N
N C C k
N N
w
w
C
N
N C C k
k
z
y
zy
z cr
Ed
y
mmLT mym zy
z
y
zy
z cr
Ed
y
mmLT mym zy
zy
The bending and axial compression verifications are:
-for the 3m column part:
160789.046281.014508.0
1
35942500080173.0
25497000052306.0
1
232265037096.0
125000
198501.092383.006119.0
1
35942500080173.0
25497000004409.1
1
232265087968.0
125000
1
,
,
1
1
,
,
1
Nmm
Nmm
N
N
M M k
N N
Nmm
Nmm
N
N
M
M k
N
N
M
Rk y
LT
Ed y zy
M
Rk z
Ed
M
Rk y
LT
Ed y
yy
M
Rk y
Ed
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-for the 6m column part:
152073.037565.014508.0
135942500063518.0
16898000050752.0
1232265037096.0
125000
179322.073204.006118.0
1
35942500063518.0
16898000098902.0
1
232265087968.0
125000
1
,
,
1
1
,
,
1
Nmm
Nmm
N
N
M
M k
N
N
Nmm
Nmm
N
N
M
M k
N
N
M
Rk y
LT
Ed y
zy
M
Rk z
Ed
M
Rk y
LT
Ed y
yy
M
Rk y
Ed
Finite elements modeling
■
■
■
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Ratio of the design normal force to design compresion resistance
Ratio of the design share force to design share resistance
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Ratio of the design share force to design share resistance
y coefficient corresponding to non-dimensional slenderness y
y
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z z
z
Internal factor, yyk
yyk
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Internal factor, yz k
yz k
13.52.2.11Reference results
Result name Result description Reference value
y y y
z z z
yyk yyk
yy
k yy
k
zyk zyk
zyk zyk
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13.52.3Calculated results
Result name Result description Value Error
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13.53 EC3 Test 16: Verifying a simply supported rectangular hollow section beam subjected tobiaxial bending
Passed
13.53.1Description
13.53.2Background
13.53.2.1Model description
■
■
■
■
►
►
■
■
Units
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Geometry
■
■
■
■
■
■
■
■
■ 0.10 M
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
►
■
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13.53.2.2Reference results for calculating the beam subjected to bi-axial bending
Cross section class
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2510
10215230022
mm
mmmmmm
t
t r b
t
c
0.1235
y f
727225 t
c
Determining the design plastic moment resistance
■
Rd pl
Ed
N
N n
, as
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■
5.05.02
ww a A
t b Aa
■
3678.05.02
f f a A
t h Aa
■
5.05.02
ww a
A
t b Aa
■
Rd y pl
w
Rd y pl
Rd y N M a
n M M ,,
,,
,,5.01
1
Nmm MPamm f W
M M M
y y pl
Rd y pl Rd y N
43
0
,
,,,, 10224660.1
235956000
■
Rd z pl
f
Rd z pl
Rd z N M a
n M M ,,
,,
,,5.01
1
Nmm MPamm f W
M M M
y z pl
Rd z pl Rd z N
43
0
,
,,,, 105.169430.1
235721000
Verifying for bi-axial bending
0.1,,
,
,,
,
Rd z N
Ed z
Rd y N
Ed y
M
M
M
M
■
66.1613.11
66.12
n
■
Nmmmmmm N Lq
M Ed y
4222
, 1031258
5000/10
8
1
Nmmmmmm N Lq
M Ed z
4222
, 1031258
5000/10
8
2
■
0.1098.00604.00378.0105.16943
103125
1022466
10312566.1
4
466.1
4
4
Nmm
Nmm
Nmm
Nmm
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%8.91000.1
098.0
Verifying for simple bending about y-y axis
0.1139.01022466
1031254
4
,,
,
Nmm
Nmm
M
M
Rd y pl
Ed y
%9.131000.1
139.0
Verifying for simple bending about z-z axis
0.11844.0105.16943
1031254
4
,,
,
Nmm
Nmm M
M
Rd z pl
Ed z
%44.181000.1
1844.0
Finite elements modeling
■
■
■
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Finite elements results
Work ratio of the design resistance for bending
13.53.2.3Reference results
Result name Result description Reference value
13.53.3Calculated results
Result name Result description Value Error
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13.54 EC3 Test 31: Verifying IPE450 column fixed on base subjected to axial compression andbending moment, both applied on top
Passed
13.54.1Description
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13.55 EC3 Test 20: Verifying the buckling resistance of a RC3020100 column
Passed
13.55.1Description
13.55.2Background
13.55.2.1Model description
■■
■
■
■
■
Units
Geometrical properties
■
■
■
Materials properties
■
■
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Boundary conditions
■
►
►
■■
Loading
■
■
13.55.2.2Buckling in the strong inertia of the profile (along Y-Y)
%100100,
Rd b
Ed
N
N
1
,
M
y
Rd b
f A N
11
22
cr
y
N
f A*
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N
mm
mm MPa
L
I E N
fy
z cr 943.2449625
10000
1011819210000²
²
²2
44
954.0943.2449625
/2359490 22
N
mm N mm
N
f A
cr
y
²)2.0(15.0
034.1954.02.0954.021.015.0²)2.0(15.0 2
1698.0954.0034.1034.1
112222
1 M 11 M
N mm N mm
N Rd b 7.15566441
/2359490698.0 22
,
N N Ed 200000
%848.121007.1556644
200000
100, N
N
N
N
Rd b
Ed
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13.55.2.3Buckling in the weak inertia of the profile (along Z-Z)
%100100,
Rd b
Ed
N N
1
,
M
y
Rd b
f A N
11
22
cr
y
N
f A*
N
mm
mm MPa
L
I E N
fy
z cr 905.1301188
10000
106278210000²
²
²2
44
309.1905.1301188
/2359490 22
N
mm N mm
N
f A
cr
y
²)2.0(15.0
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473.1309.12.0309.121.015.0²)2.0(15.0 2
1465.0309.1473.1473.1
11
2222
1 M 11 M
N mm N mm
N Rd b 75.10370191
/2359490465.0 22
,
N N Ed 200000
%286.1910075.1037019
200000100
, N
N
N
N
Rd b
Ed
Finite elements modeling
■
■
■
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Finite elements results
Coefficient corresponding to non-dimensional slenderness after Y-Y axis
Coefficient corresponding to non-dimensional slenderness after Z-Z axis
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Ratio of the design normal force to design buckling resistance (strong inertia)
Ratio of the design normal force to design buckling resistance (weak inertia)
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13.55.2.4Reference results
Result name Result description Reference value
y
z
13.55.3Calculated results
Result name Result description Value Error
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13.56 EC3 Test 41: Determining lateral torsional buckling parameters for a I-shaped laminated beamconsidering the load applied on the lower flange
Passed
13.56.1Description
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13.57 EC3 test 11: Cross section classification and compression resistance verification of arectangular hollow section column
Passed
13.57.1Description
13.57.2Background
13.57.2.1Model description
■
■
■
■
►
■
■
Units
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Geometry
■
■
■
■
■
■
■
Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
■
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13.57.2.2Reference results for calculating the cross section class
2510
10215230022
mm
mmmmmm
t
t r h
t
c
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0.1235
y f
333325 t
c
13.57.2.3Reference results for calculating the compression resistance of the cross-section
Compression resistance of the cross section
N MPamm f A N M
y Rd c 2230150
0.12359490
2
0
,
Work ratio
%48.41002230150
100000100
,
N
N
N
N
Rd c
Finite elements modeling
■
■
■
Finite elements results
Work ratio of the design resistance for uniform compression
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13.57.2.4Reference results
Result name Result description Reference value
13.57.3Calculated results
Result name Result description Value Error
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13.58 EC3 Test 36: Verifying RHS300x150x9H class 1 simply supported beam, loaded with centriccompression, uniform linear horizontal efforts and a vertical punctual load in the middle
Passed
13.58.1Description
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13.59 EC3 Test 27: Verifying an user defined I section class 3 beam simply supported with adisplacement restraint
Passed
13.59.1Description
13.59.2Background
13.59.2.1Model description
■
■
■
■
■
■
Units
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Geometrical properties
■
■ 2
14264mm A ■ mmb 220
■ mmt f 19■ mmr 0
■ mmt w 12
■ mmhw 530
■ 3
, 11.2509773 mmW yel
■ 300.2862172 mmW y
■ 3
, 41.307177 mmW z el
■ 3, 00.477512 mmW z pl
■ 467.665089874 mm I y
■ 467.33789514 mm I z
■ 473.1269555 mm I t
■ 667.6662201162989 mm I w
Materials properties
■
■
■
Boundary conditions
■
►
►
■
►
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Loading
■
■
■
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13.59.2.2Cross section Class
194.246
94.246
sup
inf Mpa
Mpa
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1235
235235
y f
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4242413838
1
4112
192530
t
cmm
mmmm
t
c
91957.4
1
47.519
2
12220
t
c
t
c
13.59.2.3Buckling verification
a) over the strong axis of the section, y-y:
α
34.0
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y y
11
22
y y y
y
y
ycr
y
y N
f A
,
*
214264mm A 467.665089874 mm I y
kN N
mm
mmmm N
l
I E N
fy
y
ycr 06.5513921.55139061²5000
67665089874./210000²
²
² 42
,
247.021.55139061
/23514264 22
,
N
mm N mm
N
f A
ycr
y y
538.0247.02.0247.034.015.0²)2.0(15.0 2 y y y
984.0
1
984.0247.0538.0538.0
11
2222
y
y
y y y
y
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b) over the weak axis of the section, z-z
α
49.0
z z
1122
z z z
z
z
z cr
y z
N
f A
,
*
ml fz 50.2
467.33789514 mm I z
214264mm A
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kN N
mm
mmmm N
l
I E N
fz
z z cr 235.1120519.11205235
²2500
67.33789514/210000²
²
² 42
,
547.019.11205235
/23514264 22
,
N
mm N mm
N
f A
z cr
y z
735.0547.02.0547.049.015.0²)2.0(15.0 2 z z z
816.0
1
819.0547.0735.0735.0
11
2222
z
z
z z z
z
13.59.2.4Lateral torsional buckling verification
25.0
1025.318
²2500/10000
8
²3
Nm
mmm N
M
Lq
z
t
z
w z cr
I E
I G L
I
I
L
I E C M
²
²
²
²1
0
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31.11 C
467.665089874 mm I y
467.33789514 mm I z
2/210000 mm N E
473.1269555 mm I t
667.6662201162989 mm I w
2
/80800 mm N G mm L 2500
3
, 11.2509773 mmW yel
Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
4001163141
58.27219.1120523531.167.33789514/210000
73.1269555/808002500
67.33789514
1001163.22
2500
67.33789514/21000031.1
²
²
²
²
422
422
4
611
2
422
1
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384.04001163141
/23511.2509773 23,
Nmm
mm N mm
M
f W
cr
y yeff LT
LT LT
36
1081.74001163141
1025.31
Nmm
M
M
cr
Ed
0156.000781.0
0156.0125.0530
2203.03.0
04.01081.7
20.0384.0
2
0,
20,0,
2
0,3
LT
cr
Ed
LT LT
LT
cr
Ed
cr
Ed
LT
M
M
h
b
M
M
M
M
2
0, LT
cr
Ed
M
M
1 LT
13.59.2.5Internal factor, yyk , calculation
yyk
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ycr
Ed
y
mLT my yy
N
N C C k
,
1
ycr
Ed y
ycr
Ed
y
N
N N
N
,
,
1
1
984.0 y
kN N Ed 3700
N
mm
mmmm N
l
I E N
fy
y
ycr 21.55139061²5000
67665089874./210000²
²
² 42
,
999.0
21.55130961
3700000984.01
21.55130961
37000001
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
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myC
0
0
,0
cr
y yeff
M
f W
34
max
, 11.2509773265
67.665089874mm
mm
mm
z
I W
y
yel
0cr M 11 C 02 C
Nmm
mm N mmmm N
mmmm N mm
mm
mm
mm
mmmm N
I E
I G L
I
I
L
I E C M
z
t
z
w z cr
3054323008
58.27219.11205235167.33789514/210000
73.1269555/808002500
67.33789514
1001163.22
2500
67.33789514/2100001
²
²
²
²
422
422
4
611
2
422
10,
439.03054323008
/23511.2509773 23
0,
,0
Nmm
mm N mm
M
f W
cr
y yeff
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4
,,
1 1120.0
TF cr
Ed
z cr
Ed
N
N
N
N C
T cr TF cr N N ,,
²
1
,
2
2
0
,
T cr
wt T cr
L
I E I G
i N
0491.02
0
2
0
222
0 z yiii z y
473.1269555 mm I t
667.6662201162989 mm I w
T cr L ,
m L T cr 50.2,
mm
mmmm N mmmm N
mm
mm N T cr
13
62242
4
2
,
10696.1
²2500
67.6662201162989/21000073.1269555/80800
0491.0
1
N N Ed 3700000
N N N T cr TF cr
13
,, 10696.1
N
mm
mmmm N
l
I E N
fz
z z cr 19.11205235
²2500
67.33789514/210000²
²
² 42
,
181.0
10696.1
37000001
19.11205235
37000001120.01120.0 4
134
,,
1
N
N
N
N
N
N
N
N C
TF cr
Ed
z cr
Ed
1
11
11
181.01120.0439.0
181.01120.0
439.0
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
T cr
Ed
z cr
Ed
LT mymLT
mz mz
LT y
LT y
mymymy
TF cr
Ed
z cr
Ed
TF cr
Ed
z cr
Ed
N
N
N
N
aC C
C C
a
aC C C
N
N
N
N C
N
N
N
N C
myC
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LT y
LT y
mymymya
aC C C
11 0,0,
yeff
eff
Ed
Ed y
y
W
A
N
M
,
,
3
,, 11.2509773 mmW W yeff yel
048.011.2509773
14264
3700000
1025.313
26
,
,
mm
mm
N
Nmm
W
A
N
M
yeff
eff
Ed
Ed y
y
998.067.665089874
73.126955511
4
4
mm
mm
I
I a
y
t LT
0mC
ycr
Ed my
N
N C
,
0, 03.01
N
mm
mmmm N
l
I E N
fy
y
ycr 21.55139061²5000
67665089874./210000²
²
² 42
,
N N Ed 3700000
002.121.55139061
370000003.010,
N
N C my
002.1100481
1048.0002.11002.1
11 0,0,
LT y
LT y
mymymya
aC C C
mLT C
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T cr
Ed
z cr
Ed
LT mymLT
N
N
N
N
aC C
,,
2
11
m z
0001396.0
999.0
21.55139061
37000001
1025.312500
0001396.067665089874./2100001
11
62
422
,,
2
2
0,
N
N
Nmmmm
mmmm N
N
N
M L
I E C
ycr
Ed
Ed y
y
my
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999.0998.0219.01
998.0219.0999.01999.0
11 0,0,
LT y
LT y
mymymya
aC C C
377.1
10696.137000001
19.1120523537000001
998.0²999.0
11 13
,,
2
N N
N N
N
N
N
N
aC C
T cr
Ed
z cr
Ed
LT
mymLT
yyk
476.1
21.55130961
37000001
1377.1002.1
1,
N
N
N
N C C k
ycr
Ed
y
mLT my yy
13.59.2.6Internal factor, yz k , calculation
z cr
Ed
y
mz yz
N
N C k
,
1
010.119.11205235
370000003.0103.01
,0,
N
N
N
N C C
z cr
Ed
mz mz
506.1
19.11205235
37000001
1776.0
1,
N
N
N
N C k
z cr
Ed
y
mz yz
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13.59.2.7Internal factor, zyk , calculation
ycr
Ed
z mLT my zy
N
N C C k
,1
917.0
19.11205235
3700000816.01
19.11205235
37000001
1
1
,
,
N
N N
N
N
N
N
N
z cr
Ed z
z cr
Ed
z
816.0 z
355.1
21.55130961
37000001
917.0377.1002.1
1,
N
N
N
N C C k
ycr
Ed
z mLT my zy
13.59.2.8Internal factor, zz k , calculation
383.1
19.11205235
37000001
917.0776.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
13.59.2.9Bending and axial compression verification
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rk z
Rd z Ed z
zz
M
Rk y
LT
Rd y Ed y
zy
M
Rk z
Ed
M
Rk z
Rd z Ed z
yz
M
Rk y
LT
Rd y Ed y
yy
M
Rk y
Ed
M
M M k
M
M M k
N
N
M
M M k
M
M M k
N
N
i y Rk A f N
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72.130.007.035.1
1
/23541.307177
1062.15383.1
1
/23511.25097731
1025.31355.1
1
/23514264816.0
3700000
53.033.008.012.0
1/23541.307177
1062.15506.1
1
/23511.25097731
1025.31476.1
1
/23514264984.0
3700000
23
6
23
6
22
23
6
23
6
22
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
mm N mm
Nmm
mm N mm
Nmm
mm N mm
N
Finite elements modeling
■
■
■
y coefficient corresponding to non-dimensional slenderness y
y
z coefficient corresponding to non-dimensional slenderness z
z
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Internal factor, yyk
yyk
Internal factor, yz k
yz k
Internal factor, zyk
zyk
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Internal factor, zz k
zz k
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13.59.2.10Reference results
Result name Result description Reference value
y y y
z z z
yyk yyk
yz k yz k
zyk zyk
zz k zz k
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13.59.3Calculated results
Result name Result description Value Error
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13.60 EC3 Test 37: Verifying RHS350x150x8.5H class 3 column, loaded with centric compression,punctual lateral load and bending moment, all applied to the top of the column
Passed
13.60.1Description
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13.61 EC3 Test 5: Class section classification and combined axial force with bending momentverification of an IPE300 column
Passed
13.61.1Description
13.61.2Background
13.61.2.1Model description
■
■
■
■
■
Units
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Materials properties
Boundary conditions
■
►
►
■
13.61.2.2Reference results for calculating the cross section class
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Finite elements modeling
■
■
■
Finite elements results
13.61.2.3Reference results
Result name Result description Reference value
13.61.3Calculated results
Result name Result description Value Error
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13.62 Verifying the "Shape sheet" command for elements which were excluded from the specializedcalculation (TTAD #12389)
Passed
13.62.1Description
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13.63 EC3 test 4: Class section classification and bending moment verification of an IPE300 column
Passed
13.63.1Description
13.63.2Background
13.63.2.1Model description
■
■
■
■
■
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Units
Materials properties
Boundary conditions
■
►
►
■
13.63.2.2Reference results for calculating the cross section class
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Finite elements results
Combined oblique bending
13.63.2.4Reference results
Result name Result description Reference value
13.63.3Calculated results
Result name Result description Value Error
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14 Timber design
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14.1 EC5: Verifying a timber purlin subjected to oblique bending
Passed
14.1.1Description
14.1.2Background
14.1.2.1Model description
■
■
■
■
■
■
■
Units
Geometry
■
■
■
■
■ 3
22
000666.06
20.01.0
6m
hbW y
■
322
000333.06
20.01.0
6 m
hb
W z
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Materials properties
■
■
Boundary conditions
■
►
►
■
Loading
■
►
►
■
14.1.2.2Reference results in calculating the timber purlin subjected to oblique bending
Reference solution for ultimate limit state verification
■
■
■
■
■
■
Pam
mm
N
W
Lq
W
M
y
z
y
y 6
3
222
102814.8000666.08
5.392.3601
8
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■
Pam
mm
N
W
Lq
W
M
z
y
z
z 6
3
222
100638.5000333.08
5.322.1101
8
■
Pak k k
f h sys
M
k m
66mod, 10615.160.10.1
3.1
9.01024
■
1max
,,
,,
,,
,,
,,
,,
,,
,,
d z m
d z m
d z m
d z m
m
d ym
d ym
m
d ym
d ym
f
f k
f k
f
Finite elements modeling
■
■
■
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Stress SMy diagram
Stress SMz diagram
Maximum work ratio for strength verification
14.1.2.3Reference results
Result name Result description Reference value
14.1.3Calculated results
Result name Result description Value Error
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14.2 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending
Passed
14.2.1Description
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14.3 EC5: Verifying a C24 timber beam subjected to shear force
Passed
14.3.1Description
14.3.2Background
of
14.3.2.1Model description
■
■
■
■
■
■
Units
Geometry
■
■
■
■
■
Materials properties
■
■
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Boundary conditions
■
►
►
■
Loading
■
►
■
14.3.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads
Reference solution for ultimate limit state verification
■
■
■
■
■
■
■
Pa
mm
N
hb
F k
eff eff
d v f 6,10485075.0
225.005025.0
25.36562
3
■
Pa Pak
f M
k v
66mod, 10538.1
3.1
8.0105.2
■
0.1,
d v
d
f
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Finite elements modeling
■
■
■
Shear force, Fz, diagram
Design shear stress diagram
Shear strength work ratio diagram
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14.3.2.3Reference results
Result name Result description Reference value
14.3.3Calculated results
Result name Result description Value Error
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■
62.602.0
5.31121212
m
m
h
l m
h
l g c y
Relative slenderness ratios
■
056.21074.0
102124.12110
6
05,0
,0,
,
Pa
Pa
E
f k c z z rel
■
028.11074.0
102162.60
, 10
6
050
,0,
,
Pa
Pa
E
f k c y
yrel
■
056.2,max ,,max, yrel z rel rel
Instability factors
2
,
2,
1
z rel z z
z c
k k k
2
,
2,
1
yrel y y
yc
k k k
Reference solution for ultimate limit state verification
■
Pa
m
N
A
N 675000
02.0
135002
■
Pam
mm
N
W
Lq
W
M
y
z
y
y 6
3
222
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■
Pam
mm
N
W
Lq
W
M
z
y
z
z 6
3
222
100638.5000333.08
50.322.1101
8
■
■
■
■
■
Pak
f M
k c
66mod,0, 10769.17
3.1
1.11021
■
Pak k k
f h sys
M
k m
66mod, 10308.200.10.1
3.1
1.11024
■
1max
,,
,,
,,
,,
,0,,
,0,
,,
,,
,,
,,
,0,,
,0,
d z m
d z m
d ym
d ym
md c z c
d c
d z m
d z m
m
d ym
d ym
d c yc
d c
f f k f k
f k
f f k
Finite elements modeling
■
■
■
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Instability factor, kcy
Instability factor, kcz
Maximum work ratio for stability verification
14.10.2.3Reference results
Result name Result description Reference value
14.10.3Calculated results
Result name Result description Value Error
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14.11 EC5: Verifying a timber column subjected to tensile forces
Passed
14.11.1Description
14.11.2Background
14.11.2.1Model description
■
■
■
Units
Geometry
■
■
■
■
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Materials properties
■
■
■
Boundary conditions
■
■
Loading
■
■
14.11.2.2Reference results in calculating the timber column subjected to tension force
Reference solution
■
■
■
3.1
150 2.0
h
■
A
N
■
h
M
k t k k
f
mod,0,
■
0.1,0,
,0, d t
d t
f
Finite elements modeling
■
■
■
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Work ratio SFx diagram
14.11.2.3Reference results
Result name Result description Reference value
14.11.3Calculated results
Result name Result description Value Error
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14.12 Verifying the timber elements shape sheet (TTAD #12337)
Passed
14.12.1Description
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14.13 EC5: Verifying a timber beam subjected to simple bending
Passed
14.13.1Description
14.13.2Background
14.13.2.1Model description
■
■
■
■
■
■
■
■
Units
Geometry
■
■
■
■
■
■
322
0005.06
20.0075.0
6m
hbW y
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Materials properties
■
■
■
Boundary conditions
■
►
►
■
Loading
■
■
14.13.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads
Reference solution for ultimate limit state verification
■
■
■
■
■
■
Pahb
Lq
W
M
y
y 6
2
2
2
2
104039.72.0075.08
5.44625.16
8
6
■
Pak k k
f h sys
M
k m
66mod, 10769.140.10.1
3.1
8.01024
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■
0.1,
, d m
d m
f
■
0.1,
, d m
d m
m f
k
Reference solution for serviceability limit state verification
300)(
LQwinst
125
Lw fin
200,
Lw finnet
■
8.600)(00749.0)(
LQwmQw inst inst
■
45.45000999.0
Lwmd w
inst CQinst
■
6.0def k
95.157800285.000475.06.06.0
Lwmmd w creepQP creep
■
47.35001284.000285.000999.0
Lwmmmwww fincreepinst fin
■
47.35001284.0001284.0 ,,
Lwmmmwww finnet c fin finnet
Finite elements modeling
■
■
■
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Work ratio diagram
14.13.2.3Reference results
Result name Result description Reference value
14.13.3Calculated results
Result name Result description Value Error
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14.14 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending andaxial compression
Passed
14.14.1Description
14.14.2Background
14.14.2.1Model description
■
■
■
■
■
■
■
Units
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Geometry
■
■
■
■
■
322
000333.06
20.005.0
6m
hbW y
Materials properties
■
■
■
■
■
Boundary conditions
■
►
►
■
Loading
■
►
►
■
14.14.2.2Reference results in calculating the timber beam subjected to combined stresses
Slenderness ratios
■
4.34605.0
51121212
m
m
b
l m
b
l g c z
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1205.11505.0
667.11121212
m
m
b
l m
b
l g c z
■
6.862.0
51121212
m
m
h
l m
h
l g c y
Relative slenderness ratios
■
958.11074.0
10215.115
,10
6
050
,0,
,
Pa
Pa
E
f k c z z rel
■
468.11074.0
10216.86
,10
6
050
,0,
,
Pa
Pa
E
f k c y
yrel
■
958.1,max ,,max, yrel z rel rel
Relative slenderness for bending
■
mmmhl l eff 9.42.020.59.029.0
■
Pamm
Pam
l h
E b
ef
61022
05.0
2
10724.149.42.0
1074.005.078.078.0
■
277.110724.14
10246
6
,
,
Pa
Pa f
crit m
k m
Instability factors
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2
,
2,
1
z rel z z
z c
k k
k
2
,
2,
1
yrel y y
yc
k k k
Reference solution for ultimate limit state verification
■
■
■
■
■
■
Pa
m N
A N 219122
101022.2191
23
■
Pak
f M
k c
66mod,0, 10538.14
3.1
9.01021
■
Pam
mm
N
W
Lq
W
M
y y
y 6
3
222
10213.8000333.08
00.515.875
8
■
Pak k k
f h sys
M
k m
66mod, 10615.160.10.1
3.1
9.01024
■
1,0,,
,
2
,,
,
d c z c
d c
d ymcrit
d m
f k f k
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Finite elements modeling
■
■
■
Stress SFx diagram
Stress SMy diagram
Lateral-torsional stability work ratio
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14.14.2.3Reference results
Result name Result description Reference value
14.14.3Calculated results
Result name Result description Value Error