ad validation guide 2014 en

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



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



TABLE OF CONTENTS

1 FINITE ELEMENT METHOD .................................................................................................23



2 CAD, RENDERING AND VISUALIZATION ........................................................................ 335



3 CLIMATIC GENERATOR ....................................................................................................343



4 COMBINATIONS................................................................................................................. 357



5 CONCRETE DESIGN ..........................................................................................................365



6 GENERAL APPLICATION.................................................................................................. 679

7 IMPORT / EXPORT ............................................................................................................. 685



8 JOINT DESIGN....................................................................................................................689

9 MESH...................................................................................................................................691

10 OLD CLIMATIC GENERATOR............................................................................................693

11 REPORTS GENERATOR....................................................................................................695



12 SEISMIC ANALYSIS ........................................................................................................... 703



13 STEEL DESIGN...................................................................................................................709



14 TIMBER DESIGN...............................................................................................................1071

15 XML TEMPLATE FILES ....................................................................................................1115

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1 Finite element method



1.1 Slender beam with variable section (fixed-free) (01-0004SDLLB_FEM)

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1.1.1 Description

1.1.2 Background

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1.1.2.3 Theoretical results

Solver Result name Result description Reference value



1.1.3 Calculated results

Result name Result description Value Error



1.2 Thin lozenge-shaped plate fixed on one side (alpha = 15 °) (01-0008SDLSB_FEM)

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1.2.1 Description

1.2.2 Background


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1.2.2.2 Eigen modes frequencies function by angle

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1.3 Bimetallic: Fixed beams connected to a stiff element (01-0026SSLLB_FEM)

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1.3.1 Description

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1.3.2.3 Vertical reaction at A and C

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1.3.2.4 Bending moment at A and C

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1.4 Cantilever rectangular plate (01-0001SSLSB_FEM)

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1.4.1 Description

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1.4.2.2 Displacement of the model in the linear elastic range

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1.5.1 Description

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1.5.2.2 Eigen mode frequencies relative to the angle

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1.6.1 Description

1.6.2 Background


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1.6.2.2 Eigen mode frequencies relative to the angle

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Eigen mode shapes







1.7 Portal frame with lateral connections (01-0030SSLLB_FEM)

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1.7.1 Description

1.7.2 Background


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1.7.2.2 Displacements at A

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1.7.2.3 Moments in A

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1.8 Circular plate under uniform load (01-0003SSLSB_FEM)

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1.8.1 Description

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1.8.2.2 Vertical displacement of the model at the center of the plate

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1.9 Thin circular ring fixed in two points (01-0006SDLLB_FEM)

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1.9.1 Description

1.9.2 Background


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1.9.2.3 Theoretic results






1.10 Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM)

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1.10.1Description

1.10.2Background

1.10.2.1Model description

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1.10.2.2Critical load on node 5

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1.10.2.3Theoretical results


1.10.3Calculated results




1.11 Vibration mode of a thin piping elbow in plane (case 1) (01-0011SDLLB_FEM)

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1.11.1Description

1.11.2Background


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1.12 EDF Pylon (01-0033SFLLA_FEM)

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1.12.1Description

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1.12.2.2Displacement of the model in the linear elastic range

Reference solution



Software ANSYS 5.3 NE/NASTRAN 7.0


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Buckling modal deformation (dominating mode)







1.13 Tied (sub-tensioned) beam (01-0005SSLLB_FEM)

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1.13.1Description

1.13.2Background


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1.13.2.2Compression force in CE bar

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Force diagrams

1.13.2.3Bending moment at point H

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1.13.2.4Vertical displacement at point D

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1.14.1Description

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1.15.1Description

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1.15.2.2Eigen mode frequencies relative to the angle

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1.16 Caisson beam in torsion (01-0037SSLSB_FEM)

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1.16.1Description

1.16.2Background


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1.16.2.2Displacement and stress at two points

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1.17 Thin cylinder under a uniform radial pressure (01-0038SSLSB_FEM)

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1.17.1Description

1.17.2Background


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1.17.2.2Stresses in all points

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1.17.2.3Cylinder deformation in all points

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1.18.1Description

1.18.2Background


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1.19 Double fixed beam with a spring at mid span (01-0015SSLLB_FEM)

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1.19.1Description

1.19.2Background


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Reference solution

EI EI EI EI

EI EI EI EI

EI EI EI EI

EI EI EI EI

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EI

33

333

333

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33



U

U EI k

v

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EI K

v v

v v v v

y)unnecessar (usually026826

244423222 vvvv

EI l

P v

EI

P v v v


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1.20 Beam on two supports considering the shear force (01-0041SSLLB_FEM)

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1.20.1Description

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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

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shear

y

flexion

z


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1.21 Torus with uniform internal pressure (01-0045SSLSB_FEM)

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1.21.1Description

1.21.2Background


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Reference solution

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1.22 Short beam on simple supports (on the neutral axis) (01-0017SDLLB_FEM)

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1.22.1Description

1.22.2Background

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1.22.2.1Eigen modes frequencies

Reference solution

E


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1.23 Annular thin plate fixed on a hub (repetitive circular structure) (01-0022SDLSB_FEM)

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1.23.1Description

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1.24 Double fixed beam (01-0016SDLLB_FEM)

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1.24.1Description

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E EI

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1.24.2.3Eigen mode frequencies of the model in the linear elastic range

Reference solution

S

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Lf nn


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1.25 Thin circular ring hanged on an elastic element (01-0014SDLLB_FEM)

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1.25.1Description

1.25.2Background


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1.26 Short beam on simple supports (eccentric) (01-0018SDLLB_FEM)

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1.26.1Description

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1.27 Thin cylinder under a uniform axial load (01-0042SSLSB_FEM)

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1.27.1Description

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1.27.2.2Stress in all points

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Reference solution

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1.28 Spherical shell under internal pressure (01-0046SSLSB_FEM)

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1.28.1Description

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1.29 Simply supported square plate (01-0036SSLSB_FEM)

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1.29.1Description

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1.29.2.2Vertical displacement at O

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1.30 Square plate under planar stresses (01-0039SSLSB_FEM)

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1.30.1Description

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1.31 Fixed thin arc in planar bending (01-0027SSLLB_FEM)

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1.31.1Description

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1.31.2.3Theoretical resultsSolver Result name Result description Reference value





1.32 Simply supported rectangular plate under a uniform load (01-0052SSLSB_FEM)

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1.32.1Description

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1.32.2.2Vertical displacement and bending moment at the center of the plate

Reference solution


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1.32.3Theoretical Background






1.33 Shear plate perpendicular to the medium surface (01-0055SSLSB_FEM)

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1.33.1Description

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1.34 Triangulated system with hinged bars (01-0056SSLLB_FEM)

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1.34.1Description

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1.34.2.2Tension force in BD bar

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1.35 A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0061SSLSB_FEM)

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1.35.1Description

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1.36 Thin cylinder under a hydrostatic pressure (01-0043SSLSB_FEM)

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1.37 Slender beam on three supports (01-0025SSLLB_FEM)

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1.37.2.3Reaction in B

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1.38 Bending effects of a symmetrical portal frame (01-0023SDLLB_FEM)

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1.38.1Description

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1.40 A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (01-0066SSLSB_FEM)

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1.40.2.2Vertical displacement at point C (center of the plate)

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1.41 Reactions on supports and bending moments on a 2D portal frame (Rafters) (01-0077SSLPB_FEM)

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1.41.1Description

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1.41.2.2Moments and actions on supports M.R. calculation on a 2D portal frame.




Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the chords







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1.43 Slender beam on two fixed supports (01-0024SSLLB_FEM)

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1.43.2.2Shear force at G

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1.43.2.3Bending moment in G

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1.43.2.4Vertical displacement at G

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1.43.2.5Horizontal reaction at A

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1.44 Slender beam of variable rectangular section with fixed-free ends (ß=5) (01-0085SDLLB_FEM)

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1.44.2.2Reference results

Reference values




TheoreticalFrequency

Result name Result description Reference value



1.45 Vibration mode of a thin piping elbow in space (case 3) (01-0069SDLLB_FEM)

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1.45.1Description


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1.46 Fixed thin arc in out of plane bending (01-0028SSLLB_FEM)

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1.46.2.2Displacements at B

Reference solution


■

■

■

1.46.2.3Moments at = 15°

Reference solution

■

■


■

■

■





1.47 Fixed/free slender beam with eccentric mass or inertia (01-0096SDLLB_FEM)

Passed

1.47.1Description

1.47.2Background


■

■

■■

1.47.2.2Problem data

Units

Geometry

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■




■

■

■

■■

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


■

■

■

■

Modal deformations



1.48 Membrane with hot point (01-0099HSLSB_FEM)

Passed

1.48.1Description

1.48.2Background

■ NAFEMS, Test T1

Analysis

■




Units

Geometry / meshing


■

■

■

Boundary conditions

■

►

►

■

Loading

■

■



1.48.3 yy stress at point A:

Reference solution:


■

■

■







1.49 Double cross with hinged ends (01-0097SDLLB_FEM)

Passed

1.49.1Description

1.49.2Background

■

■


Units

Geometry

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■

Boundary conditions

■■

Loading



■

■

Modal deformations



1.50 Stiffen membrane (01-0040SSLSB_FEM)

Passed

1.50.1Description

1.50.2Background


■

■

■

1;1,

Units

Geometry

■

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■


■

■

Boundary conditions

■

■

Loading

■

■



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


■

■

■

Deformed shape




Passed

1.51.1Description

1.51.2Background

■ Software

■

■


Units

Geometry

■

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■




■

■

■

Boundary conditions

■

■

Loading

■

■

1.51.2.2Vertical displacement at point C (center of the plate)

Reference solution


■

■

■




Passed

1.52.1Description

1.52.2Background


■

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Units

Geometry

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Boundary conditions

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■

Loading

■■

1.52.2.2Vertical displacement at point C (the center of the plate)

Reference solution


■

■■







1.53 Beam on elastic soil, hinged ends (01-0034SSLLB_FEM)

Passed

1.53.1Description

1.53.2Background


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Units

Geometry

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Boundary conditions

■

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Loading

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1.53.2.2Displacement and support reaction at A

Reference solution

■

■


■

■

■



Deformed shape

1.53.2.3Displacement and bending moment at D

Reference solution

■

■


■

■

■



Bending moment diagram







1.54 Beam on 3 supports with T/C (k = -10000 N/m) (01-0102SSNLB_FEM)

Passed

1.54.1Description

1.54.2Background


■ internal GRAITEC

■

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Units

Geometry

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■

■

Boundary conditions

■

►

►

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Loading

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■



1.54.2.2References solutions

Displacements

Mz Moments


■

■

■

Deformed shape



Moment diagram







1.55 Design of a concrete floor with an opening (03-0208SSLLG_BAEL91)

Passed

1.55.1Description

1.55.2Background


Calculation model: 2D

►

►

►►

►

►

■

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►

►

►

►

■

Slab geometry



Support positions

Positions of punctual loads



Global loading overview

Load Combinations

Code Numbers Type Title



1.55.2.2Effel Structure Results

SLS max displacements (load combination 101)

Mx bending moment for ULS load combination



My bending moment for ULS load combination

Mxy bending moment for ULS load combination

1.55.2.3Effel RC Expert Results

Main hypothesis

■

■

■

■

■



Axi reinforcements

Ayi reinforcements



Axs reinforcements

Ays reinforcements




Solver Resultname

Result description Referencevalue





1.56 Spherical dome under a uniform external pressure (01-0050SSLSB_FEM)

Passed

1.56.1Description

1.56.2Background


■

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Units

Geometry

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■

Boundary conditions

■■

Loading

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■

1.56.2.2Horizontal displacement and exterior meridian stress

Reference solution


■

■

■

Deformed shape




Passed

1.57.1Description

1.57.2Background


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Units

Geometry

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Boundary conditions

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Loading

■■

1.57.2.2Vertical displacement at point C (the center of the plate)

Reference solution


■

■■







1.58 System of two bars with three hinges (01-0002SSLLB_FEM)

Passed

1.58.1Description

1.58.2Background


■

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Units

Geometry

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■



Boundary conditions

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■

Loading

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■

1.58.2.2Displacement of the model in C

Reference solution


■

■

■

Displacement shape

1.58.2.3Bars stresses

Reference solutions


■

■

■



1.58.2.4Shape of the stress diagram







1.59 Rectangular thin plate simply supported on its perimeter (01-0020SDLSB_FEM)

Passed

1.59.1Description

1.59.2Background


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Units

Geometry

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►

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Boundary conditions

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Loading

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Reference solution


■■

■



Eigen mode shapes





1.60 Simply supported square plate under a uniform load (01-0051SSLSB_FEM)

Passed

1.60.1Description

1.60.2Background


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Units

Geometry

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Boundary conditions

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Loading

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■


Reference solution


■

■

■

1.60.2.3Theoretical result






1.61 Plane portal frame with hinged supports (01-0089SSLLB_FEM)

Passed

1.61.1Description

1.61.2Background


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■

Units

Geometry

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■

■

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■



Boundary conditions

Loading

■

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1.61.2.2Calculation method used to obtain the reference solution

■

■

■

■

■

■

■

■

1.61.2.3Reference values







1.62 Double hinged thin arc in planar bending (01-0029SSLLB_FEM)

Passed

1.62.1Description

1.62.2Background


■

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Units

Geometry

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►

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►




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Boundary conditions

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Loading

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■

1.62.2.2Displacements at A, B and C

Reference solution

■

■


■■

■



Displacements shape







1.63 A 3D bar structure with elastic support (01-0094SSLLB_FEM)

Passed

1.63.1Description


■ Internal GRAITEC;

Analysis

■

Units

Geometry

■

■■


■

■



Boundary conditions

■

■

Loading

■■


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



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



■

■

■

■

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



Reference values

7182/1015


■

■

■

Deformed shape



Normal forces diagram

1.63.1.3Reference values




1.64 Cantilever beam in Eulerian buckling with thermal load (01-0092HFLLB_FEM)

Passed

1.64.1Description

1.64.2Background


■ Euler theory;

■

■

Units

Geometry

■

■

■


■

■

■

Boundary conditions

■■

Loading

■

■

)




Reference solution

98696.0100000

98696

986964P 2

2

critical

N L

EI


■

■

■

Deformed shape







1.65 Beam on elastic soil, free ends (01-0032SSLLB_FEM)

Passed

1.65.1Description

1.65.2Background


■

■

■

Units

Geometry

■

■




Boundary conditions

■

►

►

■

Loading

■

■

1.65.2.2Bending moment and displacement at C

Reference solution

■

■


■

■

■

Bending moment diagram



1.65.2.3Displacements at A

Reference solution

■

■


■

■

■








Passed

1.66.1Description

1.66.2Background


■

■

■

Units



Geometry

■

■

■

■

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■

►

►

►

►

►


■

■

■

Boundary conditions

■

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►

►

■

Loading

■

■


Reference solution

■


■

■

■



Eigen mode shapes







1.67 Thin square plate fixed on one side (01-0019SDLSB_FEM)

Passed

1.67.1Description

1.67.2Background


■

■

■

Units

Geometry

■

■

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►

►

►

►




■

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■

Boundary conditions

■

■

Loading

■

■


Reference solution


■

■

■



Eigen mode shapes



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

■

■

■

■

■

■

■■




■

■

■

Boundary conditions

■

■

Loading


Reference frequency

Uncertainty about the reference frequencies

■

■


■

■



Modal deformations



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

Passed

1.69.1Description

1.69.2Background


■

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■

Units

Geometry

■

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►

■

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Boundary conditions

■

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Loading

■

■




Calculation method used to obtain the reference solution

Reference values

Eigen modes



1.70 Truss with hinged bars under a punctual load (01-0031SSLLB_FEM)

Passed

1.70.1Description

1.70.2Background

■ Software

■

■


Units

Geometry




Boundary conditions

■

■

Loading

■

■

1.70.2.2Displacements at C and D

Reference solution


■■

■

Displacements shape



1.71 Beam on 3 supports with T/C (k = 0) (01-0100SSNLB_FEM)

Passed

1.71.1Description

1.71.2Background



■

■

Units

Geometry

■

■


■

■

Boundary conditions

■

►

►

►

■

Loading

■

■




Displacements

Mz Moments


■

■

■

Deformed shape



Moment diagrams







1.72 Linear element in simple bending - without compressed reinforcement (02-0162SSLLB_B91)

Passed

1.72.1Description


■ J. Perchat

■

■

Units

■

■■

■

Geometry

■

■


■

■

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





■

■

■


Reference




1.73 Non linear system of truss beams (01-0104SSNLB_FEM)

Passed

1.73.1Description

1.73.2Background



■

■

Units

Geometry■

■


Boundary conditions

■

►

►

■



Loading

■

■


Displacements


■

■

■

Deformed shape



Normal forces





1.74 Simple supported beam in free vibration (01-0098SDLLB_FEM)

Passed

1.74.1Description

1.74.2Background

■ NAFEMS, FV5

Analysis type:

■


Units

Geometry

■

■

■


■

■

■



Boundary conditions

■

►

►

■

Loading



■

■

Modal deformations



1.75 Study of a mast subjected to an earthquake (02-0112SMLLB_P92)

Passed

1.75.1Description



■

■

1.75.1.2Material strength model

Units

Geometry

■

■

■

■

■

■

Masses

■

■


■

■

■



Boundary conditions

■

Loading

■


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

;



Modal deformations

1.75.1.5Spectral study

Design spectrum



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



Moment at the base


Reference






1.76 Design of a Steel Structure according to CM66 (03-0206SSLLG_CM66)

Passed

1.76.1Description

1.76.2Background


Calculation model: Simple

■

►

►

►

►

■

Model preview



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)



1.76.2.3CM66 Effel Expertise results

Hypotheses

■

■

■

■

■

■

■

■

■

■

■

■

The results of the optimization given below correspond to an iteration of the finite elements calculation.



Deflection verification

Ratio

Max values on the element

■

■

■

CM Stress diagrams

Work ratio



Stresses

Max values on the element■

■

■

Buckling lengths

Lfy



Lfz

Lateral-torsional buckling lengths

Ldi



Lds

Optimization







Solver Result name Result description Rate (%) Final section



1.77 Linear element in combined bending/tension - without compressed reinforcements - Partiallytensioned section (02-0158SSLLB_B91)

Passed

1.77.1Description


■ J. Perchat

■

■

Units

■

■

■

■

Geometry

■

■


■

■

Boundary conditions

■

►

►

■

Loading

■

►

►



►

►

■

Reinforced concrete calculation hypothesis:



1.77.1.2Reinforcement calculation

Reference solution






■

■

■



1.78 Thin cylinder under its self weight (01-0044SSLSB_MEF)

Passed

1.78.1Description

1.78.2Background


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Units

Geometry

■

■

■


■

■ ■



Boundary conditions

■

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►

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Loading

■

■

1.78.2.2Stresses

Reference solution


■

■

■


Reference solution

■

■







1.79 Pinch cylindrical shell (01-0048SSLSB_FEM)

Passed

1.79.1Description

1.79.2Background

■ Software

■

■


Units

Geometry

■■

■




■

■

Boundary conditions

■

■

Loading

■

■

1.79.2.2Vertical displacement at point A

Reference solution


■

■

■

1.79.2.3Theoretical result






1.80 Spherical shell with holes (01-0049SSLSB_FEM)

Passed

1.80.1Description

1.80.2Background

■ Software

■

■


Units

Geometry

■

■

■




■

■

Boundary conditions

■

■

Loading

■

■

1.80.2.2Horizontal displacement at point A

Reference solution


■

■

■

Deformed shape



1.80.3Theoretical Background






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

Passed

1.81.1Description

1.81.2Background


■

■

■

Units

Geometry

■

■

■


■

■

Boundary conditions

■

■



Loading

■

►

►

►

►■


Reference solution


■

■

■

Deformed shape







1.82 Simply supported rectangular plate under a uniform load (01-0053SSLSB_FEM)

Passed

1.82.1Description

1.82.2Background


■

■

■

Units

Geometry

■

■

■


■

■

Boundary conditions

■

■



Loading

■

■


Reference solution


■

■

■








Passed

1.83.1Description

1.83.2Background


■

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■

Units

Geometry

■

■

■


■

■

■

Boundary conditions

■

■



Loading

■

■


Reference solution


■

■

■

1.83.2.3Theoretical resultsSolver Result name Result description Reference value






Passed

1.84.1Description

1.84.2Background


■

■

■

Units

Geometry

■

■

■


■

■

■

Boundary conditions

■

■



Loading

■

■


Reference solution


■

■

■







1.85 Short beam on two hinged supports (01-0084SSLLB_FEM)

Passed

1.85.1Description

1.85.2Background


■

■

■

Units

Geometry

■

■

■

■


■

■

Boundary conditions

■

■



Loading


Reference values

-1.25926 x 10-3







1.86 Reactions on supports and bending moments on a 2D portal frame (Columns) (01-0078SSLPB_FEM)

Passed

1.86.1Description

1.86.2Background


■ Design and

■

■



1.86.2.2Moments and reactions on supports M.R. calculation on a 2D portal frame.

h H qh

M E B 2

²








Passed

1.87.1Description

1.87.2Background


■

■

■

Units

Geometry

■

■

■

■

■

■

■

■

■

►

►

►




■

■

■

Boundary conditions

■

■

Loading

■

■


Reference solution

■


■

■

■

Eigen mode shapes







1.88 Beam on 3 supports with T/C (k -> infinite) (01-0101SSNLB_FEM)

Passed

1.88.1Description

1.88.2Background



■

■

Units

Geometry

■

■


■

■

Boundary conditions

■

►

►

►

■

Loading

■

■




Displacements

Mz Moments


■

■

■

Deformed shape



Moment diagram







1.89 Linear system of truss beams (01-0103SSLLB_FEM)

Passed

1.89.1Description

1.89.2Background



■

■

Units

Geometry■

■


Boundary conditions

■

►

►

■



Loading

■

■


Displacements

N normal forces


■

■■

Deformed shape



Normal forces





1.90 Design of a 2D portal frame (03-0207SSLLG_CM66)

Passed

1.90.1Description

1.90.2Background


Calculation model: 2D

►

►

►

►

►

►

■

►

►

■

Model preview

Combinations



1.90.2.2Effel Structure Results

Ridge displacements (combination 104)

Diagram of normal force envelope



Envelope of bending moments diagram

1.90.2.3Effel Expert CM results

Main hypotheses

■

■

■

■

■

■

■

■



Deflection verification

Ratio

CM Stress diagrams

Work ratio

Stresses



Buckling lengths

Lfy

Lfz

Lateral-torsional buckling lengths

Ldi



Lds

Optimization






Solver Result name Result description Rate (%) Final section






Passed

1.91.1Description

1.91.2Background


■

■

■

Units

Geometry

■

■

■


■

■

■



Boundary conditions

■

■

Loading

■■

1.91.2.2Vertical displacement at point C center of the plate)

Reference solution


■

■■







1.92 Double fixed beam in Eulerian buckling with a thermal load (01-0091HFLLB_FEM)

Passed

1.92.1Description

1.92.2Background


■ Euler theory;

■

■

Units

Geometry


■

■

■

Boundary conditions

■

■

Loading

■

■




Reference solution

93.3724.117

925.29

k 724.117

2

P 2

2

critical

N L

EI


■

■

■

Deformed shape of mode 1







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


■■

■

Units

Geometry

■

■

■■


■

■

■

Boundary conditions

■

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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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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



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



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



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



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



2 CAD, rendering and visualization



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



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



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



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



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



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



3 Climatic generator



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



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




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



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



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



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



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



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



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


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



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



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



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



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



4 Combinations



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




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




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



4.13 Generating a set of combinations with seismic group of loads (TTAD #11889)

Passed

4.13.1Description


Passed

4.14.1Description


Passed

4.15.1Description




Passed

4.16.1Description


Passed

4.17.1Description



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


Passed

4.20.1Description



4.21 Verifying combinations for CZ localization (TTAD #12542)

Passed

4.21.1Description



5 Concrete Design



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■

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■

■

■

■

■

■

■

■

Units



Geometry

■

■

■

■


■

■

■

■

■

■

Boundary conditions

■

►

►

■

Loading

■

■

400

8

81040

344

8

8103040



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



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



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


■

■

■



SLS load combinations(kNm)

Theoretical reinforcement area(cm2 )

Minimum reinforcement area(cm2 )

5.1.2.3 Reference results






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


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Units



Geometry

■

■

■

■

■

■


■

■

■

■

■

Boundary conditions

■

►

►

■

Loading

■

■

2058

4061525

1518

406153025




)()(),( 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


Ecar

Eqp

cm

se

M

M t

E

E

*),(1 0



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


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




00130

260

8962

402

1007257028000130

10402570280500

8962260

4

4


■■

■

ULS and SLS load combinations (kNm)

Theoretical reinforcement area (cm2 )



Minimum reinforcement area (cm2 )







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


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Units



Geometry

■

■

■

■

■

■


■

■

■

■

■

■

► 671651

25

►56225300300 3232

■

► 78434

151

500

►

MPa f

E ck cm 31476

10

825*22000

10

8*22000

3.03.0

Boundary conditions

■

►

►

■



Loading

■

■

■

kNm M Ecq 94,1628

²80,5*75,38

kNm M Eqp 7.104

8

²80,5*75.24




)()(),( 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


Mpa f ck bc 1525*6,06,0

Mpa f yk s 400*8,0




1

394.040015*32.17

15*32.17

*

*1 sce

ce


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


d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


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




■

■

■






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



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


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Units

Geometry

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■

■

Boundary conditions

■

►

►

■

5.6.2.2 Reference results in calculating the concrete beam

The ULS load calculation:

■

■



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*



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:



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



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


■■

■



ULS load combinations(kNm)







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


■

■

■

■

■

■

■

■

■

■

■

Units



Geometry

■

■

■


■

■

■

■

■ 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

■

►►

■



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



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


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



■

■

■



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


■

■

■

■

■

■

■

■

■

■

■

■



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.



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,


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*

■


■

■

■



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







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


■

■

■

■

■

■

■

■

■ 30%2

■

■

■

■

■

Units



Geometry

■

■

■


■

■

■

■

2051

30

90230300300 3232

■

78434151

500

Boundary conditions

■

►

►

■



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:


3

2

1023

2

20090020200400401

2

466027727714321

3321 1086311020030635

28702090400

8631

1

33 1006255104804



43502870211251211251.

5545034350

43501 0

.

.1 0 2

4660243600455027727714321

7491

024362

20090

2003

2

22


3321 1086311020030635

372028702090400

8631 1

37202870

43502870211251211251.

74304350400190.4011

4657024367430

8631

111



00130

260

902

425

1068490040000130

10425900400500

902260

4

4


■

■

■



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


■

■

■

■

►

►

■

►

► 3,02

■

►►

► 10 02

■

■

■

■

■

■



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

MN N ED 99.03.0*5.14.0*35.1

00

2400

5002

4002 0

020201



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



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



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



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


■

■

■




Theoretical value (cm2 )







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


■

■

■

■

■

■

■

■

■

■ σ

■ σ

Units



Geometry

■

■

■

■

■

■


■

■

■

■

■ Mpa f ck 20■ Mpa f yk 400

■ ²42,9 cm A st

Boundary conditions

■

►

►

■

Loading

■

■



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



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


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*


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



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


■

■

■

SLS load combinations (kNm) and stresses (MPa)

Compressing stresses in concrete section σ c

σ



Compressing stresses in the steel reinforcement section σ s

σ



σ σ

σ σ





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


■

■

■

■

■

■

Units



Geometry

■

■

■

■

■

■

■

■

Boundary conditions

■

►

►

■



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)



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



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


■

■

■



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


■

■

■

■

■

3,02

Units



Geometry

■

■

■

■■

■


■

■

■

■

■

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,




)()(),( 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


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



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


Mpa f ck bc 1515*6,0*6,0

Mpa f yk s 400*8,0


387.040018*82.16

15*82.16

*

*1

sce

ce


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



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


d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


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


■

■

■




Minimum reinforcement area (cm2 )







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


■

■

■

■

►

►

►

■

►

►

■

■■

■

■

■

■



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

■





ml l 60.11*20

37.8050.0

60.11*32*32 0 a

l


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



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 :

■

■



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



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:



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



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




2

min, 5*002.0²81.078.434

353.0*10.0*10,0cm Acm

f

N A c

yd

Ed s


■

■

■





5.15 EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinearstress-strain diagram

Passed

5.15.1Description

5.15.2Background


■

■

■

■

■

■

■

■

■

■

■

Units



Geometry

■

■

■


■

■

■

■

■ 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

■

►

►

■



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:


■

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




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



■

■

■

ULS and SLS load combinations(kNm)




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


■

■

■

■

■

■

■

■

■

■

■

■



Units

Geometry

■■

■

■

■

■


■

■■

■

■ 671651

25

■ 56225300300 3232

■ 78434151

500

■ 3147610

82522000

10

822000

3030

Boundary conditions

■

►

►

■



Loading

■

■

6102555

8

145214

61067538

14150150

61030328

1494



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



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:



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



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*


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




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:


d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


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


■

■

■



5.17 EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinearstress-strain diagram

Passed

5.17.1Description

5.17.2Background


■

■

■

■

■

■

■

■

■

■

■

■

Units



Geometry

■

■

■

■


■

■

■

■

■

■

■

Boundary conditions

■

►

►

■

Loading

■

■

kNm M cq ser 240

8

²8*1020,

kNm M qp ser 184

8

²8*10*3.020,



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


Ecar

Eqp

cm

se

M

M t

E E

*),(1 0



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


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



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


00130

260

212

612

10612669030000130

103126690300500

212260

4

4




■

■

■

SLS load combinations(kNm)


Minimum reinforcement area (cm2

)



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


■

■

■

■

■

■

■

■

Units



Geometry

■

■

■

■

■

■


■

■

■

■

■

Boundary conditions

■

►

►

■

Loading

■

■

kNm M Ecq 28.2318

²80.5*)3025(

kNm M Eqp 97.1428

²80.5*)30*3.025(




)()(),( 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


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



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


Mpa f ck bc 1515*6,0*6,0

Mpa f yk s 400*8,0


1

389.040018*97.16

15*97.16

*

*1

sce

ce


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



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


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


■

■

■




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


■

■

■

■

■

■

■

■

■ σ

■ σ

■

■

Units



Geometry

■

■

■

■■

■


■

■

■

■

■ 25

■ 500

■ 1630

Boundary conditions

■

►

►

■

Loading

■

►

►

►

►




)()(),( 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



56.22

55.21

31476

200000

),(1 0

t

E

E

cm

se


Mpa f ck bc 1525*6,06,0

Mpa f yk s 400*8,0


0..2

1111

2.. 2

1

0 sc A

3540

16307156224021630562216305622

.2.. 2

1


443

11

31

01450350071056221016303

350040

..3


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



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,


■

■

■




σ

σ

σ

σ





σ σ

σ σ





5.20 EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement – Inclinedstress-strain diagram

Passed

5.20.1Description

5.20.2Background


■

■

■

■

■

■ 8,02 ■

■

■■

■

■

■

■



Units

Geometry

■

■

■

■

■

■


■

■

■

■

■

■

671651

25

MPa f

E ck cm 31476

10

82522000

10

822000

3.03.0

■



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




)(*)(*),( 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




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


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



■

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:

■

■


■ 705

■ 9420.

■



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*


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


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



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*


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


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



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:


d b

d b f

f

Max A

w

w yk

eff ct

s

**0013.0

***26.0

,

min,


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


■

■

■




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


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■

■

■

■

■

■

■

■

■

■

■



■

■

■

■

Units

Geometry

■

■

■

■

■ θ ˚

Boundary conditions

■

►

►

■



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,



ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²66.5

15,1

500*688.0

30*293.0

*

*.


■

■

■



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


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■

■

■

►

►

►

■

►

►

■

■

■

■

■

■■

■



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

■ 4511030750

35200





ml l 60.11*20

37.8050.0

60.11*32*32 0 a

l


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



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


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:



■

■

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

mmmm

mmmm

mmhmm

e 207.16

20max

30

50020

max

30

20max0




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


■■

■



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


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►

►

►

■

►

►

■ 3,02 ■

■■

■

■

■



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

►

►

■ 4511030750

35200





ml l 60.11*20

37.8050.0

60.11*32*32 0 a

l


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



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


17.1352.0

181.0*28.2*, 0

Ed

EQP

ef M

M t


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




■

■

■

■

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




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



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



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


35.0 ²25.40078.434

20*²50.0*35.0***cm

f

f hb A

yd

cd s




■

■

■

Theoretical reinforcement area(cm2

)




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


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■

■

■

■

■

■

■

■

■

■

■



■

■

■

■

■

■

■

■

■

■

■

■

■

■

■

■

■

■

■

Units

Geometry

■

■

■ θ ˚

Boundary conditions

■

►

►

■



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

*

*.



ml cmb f

f b

s

Aw

yk

ck

ww sw /²01.790sin*80.0*

500

30*08,0sin**

*08,0sin**min,


■

■

■







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


■

■

■

■

►

►

■►

■

■

■

■

■



Units

Geometry

■■

■

■

Boundary conditions

Loading

■

kN N ED 2280800*5.1800*35.1




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


■

■

■



Theoretical reinforcement area(cm2)







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



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


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■

Units



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




■

■

■

ULS load combinations (kNm)







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



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



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



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



5.46 EC2 Test 7: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram

Passed

5.46.1Description

5.46.2Background


■

■

■

■

■

■

■

■

■

■

■

Units



Geometry

■

■

■


■

■

■

■

■ 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

■

►

►■



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


■

■

■

kNm M kNm M btu Ed 92840.218



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


■

■

■

■

■

■



■

■

■

■

■

■

■

Units

Geometry

■

■

■

■

■

■ ■ ˚



Boundary conditions

■

►

►

■

Loading

406

76373 51239

82210 5770

0




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,


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


■

■

■




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,







5.48 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


■

■

■

■

►

►

►

■

►

►

■ 3,02 ■

■

■

■■

■



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

■ 08703530

0370500





ml l 60.11*20

37.8050.0

60.11*32*32 0 a

l


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



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


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


■

■

■

■



m N

M e

Ed

Ed 087.0353.0

03075.00

ml ei 03.0400

6.11

4000


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


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



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



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 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:



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



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:



Estimation of the nominal rigidity:


2.1cmcd

E E


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



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



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


■

■

■



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


■

■

■

■

►

►

►

■

►

►

► 3,02 ►

►

►

►

►

►

►



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

■ 1003000

03100





ml l 12*20

Calculating the slenderness of the column:

10440.0

12*32*32 0 a

l


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



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


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


■

■



m N

M e

Ed

Ed 10.0300.0

031.00

ml ei 03.0400

12

400

0


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


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



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



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



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




■

■

■

Theoretical reinforcement area (cm2

)








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


■

■

■

■

►

►

■

■

■■

■

Units



Geometry

■

■

■

■

Boundary conditions

Loading

■

kN N ED 1080800*35.1 5.50.2.2Reference results in calculating the concrete column


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


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




■

■

■

Theoretical reinforcement area(cm2

)







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


■

■

■

■

■

■

■

■

■

■ σ

■ σ

Units



Geometry

■

■

■

■

■

■


■

■

■

■

■ 25■ 500

■ 7037

■ 286

Boundary conditions

■

►

►

■

Loading

■

►

■

►

►

►




)()(),( 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



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


Mpa f ck bc 1525*6,06,0

Mpa f yk s 320*8,0


0..2

1111

413435

286470377290163522867037901628670379016

.2..1


443

11

31

0147201472097445349016286453472901670373

413435

..3


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


■

■

■




σ

σ

σ

σ



σ σ

σ σ





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


■

■

■

■

■

■

■

■

■ σ

■ σ

■

■

Units



Geometry

■

■

■

■■

■


■

■

■

■

■ 25

■ 500

■ 4615

Boundary conditions

■

►

►

■

Loading

■

►

►

►

►




)()(),( 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



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


Mpa f ck bc 1525*6,06,0

Mpa f yk s 320*8,0


0..2

1111

2.. 2

1

0 sc A

672620

46155347172024615471746154717

.2.. 2

1


443

11

31

003140267053047171046153

267020

..3


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



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,




■

■

■


σ

σ

σ

σ





σ σ

σ σ





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


■

■

■

■

►

►

►

■

►

►

■ 3,02 ■

■

■

■■

■

■



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

■ 1003000

03100





ml l 12*20

10440.0

12*32*32 0 a

l


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



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


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


■

■



m N

M e

Ed

Ed 10.0300.0

031.00

ml ei 03.0400

12

400

0


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


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



BC cu 496.0103.0


103.0cu

136,0)103,0*21(1*25,1 u

md z uc 331,0)136,0*4,01(*35,0)*4,01(*


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


Calculation of nominal rigidity:


2.1

cmcd

E E


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



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



MNm M Ed 133.0

1300.0

216.0

234.11*039.0


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



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



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


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




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



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


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


■

■

■



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


■

■

■

■

■

■

■

■

■

■

■

■



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.




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



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


■

■

■



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

■

■■




■

■

■

■

■

■

■

■

■

■

■

■



Units

Geometry

■

■

■

■

■

■


■

■

■

■

■

■ 671651

25

■ 56225300300 3232

■ 78434151

500

Boundary conditions

■

►

►

■



Loading

■

■

■

kNm M Ed 16.5218

²80.5*94.123

kNm M Ecq 68.3628

²80.5*25.86


372.0lu


■

■

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



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


d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


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


■

■

■




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


■

■

■

■

■

■

■

■

■

Units



Geometry

■

■

■

■

■

■


■

■

■

■

■ 671651

25

■ 56225300300 3232

■ 78434151

500

■ 3147610

8252200010

822000

3030

Boundary conditions

■

►

►

■



Loading

■

■

■

mkN M Ed .59,232

8

²80,5*31,55

mkN M Ecq .94,1628

²80,5*75,38


■

■


■

■

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




00130

260

562

761

10701524025000130

107615240250500

562260

4

4


■■

■





5.57 Modifying the "Design experts" properties for concrete linear elements (TTAD #12498)

Passed

5.57.1Description



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

■

■

■




■

■

■

■

■

■

■

■

■

■

■

■



Units

Geometry

■

■

■

■

■

■


■

■

■

■

■

■ 671651

25

■ 56225300300 3232

■ 78434151

500

Boundary conditions

■

►

►

■



Loading

■

■

■

kNm M Ed 16.5218

²80.5*94.123

kNm M Ecq 68.3628

²80.5*25.86


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


■

■

■



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


■

■

■

■

■

■

■

■

Units

Geometry

■

■

■

■

■




■

■

■■

■

Boundary conditions

■

►

►

■

Loading

■

4071505113535151351

■

285150135

■

1801503013530



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



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,


yk

ctmc s

f

f A A .

Mpa f f ck ctm 56,225*30.0*30.0 3/23/2



²04.020.0*20.0 m Ac

²05.2500

56.2*04.0* cm

f

f A A

yk

ctmc s


■

■

■

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)



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


■

■

■

■

■

■

■

■

Units



Geometry

■

■

■

■

■

■


■

■

■

■

■ 671651

25

■ 56225300300 3232

■ 78434151

500

■ 3147610

8252200010

822000

3030

Boundary conditions

■

►

►

■



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




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




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


■

■

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(*

■




■ 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


■

00130

260

562

►

731

10701519025000130

107315190250500

562260

4

4


■

■

■



5.61 EC2 Test 8: Verifying a rectangular concrete beam without compressed reinforcement –Inclined stress-strain diagram

Passed

5.61.1Description

5.61.2Background


■

■■

■

■

■

■

■

■

■

■

■

■

■

■



Units

Geometry

■

■

■

■

■

■


■

■

■

■

■

■ 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



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




)(*)(*),( 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




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




■

■

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


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


■

■

■



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


■

■

■

■

■

■

■

■

■

■

■

■



■

■

■

■

■

■

■

■

■

■

■

■

■

■

■

■ 80.0

■

■

■

Units

Geometry

■

■

■ θ ˚

Boundary conditions

■

►

►

■



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

*

*.



ml cmb f

f b

s

Aw

yk

ck

ww sw /²21.590sin*55.0*

500

35*08,0sin**

*08,0sin**min,


■

■

■

5.62.2.4Reference resultsResult name Result description Reference value






Passed

5.63.1Description

5.63.2Background


■

■

■

■

■

■

■

■

■

■



■

■

Units

Geometry

■

■

■

■

■

■

■ θ ˚

Boundary conditions

■

►

►

■

Loading:

2

* l P

V

u

Ed

KN V Ed 2632

5.91*75.5



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



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,


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,


■

■

■



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


■

■

■

■

►

►

■

►

■ 3,02 ■

■

■

■

■

■

■



Units

Geometry

■

■

■

■

Boundary conditions

Loading

■

■





ml l 84*2*20

28.6940.0

8*32*32 0 a

l


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



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


90.039.0*30.2*, 0 Ed

EQP

ef M

M t


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



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


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




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


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




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



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:




■

■

■

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

2 )





Passed

5.65.1Description

5.65.2Background


■

■

■

■

■

■

■

■

■

■



■

■

Units

Geometry

■

■

■

■

■

■

■ θ ˚

Boundary conditions

■

►

►

■

Loading:

2

* l P

V

u

Ed

KN V Ed 2632

5.91*75.5



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



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,


ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²67,8

15,1

500*566,0

45*211,0

*

*.


■

■

■







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


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Units

Geometry

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Boundary conditions

Loading

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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


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




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



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



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



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


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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


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Loading

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kN N ED 67505000*35.1



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■■




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


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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



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


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Loading

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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


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■



Longitudinal reinforcement for SLS load combinations

Minimum longitudinal reinforcement







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


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Units

Geometry

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Boundary conditions

Loading

■

■ me 166.0

3575.1

225.00





ml l 950.4*2*20

28.6945.0

9*32*32 0 a

l


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



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


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



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


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




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


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




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



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




0029.045.0*70.0

10*05,9 4

MN N Ed 3575.12

MNm M Ed 48.22



008.0

MN N Ed 3575.12

MNm M Ed 563.02

0095.0



MN N Ed 3575.12

MNm M Ed 499.02


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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


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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


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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ULS load combinations (kNm)







6 General application



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



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



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



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



7 Import / Export



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



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



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



8 Joint Design



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



9 Mesh



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



10 Old Climatic Generator



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



11 Reports Generator



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



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



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



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



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



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



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



12 Seismic analysis



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



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



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



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



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



13 Steel design



13.1 EC3 Test 15: Verifying a rectangular hollow section column subjected to bending and axialefforts

Passed

13.1.1Description

13.1.2Background


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Units

Geometry

■■

■

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■

■

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■ 0.10 M


■

■

Boundary conditions

■

►

►

■

Loading

■

►

►

■



13.1.2.2Reference results for calculating the column subjected to bending and axial force

Cross section class



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



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 results

Work ratio of the design resistance for uniform compression

Work ratio of the design resistance for bending





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


■

■

■

■

►

■

■

Units



Geometry

■

■

■

■

■

■

■ 0.10 M


■

■

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

■



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


■

■■




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)







13.3 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


■

■

■

■

■

■

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


■

■

■

Boundary conditions

■

►

►



Loading

■

13.3.2.2Cross section Class

1253.006.179

30.45

sup

inf

Mpa

Mpa



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



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



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



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



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



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



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,



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



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,,



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



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



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



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



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



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



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



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


■

■

■



y coefficient corresponding to non-dimensional slenderness y

y

z z

z



Internal factor, yyk

yyk

Internal factor, yz k

yz k



Internal factor, zyk

zyk

Internal factor, zz k

zz k



LT





y y y

z z z

yyk yyk

yz k yz k

zyk zyk

zz k zyk

LT





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



13.6 EC3 Test 6: Class section classification and combined biaxial bending verification of anIPE300 beam

Passed

13.6.1Description

13.6.2Background


■

■

■

■

■

Units




Boundary conditions

■

►►

■

13.6.2.2Reference results for calculating the cross section class



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 ,,



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


■

■

■




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


■

■

■

■

►

■

■

Units

Geometry

■

■

■

■

■

■

■ 0.10 M




■

■

Boundary conditions

■

►

►

■

Loading

■

►■



13.7.2.2Cross-section classification

67.1318

2/)8500(

mm

mmmm

t

c

8136.0235

y f



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



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

►

►

►



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


■

■

■








13.8 EC3 test 9: Verifying the classification and the compression resistance of a welded built-upcolumn

Passed

13.8.1Description

13.8.2Background


■

■

■

■

►

■

■

Units



Geometry

■

■

■

■

■

■

■ 0.10 M


■

■

Boundary conditions

■

►

►

■

Loading

■

►

■



13.8.2.2Cross-section classification

67.1318

2/)8500(

mm

mmmm

t

c

81.0235

y f

34.111467.13 t c



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



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,



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


■

■

■



13.9 EC3 Test 21: Verifying the buckling resistance of a CHS219.1x6.3H column

Passed

13.9.1Description

13.9.2Background


■■

■

■

■

■

Units


■

■

■

■

■

■




■

■

Boundary conditions

■

►

►

■

■

Loading

■

■



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



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




■

■

■


The appropriate non-dimensional slenderness

LT



Ratio of the design normal force to design buckling resistance in the strong inertia of the profile



y y y

ySN





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



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


■

■

■

■

■

■

Units




■

■ 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

■

■

■

■


■

■

■

Boundary conditions

■

►

►



Loading

■


1873.030.406

84.354

sup

inf

Mpa

Mpa



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



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



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



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





α

34.0

y y

11

22

y y y

y

y

ycr

yeff

y N

f A

,

*



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


α

49.0



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


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



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




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


yyk



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



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,,



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



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




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


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


844.0

905.1747336

3280001

893.07677.0

1,

N

N

N

N C k

z cr

Ed

z mz zz


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



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


■

■

■




y

z z

z




yyk


yz k




zyk


zz k





y y y

z z z

yyk yyk

yz k yz k

zyk zyk

zyk zyk





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


■

■

■

■

■

■



Units


■

■ 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

■

■

■

■


■

■

■

Boundary conditions

■

►



Loading

■


118.083.268

76.48

sup

inf

Mpa

Mpa



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



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



α

34.0



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




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



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


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



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




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


yyk



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



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

,



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



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




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


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


857.0

38.8661700

10000001

977.0776.0

1,

N

N

N

N C k

z cr

Ed

z mz zz




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


■

■■




y

z z

z




yyk


yz k




zyk


zz k





y y y

z z z

yyk yyk

yz k yz k

zyk zyk

zyk zyk





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



13.18 EC3 fire verification: Verifying the work ratios after performing an optimization for steelprofiles (TTAD #11975)

Passed

13.18.1Description



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


■

■

■

■

►

■

■



Units

Geometry

■■

■

■

■

■

■

■ 0.10 M


■

■

Boundary conditions

■

►

►

■

Loading

■

►

■



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


Cross section class



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


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


■

■

■




Work ratio of the design shear resistance




13.20 EC3 Test 19: Verifying the buckling resistance for a IPE300 column

Passed

13.20.1Description

13.20.2Background


■■

■

■

■

■

Units


■

■

■


■

■



Boundary conditions

■

►

►■

■

Loading

■

■


%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*



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



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



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


■

■

■




Coefficient corresponding to non-dimensional slenderness after Y-Y axis

Ratio of the design normal force to design buckling resistance (strong inertia)





y

z





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


■

■

■

■

►

►

■

■

Units



Geometry

■

■

■

■

■

■ 3

, 3512000 mmW y pl

■ 0.10 M


■

■

Boundary conditions

■

►

►

■

Loading

■

►►

■



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 ,



■

► 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


■

■

■




Work ratio of the design resistance for oblique bending



13.22 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow sectionbeam

Passed

13.22.1Description

13.22.2Background


■

■

■

■

►

■

■

Units

Geometry

■

■

■

■

■

■■ 0.10 M




■

■

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


■

■

■




Work ratio of the design shear resistance







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



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



13.31 EC3 Test 3: Class section classification and share and bending moment verification of anIPE300 column

Passed

13.31.1Description

13.31.2Background


■

■

■

■

■



Units


Boundary conditions

■

►

►

■




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

,



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




■

■

■


Shear z direction work ratio



Combined oblique bending







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



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


■

■

■

■

■

■



Units

Geometry

■■

■

■

■

■

■ 0.10 M


■

■

Boundary conditions

■

►

►

■

Loading

■

►

■




Cross section class



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




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


■■

■



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


■

■

■

■

■

■



Units


■

■ 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

■

■

■

■


■

■

■

Boundary conditions

■

►

►

■

►



Loading

■


1873.030.406

84.354

sup

inf

Mpa

Mpa



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



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



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



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





α

34.0

y y

11

22

y y y

y

y

ycr

yeff

y N

f A

,

*



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


α

49.0



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


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



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

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mmmm N mm

mm

mm

mm

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I E

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I

I

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z

t

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43.40224022433856

32.439626.698934731.163.26627490/210000

514614/808002810

63.26627490

10808.49

2810

63.26627490/21000031.1

²

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4

611

2

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1



34

max

, 533.317922953.455

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466.04022433856

/275533.3179229 23,

Nmm

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10808.494

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89.54345434891269

32.439626.698934777.163.26627490/210000

514614/808002810

63.26627490

10808.49

2810

63.26627490/21000077.1

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56.30703070560199

32.439626.6989347163.26627490/210000

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55.9533.3179229

2.7815328000

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56.356.171.119.0

1

/27510.242068

104.127815.0

1

/275533.3179229803.0

101274944.0

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/2752.7815812.0

328000

46.358.173.115.0

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101274952.0

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328000

23

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23

6

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mm N mm

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mm N mm

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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


■■

■




y

z z

z




yyk


yz k




zyk


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y y y

z z z

yyk yyk

yz k yz k

zyk zyk

zyk zyk





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



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


■

■

■

■

■

■



Units


■

■ 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


■

■

■

Boundary conditions

■

►

►

►



Loading

■


1253.006.179

30.45

sup

inf

Mpa

Mpa



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



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



α



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




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



kN N

mm

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l

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fz

z z cr 71.3158151.1581706

²2810

46.6025866/210000²

²

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,

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

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z z z

z



z

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251

,

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top y

botom y



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

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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

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422

422

4

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661.074.315363380

/27518.501177 23,

Nmm

mm N mm

M

f W

cr

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1²²

1

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LT




2733.1150

260

mm

mm

b

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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

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kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

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10.4266.426102243

20.15251.158170677.146.6025866/210000

97.149294/808002810

46.6025866

09362763829

2810

46.6025866/21000077.1

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569.06.426102243

/27518.501177 23,

Nmm

mm N mm

M

f W

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1²²

1

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LT

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2733.1150

260

mm

mm

b

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752.0²569.02.0569.049.015.0²2.015.0 LT LT LT LT

1804.0²569.0²752.0752.0

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ycr 33.380295.3802327²

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985.0

95.3802327

328000839.01

95.3802327

3280001

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N

N N

N

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ycr

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0

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318.501177 mmW y

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kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

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I E

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73.2408.240735730

20.15251.1581706146.6025866/210000

97.149294/808002810

46.6025866

88.19351706542

2810

46.6025866/2100001

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422

422

4

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757.08.240735730

/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 ,,

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



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



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



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



041.0

/27596.123381

10000000

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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


yy

ycr

Ed

y

mLT my yyC

N

N C C k

1̀

1,



-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



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,



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


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



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


,,

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




,,

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



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



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




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 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



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



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


,,

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




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


,,

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




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


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




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


■

■

■




y

z z

z




yyk


yz k




zyk


zz k



LT






y y y

z z z

yyk yyk

yz k yz k

zyk zyk

zz k zyk

LT



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


■

■

■

■

►

►

►

■

■



Units

Geometry

■■

■

■

■

■

■

■

■

■

■ 0.10 M


■

■

■

Boundary conditions

■

►

►

■

Loading

■

►

►

►

■



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



■

■ 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


■

■

■




C1 parameter

C2 parameter

Elastic critical moment

Reduction factor for lateral torsional buckling





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


■

■

■

■

■

Units




Boundary conditions

■

►

►

■


■

■

■



276.57.10

45.56

mm

mm

t

c

92.0



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

,



*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


■

■

■




13.45 EC3 Test 1: Class section classification and compression verification of an IPE300 column

Passed

13.45.1Description

13.45.2Background


■

■

■

■

■

Units




Boundary conditions

■

►►

■




1



276.57.10

45.56

mm

mm

t

c

1

9*9276.57.10

45.56

mm

mm

t

c



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

,


■

■



13.46 EC3 Test 7: Class section classification and compression resistance for an IPE600 column

Passed

13.46.1Description

13.46.2Background


■

■

■

■

►

■

■

Units


■

■

Boundary conditions

■

►

►

■



Loading

■

■




83.4212

224219600

mm

mmmmmm

t

c

0.1235

y

f



424283.42 t

c

21.419

2/)22412220(

mm

mmmmmm

t

c

0.1235

y f

9921.4 t

c




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


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




■

■

■









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



13.51 EC3 Test 22: Verifying the lateral torsional buckling of a IPE300 beam

Passed

13.51.1Description

13.51.2Background


■■

■

■

■

■

Units


■

■

■

■




■

■

Boundary conditions

■

►

►

■

Loading

■

■


%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



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



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



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


■

■

■




Lateral torsional buckling coefficient

Elastic critical moment for lateral-torsional buckling



LT





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


■

■

■

■

■

■

Units




■

■ 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

■

■

■

■


■

■

■

Boundary conditions

■

►

►

■



Loading

■

■



CTICM model




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



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



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



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



kNm M Rd y pl

7.359,,



α

21.0


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

,



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



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




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



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


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



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



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


α

34.0

06804.1²94040.02.094040.034.015.0 LT

163518.0²94040.0²06804.106804.1

1

²²

1

LT LT LT

LT



LT 694.0 LT


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



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



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



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



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



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


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



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


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



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



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



■ 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



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



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



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



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



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



-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


■

■

■



Ratio of the design normal force to design compresion resistance

Ratio of the design share force to design share resistance



Ratio of the design share force to design share resistance


y



z z

z


yyk




yz k



y y y

z z z

yyk yyk

yy

k yy

k

zyk zyk

zyk zyk



13.53 EC3 Test 16: Verifying a simply supported rectangular hollow section beam subjected tobiaxial bending

Passed

13.53.1Description

13.53.2Background


■

■

■

■

►

►

■

■

Units



Geometry

■

■

■

■

■

■

■

■

■ 0.10 M


■

■

Boundary conditions

■

►

►

■

Loading

■

►

►

■



13.53.2.2Reference results for calculating the beam subjected to bi-axial bending

Cross section class



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



■

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



%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


■

■

■



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



13.55 EC3 Test 20: Verifying the buckling resistance of a RC3020100 column

Passed

13.55.1Description

13.55.2Background


■■

■

■

■

■

Units


■

■

■


■

■



Boundary conditions

■

►

►

■■

Loading

■

■


%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 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



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



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


■

■

■





Coefficient corresponding to non-dimensional slenderness after Z-Z axis



Ratio of the design normal force to design buckling resistance (strong inertia)

Ratio of the design normal force to design buckling resistance (weak inertia)





y

z





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



13.57 EC3 test 11: Cross section classification and compression resistance verification of arectangular hollow section column

Passed

13.57.1Description

13.57.2Background


■

■

■

■

►

■

■

Units



Geometry

■

■

■

■

■

■

■


■

■

Boundary conditions

■

►

►

■

Loading

■

►

■




2510

10215230022

mm

mmmmmm

t

t r h

t

c



0.1235

y f

333325 t

c



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


■

■

■





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



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


■

■

■

■

■

■

Units




■

■ 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


■

■

■

Boundary conditions

■

►

►

■

►



Loading

■

■

■




194.246

94.246

sup

inf Mpa

Mpa



1235

235235

y f



4242413838

1

4112

192530

t

cmm

mmmm

t

c

91957.4

1

47.519

2

12220

t

c

t

c



α

34.0



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




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



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


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



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



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


yyk



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



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



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



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



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



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


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




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


383.1

19.11205235

37000001

917.0776.0

1,

N

N

N

N C k

z cr

Ed

z mz zz


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



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


■

■

■


y

z coefficient corresponding to non-dimensional slenderness z

z




yyk


yz k


zyk




zz k





y y y

z z z

yyk yyk

yz k yz k

zyk zyk

zz k zz k



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



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


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■

Units




Boundary conditions

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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



13.63 EC3 test 4: Class section classification and bending moment verification of an IPE300 column

Passed

13.63.1Description

13.63.2Background


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■

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■



Units


Boundary conditions

■

►

►

■





Combined oblique bending







14 Timber design



14.1 EC5: Verifying a timber purlin subjected to oblique bending

Passed

14.1.1Description

14.1.2Background


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■

■

■

■

■

■

Units

Geometry

■

■

■

■

■ 3

22

000666.06

20.01.0

6m

hbW y

■

322

000333.06

20.01.0

6 m

hb

W z




■

■

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



■

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


■

■

■



Stress SMy diagram

Stress SMz diagram

Maximum work ratio for strength verification







14.2 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending

Passed

14.2.1Description



14.3 EC5: Verifying a C24 timber beam subjected to shear force

Passed

14.3.1Description

14.3.2Background

of


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■

■

Units

Geometry

■

■

■

■

■


■

■



Boundary conditions

■

►

►

■

Loading

■

►

■

14.3.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads


■

■

■

■

■

■

■

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




■

■

■

Shear force, Fz, diagram

Design shear stress diagram

Shear strength work ratio diagram



■

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


■

Pa

m

N

A

N 675000

02.0

135002

■

Pam

mm

N

W

Lq

W

M

y

z

y

y 6

3

222



■

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


■

■

■



Instability factor, kcy

Instability factor, kcz

Maximum work ratio for stability verification







14.11 EC5: Verifying a timber column subjected to tensile forces

Passed

14.11.1Description

14.11.2Background


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Units

Geometry

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■

■




■

■

■

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


■

■

■



Work ratio SFx diagram







14.12 Verifying the timber elements shape sheet (TTAD #12337)

Passed

14.12.1Description



14.13 EC5: Verifying a timber beam subjected to simple bending

Passed

14.13.1Description

14.13.2Background


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■

Units

Geometry

■

■

■

■

■

■

322

0005.06

20.0075.0

6m

hbW y




■

■

■

Boundary conditions

■

►

►

■

Loading

■

■

14.13.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads


■

■

■

■

■

■

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



■

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


■

■

■



Work ratio diagram







14.14 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending andaxial compression

Passed

14.14.1Description

14.14.2Background


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■

■

■

■

Units



Geometry

■

■

■

■

■

322

000333.06

20.005.0

6m

hbW y


■

■

■

■

■

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



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



2

,

2,

1

z rel z z

z c

k k

k

2

,

2,

1

yrel y y

yc

k k k


■

■

■

■

■

■

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




■

■

■

Stress SFx diagram

Stress SMy diagram

Lateral-torsional stability work ratio



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