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Benchmarks Steel Code Check EN 1993
EN 1993-1-1 EN 1993-1-2 EN 1993-1-3 EN 1993-1-5
Benchmarks Steel Code Check EN 1993
Release: Scia Engineer 2010.0
Author: P. Van Tendeloo
Document: Benchmarks Steel Code Check EN 1993
Revision: 02/2010
All information in this document is subject to modification without prior notice. No part
or this document may be reproduced, stored in a database or retrieval system or
published, in any form or in any way, electronically, mechanically, by print, photo print,
microfilm or any other means without prior written permission from the publisher. SCIA
Software is not responsible for any direct or indirect damage because or imperfections in
the documentation and/or the software.
© Copyright 2010 SCIA Software. All rights reserved
Benchmarks Steel Code Check EN 1993
Benchmarks Steel Code Check EN 1993
INTRODUCTION 1
BENCHMARKS EN 1993-1-1 4
Benchmark 1: Global Imperfections .......................................................... 4 Benchmark 2: Bow Imperfections ............................................................. 6 Benchmark 3: Material Yield Strength ...................................................... 9 Benchmark 4: Effective Cross-Section Area ........................................... 12 Benchmark 5: Designer’s Guide Ex. 5.1 ................................................. 16 Benchmark 6: Designer’s Guide Ex. 6.2 ................................................. 18 Benchmark 7: Designer’s Guide Ex. 6.4 ................................................. 20 Benchmark 8: Designer’s Guide Ex. 6.5 ................................................. 21 Benchmark 9: Designer’s Guide Ex. 6.6 ................................................. 23 Benchmark 10: Designer’s Guide Ex. 6.7 ............................................... 25 Benchmark 11: Designer’s Guide Ex. 6.8 ............................................... 27 Benchmark 12: Designer’s Guide Ex. 6.9 ............................................... 32 Benchmark 13: Designer’s Guide Ex. 6.10 ............................................. 39 Benchmark 14: Designer’s Guide Ex. 13.1 ............................................. 45 Benchmark 15: Designer’s Guide Ex. 13.3 ............................................. 47 Benchmark 16: Nachweispraxis Beispiel 1 ............................................. 49 Benchmark 17: ECCS N°119 Worked Example 1 .................................. 51 Benchmark 18: ECCS N°119 Worked Example 2 .................................. 56 Benchmark 19: ECCS N°119 Worked Example 3 .................................. 64 Benchmark 20: ECCS N°119 Worked Example 4 .................................. 69 Benchmark 21: ECCS N°119 Worked Example 5 .................................. 75 Benchmark 22: ECCS N°119 Members in building frames .................... 84 Benchmark 23: Access Steel Document SX002a-EN-EU ...................... 93 Benchmark 24: Access Steel Document SX001a-EN-EU ...................... 95 Benchmark 25: Access Steel Document SX007a-EN-EU ...................... 98 Benchmark 26: Access Steel Document SX030a-EN-EU .................... 101 Benchmark 27: Access Steel Document SX029a-EN-EU .................... 114 Benchmark 28: Access Steel Document SX021a-EN-EU .................... 126
BENCHMARKS EN 1993-1-2 130
Benchmark 29: Access Steel Document SX044a-EN-EU .................... 130 Benchmark 30: Access Steel Document SX046a-EN-EU .................... 134 Benchmark 31: Access Steel Document SX047a-EN-EU .................... 138 Benchmark 32: Access Steel Document SX048a-EN-EU .................... 142 Benchmark 33: Access Steel Document SX043a-EN-EU .................... 146 Benchmark 34: Temperature Domain ................................................... 149 Benchmark 35: Combined Compression and Bending ......................... 154
BENCHMARKS EN 1993-1-3 164
Benchmark 36: Designer’s Guide Ex. 13.1 ........................................... 164 Benchmark 37: Designer’s Guide Ex. 13.2 ........................................... 167 Benchmark 38: Access Steel Document SX022a-EN-EU .................... 170 Benchmark 39: Access Steel Document SX023a-EN-EU .................... 173 Benchmark 40: Access Steel Document SX024a-EN-EU .................... 177 Benchmark 41: Access Steel Document SX025a-EN-EU .................... 180 Benchmark 42: Stiffened Cross-section ................................................ 182 Benchmark 43: Purlin Design in Uplift .................................................. 191
Benchmarks Steel Code Check EN 1993
1
Introduction
In this document, the results of Scia Engineer concerning the Steel Code Check according
to EN 1993 are compared to benchmark projects.
A total of 43 benchmarks are evaluated for EN 1993-1-1, EN 1993-1-2 and EN 1993-1-3.
In addition some benchmarks include parts of EN 1993-1-5.
An overview of supported articles as well as theoretical background on how specific code
rules have been implemented/supported within Scia Engineer can be found in the Steel
Code Check Theoretical Background document, revision 12/2009.
All checks are executed according to the regulations given in the following codes and
correction sheets:
Eurocode 3
Design of steel structures
Part 1 - 1 : General rules and rules for buildings
EN 1993-1-1:2005
Eurocode 3
Design of steel structures
Part 1 - 1 : General rules and rules for buildings
EN 1993-1-1:2005/AC:2009 Corrigendum
Eurocode 3
Design of steel structures
Part 1 - 2 : General rules - Structural fire design
EN 1993-1-2:2005
Eurocode 3
Design of steel structures
Part 1 - 2 : General rules - Structural fire design
EN 1993-1-2:2005/AC:2009 Corrigendum
Eurocode 3
Design of steel structures
Part 1-3: General rules
Supplementary rules for cold-formed members and sheeting
EN 1993-1-3:2006
Benchmarks Steel Code Check EN 1993
2
Eurocode 3
Design of steel structures
Part 1-3: General rules
Supplementary rules for cold-formed members and sheeting
EN 1993-1-3:2006/AC:2009 Corrigendum
Eurocode 3
Design of steel structures
Part 1.5 : Plated structural elements
EN 1993-1-5 : 2006
Eurocode 3
Design of steel structures
Part 1.5 : Plated structural elements
EN 1993-1-5 : 2006/AC:2009 Corrigendum
The following list gives an overview of the different benchmarks.
Benchmarks EN 1993-1-1
Benchmarks 1 to 4 concern manual calculations.
Benchmarks 5 to 15 concern examples of Designer’s Guide to EN 1993-1-1 Eurocode 3,
The Steel Construction Institute, 2005.
Benchmark 16 concerns an example of Nachweispraxis Biegeknicken und
Biegedrillknicken, Ernst & Sohn, 2002.
Benchmarks 17 to 22 concern examples of ECCS N°119 Rules for Member Stability in
EN 1993-1-1, Background documentation and design guidelines, ECCS, 2006.
Benchmarks 23 to 28 concern examples of Access Steel, which can be found on the
website http://www.access-steel.com/
Benchmarks EN 1993-1-2
Benchmarks 29 to 33 concern examples of Access Steel, which can be found on the
website http://www.access-steel.com/
Benchmarks 34 to 35 concern manual calculations.
Benchmarks Steel Code Check EN 1993
3
Benchmarks EN 1993-1-3
Benchmarks 36 to 37 concern examples of Designer’s Guide to EN 1993-1-1 Eurocode 3,
The Steel Construction Institute, 2005.
Benchmarks 38 to 41 concern examples of Access Steel, which can be found on the
website http://www.access-steel.com/
Benchmarks 42 to 43 concern manual calculations.
For each Benchmark, the reference results and the Scia Engineer output are given. Where
needed, the results are followed by comments.
More background information concerning each benchmark can be found in the specified
references.
For those benchmarks in which the verification is done using both Interaction Method 1
and 2 two Scia Engineer project files are provided (XXX_1.esa and XXX_2.esa).
Benchmark 1: Global Imperfections
4
Benchmarks EN 1993-1-1
Benchmark 1: Global Imperfections
Project file: EN_Benchmark01.esa
Scia Engineer Version 10.0.86
Introduction
In this benchmark, the equivalent sway imperfections according to EN 1993-1-1
are checked.
A portal frame is modeled as shown on the following picture. The frame has a
total height of 12m and is loaded on the top side of the columns by 100 kN point
loads. The column bases are taken as fixed, the beam-column connections as
hinged.
Benchmark 1: Global Imperfections
5
Reference Results
The results are checked by a manual calculation.
3
2577,0
12
22
hh
3
2h
816,03
115,0
115,0
mm
0027217,0816,0577,0200
10 mh
This results in a leverage arm e for the point loads at the top:
mtghe 03266,00027217,012)(
Due to this leverage arm, the expected moment at the column bases is calculated
as follows:
kNmmkNeFM 266,303266,0100
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 2: Bow Imperfections
6
Benchmark 2: Bow Imperfections
Project file: EN_Benchmark02.esa
Scia Engineer Version 10.0.86
Introduction
In this benchmark, the local bow imperfections according to EN 1993-1-1 are
checked.
A set of six Euler columns is modeled. The columns have length 4m and cross-
section IPE 240. For each column bow imperfections and normal force loading
are defined as shown in the following table:
Column Bow imperfection Normal Force [kN]
B1 According to code – elastic 100
B2 According to code – plastic 100
B3 According to code – elastic – only if required 100
B4 According to code – plastic – only if required 100
B5 According to code – elastic – only if required 1300
B6 According to code – plastic – only if required 1300
Benchmark 2: Bow Imperfections
7
Reference Results
The results are checked by a manual calculation.
IPE 240 Buckling curve y-y: a
Buckling curve z-z : b
Elastic analysis: curve a: 300
10
L
e
curve b: 250
10
L
e
Plastic analysis: curve a: 250
10
L
e
curve b: 200
10
L
e
For the imperfections ‘if required’ the critical Euler load is calculated:
2
2
2
2
4000
38920000210000,
L
EIyNcr
y5041,64 kN
25% of Ncr,y = 1260,41 kN
2
2
2
2
4000
2836000210000,
L
EIzNcr z 367,37 kN
25% of Ncr,z = 91,84 kN
With a length of 4m the imperfection value e0 can be calculated for each column
for each direction. Due to these imperfection values, the normal force loading will
cause bending moments My and Mz in the columns. The expected results are
shown in the following table.
Benchmark 2: Bow Imperfections
8
Column Buckling
axis
e0 [mm] N [kN] My [kNm] Mz [kNm]
B1 y-y
z-z
13,33
16
100 1,33
1,6
B2 y-y
z-z
16
20
100 1,6
2
B3 y-y
z-z
0
16
100 0
1,6
B4 y-y
z-z
0
20
100 0
2
B5 y-y
z-z
13,33
16
1300 17,33
20,8
B6 y-y
z-z
16
20
1300 20,8
26
For columns B3 and B4 the normal force loading is lower then the limit for
buckling around the y-y axis so no imperfection has to be applied in that case. For
buckling around the z-z axis the imperfection is required.
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 3: Material Yield Strength
9
Benchmark 3: Material Yield Strength
Project file: EN_Benchmark03.esa
Scia Engineer Version 10.0.86
Introduction
In this benchmark, two items are checked:
- Reduction of the yield strength in function of the thickness for rolled
sections, according to EN 1993-1-1.
- Calculation of the average yield strength for cold-formed sections
according to EN 1993-1-3.
Two sections are modeled: hot rolled HE1000X393 fabricated from S235 and a
cold-formed RHSCF300/100/12.5 fabricated from S275.
Reference Results
The results are checked by a manual calculation.
CS1 - HE1000X393 – S235
tf = 43,9 mm > 40 mm fy = 215 N/mm²
With area A = 50020 mm² and M0 =1,00 the compression capacity will be:
kNfyA
NRdM
3,1075400,1
21550020
0
Benchmark 3: Material Yield Strength
10
CS2 - RHSCF300/100/12.5 – S275
The average yield strength is calculated as follows:
2
2 fybfufybfu
A
kntfybfya
g
With: fyb = 275 N/mm²
fu = 430 N/mm²
Ag = 8700 mm²
k = 7 for cold rolling
n = 4 (90° bends)
t = 12,5 mm
2
275430275430
8700
5,1247275
2
fya
5,35295,352fya
fya = 352,5 N/mm²
With M0 =1,00 the compression capacity will be:
kNfyA
NRdM
75,306600,1
5,3528700
0
Scia Engineer Results
Results for CS1 - HE1000X393 – S235
Benchmark 3: Material Yield Strength
11
Results for CS2 - RHSCF300/100/12.5 – S275
Comments
The results correspond to the benchmark results.
Benchmark 4: Effective Cross-Section Area
12
Benchmark 4: Effective Cross-Section Area
Project file: EN_Benchmark04.esa
Scia Engineer Version 10.0.86
Introduction
In this benchmark, the effective cross-section is calculated for a rolled section
with class 4 web. The cross-section is of type IPE 600, fabricated from S355 and
loaded by uniform compression.
The classification is done according to EN 1993-1-1, the calculation of the
effective cross-section area is done according to EN 1993-1-5.
Reference Results
The results are checked by a manual calculation.
S355 = 0,81
IPE 600 H = 600 mm
B = 220 mm
tf = 19 mm
tw = 12 mm
r = 24 mm
A = 15600 mm²
Classification for outstand flanges
242
12
2
220
22r
twBc 80 mm
19
80
tf
c4,21
Limit for class 1: 9 = 7,32
4,21 < 7,32
The flanges are classified as class 1
Benchmark 4: Effective Cross-Section Area
13
Classification for internal compression parts
24219260022 rtfHc 514 mm
12
514
tw
c42,83
Limit for class 3: 42 = 34,17
42,83 > 34,17
The web is classified as class 4
Calculation of effective area
b c = 514 mm
= 1,0
k = 4,0
Benchmark 4: Effective Cross-Section Area
14
481,04,28
12514
p 0,9369
29369,0
13055,09369,00,8228
5148228,0effb 422,83 mm
be1 = be2 = 211,46 mm
220
12
19
211
24
Aeff = 220 x 19 x 2 + 211,46 x 12 x 2 + 2 x 24x 12 = 14011,16 mm²
With M0 =1,00 the compression capacity will be:
kNfyA
NRdM
eff96,4973
00,1
35516,14011
0
Benchmark 4: Effective Cross-Section Area
15
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 5: Designer’s Guide Ex. 5.1
16
Benchmark 5: Designer’s Guide Ex. 5.1
Project file: EN_Benchmark05.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 5.1: Cross-section classification under
combined bending and compression of Designer’s Guide to EN 1993-1-1
Eurocode 3, The Steel Construction Institute, 2005.
A member is to be designed to carry combined bending and axial load. In the
presence of a major axis bending moment and an axial force of 300 kN, the cross-
section classification is determined of a 406 x 178 x 54 UB in grade S275 steel.
Reference Results
The reference gives following results:
Classification under pure compression
Flanges c/tf 6,86
Class 1 limit 8,32
Flanges Class 1
Web c/tw 46,81
Class 3 limit 38,8
Web Class 4
Classification under combined loading
Flanges c/tf 6,86
Class 1 limit 8,32
Flanges Class 1
Web c/tw 46,81
Class 2 limit 52,33
Web Class 2
Benchmark 5: Designer’s Guide Ex. 5.1
17
Scia Engineer Results
Classification under pure compression
Classification under combined loading
Comments
The results correspond to the benchmark results.
Benchmark 6: Designer’s Guide Ex. 6.2
18
Benchmark 6: Designer’s Guide Ex. 6.2
Project file: EN_Benchmark06.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.2: Cross-section resistance in compression
of Designer’s Guide to EN 1993-1-1 Eurocode 3, The Steel Construction Institute,
2005.
A 254 x 254 x 73 UC is to be used as a short compression member. The resistance
of the cross-section in compression is calculated assuming grade S355 steel.
Reference Results
The reference gives following results:
Classification
Flanges c/tf 7,77
Class 2 limit 8,14
Flanges Class 2
Web c/tw 23,29
Class 1 limit 26,85
Web Class 1
Compression resistance
Nc,Rd 3305 kN
Benchmark 6: Designer’s Guide Ex. 6.2
19
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 7: Designer’s Guide Ex. 6.4
20
Benchmark 7: Designer’s Guide Ex. 6.4
Project file: EN_Benchmark07.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.4: Shear resistance of Designer’s Guide to
EN 1993-1-1 Eurocode 3, The Steel Construction Institute, 2005.
The shear resistance is determined of a 229 x 89 rolled channel section in grade
S275 steel loaded parallel to the web.
Reference Results
The reference gives following results:
Shear resistance
Av 2092 mm²
Vpl,Rd 332 kN
Shear buckling does not need to be considered
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 8: Designer’s Guide Ex. 6.5
21
Benchmark 8: Designer’s Guide Ex. 6.5
Project file: EN_Benchmark08.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.5: Cross-section resistance under combined
bending and shear of Designer’s Guide to EN 1993-1-1 Eurocode 3, The Steel
Construction Institute, 2005.
A short span (1,4m), simply supported, laterally restrained beam is to be designed
to carry a central point load of 1050 kN. A 406 x 178 x 74 UB in grade S275 steel
is assessed for its suitability for this application.
Reference Results
The reference gives following results:
Classification
Flanges c/tf 4,68
Class 1 limit 8,32
Flanges Class 1
Web c/tw 37,94
Class 1 limit 66,56
Web Class 1
Bending resistance
Mc,y,Rd 412 kNm
Shear resistance
Av 4184 mm²
Vpl,Rd 689,2 kN
Shear buckling does not need to be considered
Benchmark 8: Designer’s Guide Ex. 6.5
22
Resistance to combined bending and shear
My,V,Rd 386,8 kNm
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 9: Designer’s Guide Ex. 6.6
23
Benchmark 9: Designer’s Guide Ex. 6.6
Project file: EN_Benchmark09.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.6: Cross-section resistance under combined
bending and compression of Designer’s Guide to EN 1993-1-1 Eurocode 3, The
Steel Construction Institute, 2005.
A member is to be designed to carry a combined major axis bending moment and
an axial force. In this example, a cross-section check is performed to determine
the maximum bending moment that can be carried by a 457 x 191 x 98 UB in
grade S235 steel in the presence of an axial force of 1400 kN.
Reference Results
The reference gives following results:
Classification
Flanges c/tf 4,11
Class 1 limit 9,0
Flanges Class 1
Web c/tw 35,75
Class 2 limit 38,0
Web Class 2
Compression resistance
Npl,Rd 2937,5 kN
Bending resistance
Mpl,y,Rd 524,5 kNm
Benchmark 9: Designer’s Guide Ex. 6.6
24
Resistance to combined bending and axial force
MN,y,Rd 342,2 kNm
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 10: Designer’s Guide Ex. 6.7
25
Benchmark 10: Designer’s Guide Ex. 6.7
Project file: EN_Benchmark10.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.7: Buckling resistance of a compression
member of Designer’s Guide to EN 1993-1-1 Eurocode 3, The Steel Construction
Institute, 2005.
A circular hollow section member is used as an internal column in a multi-storey
building. The column has pinned boundary conditions at each end, and the inter-
storey height is 4m. The critical combination of actions results in a design axial
force of 1630 kN. The suitability of a hot rolled 244,5 x 10 CHS in grade S275
steel is assessed for this application.
Reference Results
The reference gives following results:
Classification
Tube d/t 24,5
Class 1 limit 42,7
Tube Class 1
Compression resistance
Nc,Rd 2026,8 kN
Member Buckling resistance in compression
Ncr 6571 kN
red 0,56
curve a
0,21
0,91
Nb,Rd 1836,5 kN
Benchmark 10: Designer’s Guide Ex. 6.7
26
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 11: Designer’s Guide Ex. 6.8
27
Benchmark 11: Designer’s Guide Ex. 6.8
Project file: EN_Benchmark11.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.8: Lateral Torsional Buckling resistance of
Designer’s Guide to EN 1993-1-1 Eurocode 3, The Steel Construction Institute,
2005.
A simply supported primary beam is required to span 10,8m and to support two
secondary beams. The secondary beams are connected through fin plates to the
web of the primary beam, and full lateral restraint may be assumed at these points.
A 762 x 267 x 173 UB section is considered in grade S275 steel.
For Lateral Torsional Buckling the general case is used.
Reference Results
The reference gives following results:
Classification
Flanges c/tf 5,08
Class 1 limit 8,32
Flanges Class 1
Web c/tw 48,0
Class 1 limit 66,6
Web Class 1
Benchmark 11: Designer’s Guide Ex. 6.8
28
Bending resistance
Mc,y,Rd 1704 kNm
Shear resistance
Av 9813 mm²
Vpl,Rd 1959 kN
Shear buckling does not need to be considered
Resistance to combined bending and shear
My,V,Rd 1704 kNm
Lateral torsional buckling: segment BC
C1 1,052
Mcr 5699 kNm
red LT 0,55
LT 0,34
LT 0,86
Mb,Rd 1469 kNm
Lateral torsional buckling: segment CD
C1 1,879
Mcr 4311 kNm
red LT 0,63
LT 0,34
LT 0,82
Mb,Rd 1402 kNm
Benchmark 11: Designer’s Guide Ex. 6.8
29
Scia Engineer Results
Benchmark 11: Designer’s Guide Ex. 6.8
30
LTB for segment BC:
LTB for segment CD:
Benchmark 11: Designer’s Guide Ex. 6.8
31
Comments
- The results correspond to the benchmark results.
- The benchmark gives a wrong moment diagram. In Scia Engineer the loading has
been adapted to obtain the same diagram since the values of the end moments
influence the calculation of the C1 factor.
- A small difference in the values for Mcr is caused by a different Iw section
property: Reference Iw = 9390 x 10^9 mm^6 Scia Engineer Iw = 9551,7 x
10^9 mm^6.
Benchmark 12: Designer’s Guide Ex. 6.9
32
Benchmark 12: Designer’s Guide Ex. 6.9
Project file: EN_Benchmark12.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.9: Member resistance under combined
major axis bending and axial compression of Designer’s Guide to EN 1993-1-1
Eurocode 3, The Steel Construction Institute, 2005.
A rectangular hollow section member is to be used as a primary floor beam of a
7,2 m span in a multi-storey building. Two design point loads of 58 kN are
applied to the primary beam from secondary beams. The secondary beams are
connected through fin plates to the webs of the primary beam, and full lateral and
torsional restraint may be assumed at these points. The primary beam is also
subjected to a design axial force of 90 kN.
The suitability of a hot rolled 200 x 100 x 16 RHS in grade S355 steel is assessed
for this application.
For Lateral Torsional Buckling the general case is used.
The interaction factors kij for combined bending and compression are determined
using alternative method 1 (Annex A).
Benchmark 12: Designer’s Guide Ex. 6.9
33
Reference Results
The reference gives following results:
Classification (under pure compression)
Web c/tw 9,50
Class 1 limit 26,85
Web Class 1
Compression resistance
Nc,Rd 2946,5 kN
Shear resistance
Av 5533,3 mm²
Vpl,Rd 1134 kN
Shear buckling does not need to be considered
Bending resistance
Mc,y,Rd 174,3 kNm
Resistance to combined bending, shear and axial force
My,NV,Rd 174,3 kNm
Member Buckling resistance in compression
Ncr,y 1470 kN Ncr,z 4127 kN
red ,y 1,42 red ,z 0,84
y 0,21 z 0,21
y 0,41 z 0,77
Nb,y,Rd 1209 kN Nb,z,Rd 2266 kN
Benchmark 12: Designer’s Guide Ex. 6.9
34
Member Buckling resistance in bending: segment BC
C1 1,0
Mcr 3157 kNm
red LT 0,23
LT 0,76
LT 0,97
Mb,Rd 169,5 kNm
Verification according to Method 1
red ,0 0,23
Cmy,0 1,01
aLT 0,189
bLT 0
dLT 0
Cmy 1,01
CmLT 1,00
y 0,96
z 0,99
wy 1,33
wz 1,27
npl 0,03
Cyy 0,98
Czy 0,95
kyy 1,06
kzy 0,69
eq. (6.61) 0,94
eq. (6.62) 0,61
Benchmark 12: Designer’s Guide Ex. 6.9
35
Scia Engineer Results
Benchmark 12: Designer’s Guide Ex. 6.9
36
Benchmark 12: Designer’s Guide Ex. 6.9
37
Comments
- The results correspond to the benchmark results.
- In Scia Engineer an RRW section was used to obtain the same Wpl.
- There is a slight difference in Mcr due to the fact the reference ignores the
warping contribution.
Benchmark 12: Designer’s Guide Ex. 6.9
38
- According to EN 1993-1-1 art. 6.3.2.1(4) the effect of lateral-torsional buckling
may be ignored ( LT = 1,00) in case:
with = 0,40 by default
0,23 < 0,40 => LT = 1,00
The reference does not take this into account and thus has LT = 0,97.
- The critical check is at 2,4m. To obtain the shear check and classification for pure
compression, member data are used for checking the position at 0m.
Benchmark 13: Designer’s Guide Ex. 6.10
39
Benchmark 13: Designer’s Guide Ex. 6.10
Project file: EN_Benchmark13.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 6.10: Member resistance under combined bi-
axial bending and axial compression of Designer’s Guide to EN 1993-1-1
Eurocode 3, The Steel Construction Institute, 2005.
An H section member of length 4,2m is to be designed as a ground floor column
in a multi-storey building. The frame is moment resisting in-plane and pinned out-
of-plane, with diagonal bracing provided in both directions. The column is
subjected to major-axis bending due to horizontal forces and minor axis bending
due to eccentric loading from the floor beams. From the structural analysis, the
design effects are shown in following figure. The suitability of a hot rolled 305 x
305 x 240 H section in grade S275 steel is assessed for this application.
For Lateral Torsional Buckling the general case is used.
The interaction factors kij for combined bending and compression are determined
using alternative method 2 (Annex B).
Reference Results
The reference gives following results:
Classification
Flanges c/tf 3,51
Class 1 limit 8,32
Flanges Class 1
Benchmark 13: Designer’s Guide Ex. 6.10
40
Web c/tw 10,73
Class 1 limit 30,51
Web Class 1
Compression resistance
Nc,Rd 8415 kN
Bending resistance
Mc,y,Rd 1168 kNm
Mc,z,Rd 536,5 kNm
Shear resistance
Av,z 8605,82 mm²
Vpl,z,Rd 1366,36 kN
Av,y 24227 mm²
Vpl,y,Rd 3847 kN
Shear buckling does not need to be considered
Resistance to combined bending, shear and axial force
My,NV,Rd 773,8 kNm
Mz,NV,Rd 503,9 kNm
2
2,04
Member Buckling resistance in compression
Ncr,y 153943 kN Ncr,z 23863 kN
red ,y 0,23 red ,z 0,59
y 0,34 z 0,49
y 0,99 z 0,79
Nb,y,Rd 8314 kN Nb,z,Rd 6640 kN
Benchmark 13: Designer’s Guide Ex. 6.10
41
Member Buckling resistance in bending
C1 2,752
Mcr 17114 kNm
red LT 0,26
LT 0,21
LT 0,99
Mb,Rd 1152 kNm
Verification according to Method 2
Cmy 0,40
Cmz 0,60
CmLT 0,40
kyy 0,41
kzz 0,78
kyz 0,47
kzy 0,79
eq. (6.61) 0,66
eq. (6.62) 0,97
Scia Engineer Results
Benchmark 13: Designer’s Guide Ex. 6.10
42
Benchmark 13: Designer’s Guide Ex. 6.10
43
Benchmark 13: Designer’s Guide Ex. 6.10
44
Comments
- The reference applies a wrong formula for Av,z in the shear resistance check. The
results shown above for Av,z and Vpl,z,Rd are those corrected by manual
calculation.
- There is a slight difference in Mcr due to a different C1 factor. Reference C1 =
2,752 Scia Engineer C1 = 2,70.
In Scia Engineer the C1 factor for end-moment loading is calculated according to
the approximate formula (F.3) of informative annex F of ENV 1993-1-1:1992.
This formula is limited to 2,70.
- According to EN 1993-1-1 art. 6.3.2.1(4) the effect of lateral-torsional buckling
may be ignored ( LT = 1,00) in case:
with = 0,40 by default
0,26 < 0,40 => LT = 1,00
The reference does not take this into account and thus has LT = 0,99.
- To determine the interaction factors kij using alternative method 2 (Annex B) a
distinction is made between members not susceptible to torsional deformations
(Table B.1) and members susceptible to torsional deformations (Table B.2).
The reference concludes that the member is susceptible to torsional deformations
and uses Table B.2 leading to a kzy value of 0,79.
However, due to the previous point, since LT = 1,00 the member is considered
within Scia Engineer as being non-susceptible to LT-buckling and thus Table B.1
is applied leading to a kzy value of 0,6 kyy = 0,6 * 0,406 = 0,2436
Benchmark 14: Designer’s Guide Ex. 13.1
45
Benchmark 14: Designer’s Guide Ex. 13.1
Project file: EN_Benchmark14.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 13.1: Calculation of section properties for
local buckling of Designer’s Guide to EN 1993-1-1 Eurocode 3, The Steel
Construction Institute, 2005.
The effective area and the horizontal shift in neutral axis due to local buckling are
calculated for a 200 x 65 x 1,6 lipped channel in zinc-coated steel with a nominal
yield strength of 280 N/mm². The section is subjected to pure compression.
The properties are calculated from the idealized section given in the reference.
Reference Results
The reference gives following results:
Effective section properties
Aeff 341,5 mm²
eN 8,66 mm
Benchmark 14: Designer’s Guide Ex. 13.1
46
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 15: Designer’s Guide Ex. 13.3
47
Benchmark 15: Designer’s Guide Ex. 13.3
Project file: EN_Benchmark15.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 13.3: Member resistance in compression
(checking flexural, torsional and torsional-flexural buckling) of Designer’s Guide
to EN 1993-1-1 Eurocode 3, The Steel Construction Institute, 2005.
The member resistance of a 100 x 50 x 3 plain channel section column subjected
to compression is calculated. The column length is 1,5m, with pinned end
conditions, so the effective length is assumed equal to the system length. The steel
has yield strength 280 N/mm².
Reference Results
The reference gives following results:
Member resistance in compression
A 555 mm²
Ncr,y 787 kN
Ncr,z 127 kN
Ncr,T 121 kN
Sigma,cr,T 218 N/mm²
Ncr,TF 114 kN
Sigma,cr,TF 205 N/mm²
red 1,16
0,49
0,45
Nb,Rd 69,17 kN
Benchmark 15: Designer’s Guide Ex. 13.3
48
Scia Engineer Results
Comments
- The results correspond to the benchmark results.
- The reference calculates a wrong formula for Nb,Rd. The result shown above for
Nb,Rd is that corrected by manual calculation.
Benchmark 16: Nachweispraxis Beispiel 1
49
Benchmark 16: Nachweispraxis Beispiel 1
Project file: EN_Benchmark16.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Beispiel 1: Schnittgrössenberechnung und
Spannungsberechnung bei zweiachsiger Biegung mit Torsion of Nachweispraxis
Biegeknicken und Biegedrillknicken, Ernst & Sohn, 2002.
A member with forked end supports is loaded in axial compression, bi-axial
bending and torsion. The member concerns an IPE 200 of steel grade S235. A
direct stress check is performed according to EN 1993-1-3 in the middle of the
member which includes the direct stress due to warping.
Reference Results
The reference gives following results:
I
My
Iz
Mzz
Iy
My
A
N
23,18)9,47(
12990
78,624)5(
142
270)10(
1940
1170
5,28
6,12
cm
kN
284,18204,2307,9531,6042,4
mm
N
With M0 =1,00 the Stress check according to EN 1993-1-3 formula (6.11a) is:
0
,
M
Edtot
fya
00,1
23584,182 Unity check: 0,78
Benchmark 16: Nachweispraxis Beispiel 1
50
Scia Engineer Results
Comments
The results correspond to the benchmark results.
Benchmark 17: ECCS N°119 Worked Example 1
51
Benchmark 17: ECCS N°119 Worked Example 1
Project file: EN_Benchmark17_1.esa & EN_Benchmark17_2.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Worked Example 1 of ECCS N°119 Rules for Member
Stability in EN 1993-1-1, Background documentation and design guidelines,
ECCS, 2006.
This first worked example deals with the basic case of in-plane behaviour. The
beam-column is subjected to compression and triangular major axis bending
moment. The member is so restrained that both lateral and lateral torsional
displacements are prevented.
The interaction factors kij for combined bending and compression are determined
using both alternative method 1 (Annex A) and alternative method 2 (Annex B).
Reference Results
The reference gives following results:
Classification
Flanges c/tf 4,1
Class 1 limit 9,0
Flanges Class 1
Web c/tw 28,39
Class 1 limit 33,00
Web Class 1
Benchmark 17: ECCS N°119 Worked Example 1
52
Compression resistance
Nc,Rd 669 kN
Bending resistance
Mc,y,Rd 51,8 kNm
Shear resistance
Av,z 1400 mm²
Vpl,z,Rd 190 kN
Shear buckling does not need to be considered
Resistance to combined bending, shear and axial force
My,NV,Rd 44,7 kNm
Member Buckling resistance in compression
Ncr,y 3287 kN
red ,y 0,451
y 0,21
y 0,939
Verification according to Method 1
Cmy,0 0,782
bLT 0
Cmy 0,782
y 0,996
wy 1,135
Cyy 1,061
eq. (6.61) 0,985
Benchmark 17: ECCS N°119 Worked Example 1
53
Verification according to Method 2
Cmy 0,6
kyy 0,65
eq. (6.61) 0,874
Scia Engineer Results
Benchmark 17: ECCS N°119 Worked Example 1
54
Verification according to Method 1
Benchmark 17: ECCS N°119 Worked Example 1
55
Verification according to Method 2
Comments
- The results correspond to the benchmark results.
- The reference calculates a wrong value for c in the classification of the web. The
result shown above for c is that corrected by manual calculation.
- The reference calculates a wrong value for Av,z in the shear resistance check. The
result shown above for Av,z is that corrected by manual calculation.
Benchmark 18: ECCS N°119 Worked Example 2
56
Benchmark 18: ECCS N°119 Worked Example 2
Project file: EN_Benchmark18_1.esa & EN_Benchmark18_2.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Worked Example 2 of ECCS N°119 Rules for Member
Stability in EN 1993-1-1, Background documentation and design guidelines,
ECCS, 2006.
This second worked example deals with spatial behaviour. The beam-column is
subjected to compression, transverse forces and major axis end moments. the
transverse load is assumed to act at the shear centre. Lateral torsional buckling is
not prevented, and may therefore occur.
The interaction factors kij for combined bending and compression are determined
using both alternative method 1 (Annex A) and alternative method 2 (Annex B).
Reference Results
The reference gives following results:
Classification
Flanges c/tf 4,6
Class 1 limit 9,0
Flanges Class 1
Web c/tw 41,8
Class 2 limit 43,00
Web Class 2
Benchmark 18: ECCS N°119 Worked Example 2
57
Compression resistance
Nc,Rd 2714 kN
Bending resistance
Mc,y,Rd 516 kNm
Shear resistance
Av,z 5990 mm²
Vpl,z,Rd 819 kN
Shear buckling does not need to be considered
Resistance to combined bending, shear and axial force
My,NV,Rd 468 kNm
Member Buckling resistance in compression
Ncr,y 81549 kN Ncr,z 3624 kN
red ,y 0,182 red ,z 0,865
y 0,21 z 0,34
y 1,00 z 0,683
Member Buckling resistance in bending (General)
C1 2,15
Mcr 2179 kNm
red LT 0,486
LT 0,34
LT 0,89
kc 0,653
f 0,861
LT,mod 1,00
Benchmark 18: ECCS N°119 Worked Example 2
58
Member Buckling resistance in bending (Rolled)
red LT 0,473
LT 0,49
LT 0,959
kc 0,653
f 0,864
LT,mod 1,00
Verification according to Method 1
y 1,00
wy 1,138
z 0,918
wz 1,5
Cmy,0 0,789
Mcr0 1014 kNm
red 0 0,713
aLT 0,998
bLT 0
dLT 0
Cmy 0,919
Cyy 1,003
Czy 0,893
eq. (6.61) 0,936
eq. (6.62) 0,777
Benchmark 18: ECCS N°119 Worked Example 2
59
Verification according to Method 2
Cmy 0,495
Cm,LT 0,495
kyy 0,492
kzy 0,847
eq. (6.61) 0,628
eq. (6.62) 1,006
Scia Engineer Results
Benchmark 18: ECCS N°119 Worked Example 2
60
Verification according to Method 1
Benchmark 18: ECCS N°119 Worked Example 2
61
Benchmark 18: ECCS N°119 Worked Example 2
62
Verification according to Method 2
Comments
- The reference calculates a wrong value for Av,z in the shear resistance check. The
result shown above for Av,z is that corrected by manual calculation.
- Since it concerns a case of combined loading, the FriLo LTB solver is used to
calculate the exact Mcr through an eigenvalue solution. The reference uses an
approximate graphic for determining C1 (and thus Mcr).
Reference Mcr = 2179 kNm Scia Engineer Mcr = 2310,41 kNm.
Benchmark 18: ECCS N°119 Worked Example 2
63
- According to EN 1993-1-1 art. 6.3.2.1(4) the effect of lateral-torsional buckling
may be ignored ( LT = 1,00) in case:
with = 0,40 by default
350 / 2310,41 = 0,1515 < 0,16 => LT = 1,00
The reference does not take this into account and thus has LT = 0,89.
- In the determination of Cmy,0 for method 1 the reference assumes the moment
diagram to be linear which is not the case. The reference thus uses the linear
approximation where Scia Engineer uses the correct general method for
calculating Cmy,0. The reference is thus not consistent: for C1 the combined
loading is taken into account, but for Cmy,0 not.
- In the verification according to method 1, the reference uses the ‘General Case’
for LTB. However, the reference also applies the reduction factor f to calculate
LT,mod in this case. In EN 1993-1-1 this reduction is only specified for the
‘Rolled sections and equivalent welded sections Case’ and not for the ‘General
Case’.
Due to the differences in the LTB reduction factor and in the Cmy,0 factor, the
eventual verification formula’s have differences.
- In the verification according to method 2, the reference uses the ‘Rolled sections
and equivalent welded sections Case’ for LTB. For determination of kc, the
reference uses specific tables according to BS 5950. In Scia Engineer the default
table according to EN 1993-1-1 is used.
Reference kc = 0,653 Scia Engineer kc = 0,91
- To determine the interaction factors kij using alternative method 2 (Annex B) a
distinction is made between members not susceptible to torsional deformations
(Table B.1) and members susceptible to torsional deformations (Table B.2).
The reference concludes that the member is susceptible to torsional deformations
and uses Table B.2 leading to a kzy value of 0,847.
However, since LT = 1,00 the member is considered within Scia Engineer as
being non-susceptible to LT-buckling and thus Table B.1 is applied leading to a
kzy value of 0,6 kyy = 0,6 * 0,492 = 0,2952
Benchmark 19: ECCS N°119 Worked Example 3
64
Benchmark 19: ECCS N°119 Worked Example 3
Project file: EN_ Benchmark19_1.esa & EN_Benchmark19_2.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Worked Example 3 of ECCS N°119 Rules for Member
Stability in EN 1993-1-1, Background documentation and design guidelines,
ECCS, 2006.
This third worked example deals with spatial behaviour. The beam-column is
subjected to compression and transverse forces causing major axis bending.
Lateral torsional buckling is not a potential mode of failure because of the shape
of the cross-section.
The interaction factors kij for combined bending and compression are determined
using both alternative method 1 (Annex A) and alternative method 2 (Annex B).
Reference Results
The reference gives following results:
Classification
Web c/tw 14,0
Class 1 limit 33,00
Web Class 1
Compression resistance
Nc,Rd 1316 kN
Benchmark 19: ECCS N°119 Worked Example 3
65
Bending resistance
Mc,y,Rd 82,7 kNm
Shear resistance
Av,z 3600 mm²
Vpl,z,Rd 488 kN
Shear buckling does not need to be considered
Member Buckling resistance in compression
Ncr,y 3600 kN Ncr,z 1161 kN
red ,y 0,605 red ,z 1,065
y 0,21 z 0,21
y 0,888 z 0,62
Verification according to Method 1
y 0,969
wy 1,266
z 0,543
wz 1,184
Cmy,0 1,007
bLT 0
dLT 0
Cmy 1,007
Cyy 0,868
Czy 0,524
eq. (6.61) 0,946
eq. (6.62) 1,131
Benchmark 19: ECCS N°119 Worked Example 3
66
Verification according to Method 2
Cmy 0,95
kyy 1,213
kzy 0,728
eq. (6.61) 0,904
eq. (6.62) 1,112
Scia Engineer Results
Benchmark 19: ECCS N°119 Worked Example 3
67
Verification according to Method 1
Benchmark 19: ECCS N°119 Worked Example 3
68
Verification according to Method 2
Comments
- The results correspond to the benchmark results.
- The reference uses an RHS200x100x10 which has different properties than the
same section according to British Standard, Stahlbau Zentrum Schweiz or Voest-
Alpine Krems. In Scia Engineer the section according to British Standard has
been used. Due to differences in the cross-section properties, small differences in
the classification and verification occur.
Benchmark 20: ECCS N°119 Worked Example 4
69
Benchmark 20: ECCS N°119 Worked Example 4
Project file: EN_Benchmark20_1.esa & EN_Benchmark20_2.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Worked Example 4 of ECCS N°119 Rules for Member
Stability in EN 1993-1-1, Background documentation and design guidelines,
ECCS, 2006.
This fourth worked example deals with spatial behaviour. The beam-column is
subjected to compression and biaxial bending. Lateral torsional buckling is not a
potential mode of failure because of the shape of the cross-section.
The interaction factors kij for combined bending and compression are determined
using both alternative method 1 (Annex A) and alternative method 2 (Annex B).
Reference Results
The reference gives following results:
Classification
Web c/tw 14,0
Class 1 limit 33,00
Web Class 1
Compression resistance
Nc,Rd 1316 kN
Benchmark 20: ECCS N°119 Worked Example 4
70
Bending resistance
Mc,y,Rd 82,7 kNm
Mc,z,Rd 49,8 kNm
Shear resistance
Av,z 3600 mm²
Vpl,z,Rd 488 kN
Av,y 2000 mm²
Vpl,y,Rd 271 kN
Shear buckling does not need to be considered
Resistance to combined bending, shear and axial force
My,NV,Rd 82,7 kNm
Mz,NV,Rd 44,9 kNm
1,763
1,763
Member Buckling resistance in compression
Ncr,y 3600 kN Ncr,z 1161 kN
red ,y 0,605 red ,z 1,065
y 0,21 z 0,21
y 0,888 z 0,620
Verification according to Method 1
y 0,990
wy 1,266
z 0,883
wz 1,184
Cmy,0 0,998
Cmy 0,998
Benchmark 20: ECCS N°119 Worked Example 4
71
Cmz,0 0,759
Cmz 0,759
bLT 0
dLT 0
Cyy 0,954
Cyz 0,919
Czy 0,827
Czz 1,012
eq. (6.61) 0,923
eq. (6.62) 0,988
Verification according to Method 2
Cmy 0,933
Cmz 0,6
kyy 1,030
kyz 0,466
kzy 0,618
kzz 0,777
eq. (6.61) 0,817
eq. (6.62) 0,903
Scia Engineer Results
Benchmark 20: ECCS N°119 Worked Example 4
72
Benchmark 20: ECCS N°119 Worked Example 4
73
Verification according to Method 1
Benchmark 20: ECCS N°119 Worked Example 4
74
Verification according to Method 2
Comments
- The results correspond to the benchmark results.
- The reference uses an RHS200x100x10 which has different properties than the
same section according to British Standard, Stahlbau Zentrum Schweiz or Voest-
Alpine Krems. In Scia Engineer the section according to British Standard has
been used. Due to differences in the cross-section properties, small differences in
the classification and verification occur.
- A small difference in shear resistance occurs due to the fact that the reference uses
a formula to calculate the shear area which is different than the formula given in
EN 1993-1-1.
Benchmark 21: ECCS N°119 Worked Example 5
75
Benchmark 21: ECCS N°119 Worked Example 5
Project file: EN_Benchmark21_1.esa & EN_Benchmark21_2.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Worked Example 5 of ECCS N°119 Rules for Member
Stability in EN 1993-1-1, Background documentation and design guidelines,
ECCS, 2006.
This fifth worked example deals with spatial behaviour. The beam-column is
subjected to compression and biaxial bending. The transverse loading is assumed
to act through the shear center. Lateral torsional buckling is a potential mode of
failure according to the shape of the cross-section.
The interaction factors kij for combined bending and compression are determined
using both alternative method 1 (Annex A) and alternative method 2 (Annex B).
Reference Results
The reference gives following results:
Classification
Flanges c/tf 4,6
Class 1 limit 9
Flanges Class 1
Web c/tw 41,8
Class 1 limit 45,6
Web Class 1
Benchmark 21: ECCS N°119 Worked Example 5
76
Compression resistance
Nc,Rd 2715 kN
Bending resistance
Mc,y,Rd 516 kNm
Mc,z,Rd 78,9 kNm
Shear resistance
Av,z 5990 mm²
Vpl,z,Rd 814 kN
Av,y 6718 mm²
Vpl,y,Rd 912 kN
Shear buckling does not need to be considered
Resistance to combined bending, shear and axial force
My,NV,Rd 516 kNm
Mz,NV,Rd 78,9 kNm
2
1
Member Buckling resistance in compression
Ncr,y 71038 kN Ncr,z 3157 kN
red ,y 0,195 red ,z 0,927
y 0,21 z 0,34
y 1,00 z 0,644
Benchmark 21: ECCS N°119 Worked Example 5
77
Member Buckling resistance in bending (General)
C1 1,2
Mcr 1079 kNm
red LT 0,691
LT 0,34
LT 0,789
kc 0,907
f 0,955
LT,mod 0,826
Member Buckling resistance in bending (Rolled)
red LT 0,697
LT 0,49
LT 0,827
kc 0,907
f 0,954
LT,mod 0,867
Verification according to Method 1
y 1,00
wy 1,138
z 0,937
wz 1,5
Cmy,0 0,999
Cmy 1,00
Cmz,0 0,771
Cmz 0,771
Mcr0 899 kNm
Benchmark 21: ECCS N°119 Worked Example 5
78
red 0 0,757
aLT 0,998
bLT 0,043
cLT 0,468
dLT 0,347
eLT 0,719
Cyy 0,981
Cyz 0,863
Czy 0,843
Czz 1,014
eq. (6.61) 0,964
eq. (6.62) 0,870
Verification according to Method 2
Cmy 0,925
Cm,LT 0,925
Cmz 0,6
kyy 0,924
kyz 0,489
kzy 0,961
kzz 0,815
eq. (6.61) 0,752
eq. (6.62) 0,974
Benchmark 21: ECCS N°119 Worked Example 5
79
Scia Engineer Results
Benchmark 21: ECCS N°119 Worked Example 5
80
Verification according to Method 1
Benchmark 21: ECCS N°119 Worked Example 5
81
Benchmark 21: ECCS N°119 Worked Example 5
82
Verification according to Method 2
Benchmark 21: ECCS N°119 Worked Example 5
83
Comments
- The results correspond to the benchmark results.
- There are some small round-off differences between the cross-section properties.
In Scia Engineer the cross-section according to the Arcelor catalogue has been
used.
- The reference calculates a wrong value for the shear area in the shear resistance
check. The result shown above for the shear area is that corrected by manual
calculation.
- Since it concerns a case of combined loading, the FriLo LTB solver is used to
calculate the exact Mcr through an eigenvalue solution.
- In the verification according to method 1, the reference uses the ‘General Case’
for LTB. However, the reference also applies the reduction factor f to calculate
LT,mod in this case. In EN 1993-1-1 this reduction is only specified for the
‘Rolled sections and equivalent welded sections Case’ and not for the ‘General
Case’.
Due to the differences in the LTB reduction factor, the eventual verification
formula’s have differences.
- In the verification according to method 1, both the reference and Scia Engineer
use the modified formula for calculation of Czz.as given in correction sheet EN
1993-1-1:2005/AC:2009.
Benchmark 22: ECCS N°119 Members in building frames
84
Benchmark 22: ECCS N°119 Members in building frames
Project file: EN_Benchmark22_1.esa & EN_Benchmark22_2.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example Members in building frames of ECCS
N°119 Rules for Member Stability in EN 1993-1-1, Background documentation
and design guidelines, ECCS, 2006.
In this example, a three bay – three storey building is analysed. The building is
loaded by permanent loads, different cases of imposed loads and wind loading. A
2nd
order analysis is carried out taking into account sway imperfections. The
verification is done for one of the inner columns.
The interaction factors kij for combined bending and compression are determined
using both alternative method 1 (Annex A) and alternative method 2 (Annex B).
Reference Results
The reference gives following results:
Sway imperfection
m 0,791
h 2/3
0,00264
Benchmark 22: ECCS N°119 Members in building frames
85
Buckling ratio for in-plane buckling
ly / Lc 0,777
Classification
Flanges c/tf 5,77
Class 1 limit 9,0
Flanges Class 1
Web c/tw 17,7
Class 1 limit 33,00
Web Class 1
Compression resistance
Nc,Rd 2782 kN
Bending resistance
Mc,y,Rd 302 kNm
Shear resistance
Av,z 3755 mm²
Vpl,z,Rd 509,5 kN
Shear buckling does not need to be considered
Resistance to combined bending, shear and axial force
My,NV,Rd 232,5 kNm
Member Buckling resistance in compression
red ,y 0,258 red ,z 0,566
y 0,34 z 0,49
y 0,979 z 0,805
Benchmark 22: ECCS N°119 Members in building frames
86
Member Buckling resistance in bending (General)
C1 1,77
Mcr 2488 kNm
red LT 0,348
LT 0,21
LT 0,966
kc 0,752
f 0,927
LT,mod 1,00
Member Buckling resistance in bending (Rolled)
red LT 0,339
LT 0,34
LT 1,00
kc 0,752
f 0,929
LT,mod 1,00
Verification according to Method 1
y 1,00
z 0,978
wy 1,118
wz 1,5
Cmy,0 0,787
Cmy 0,895
Mcr0 1406 kNm
red 0 0,463
aLT 0,992
bLT 0
Benchmark 22: ECCS N°119 Members in building frames
87
dLT 0
Cyy 1,037
Czy 0,998
eq. (6.61) 0,588
eq. (6.62) 0,534
Verification according to Method 2
Cmy 0,6
CmLT 0,6
kyy 0,612
kzy 0,936
eq. (6.61) 0,508
eq. (6.62) 0,674
Benchmark 22: ECCS N°119 Members in building frames
88
Scia Engineer Results
Benchmark 22: ECCS N°119 Members in building frames
89
Verification according to Method 1
Benchmark 22: ECCS N°119 Members in building frames
90
Benchmark 22: ECCS N°119 Members in building frames
91
Verification according to Method 2
Benchmark 22: ECCS N°119 Members in building frames
92
Comments
- The results correspond to the benchmark results.
- The reference assumes that, during 2nd
Order analysis, the bending moment
remains linear. An exact 2nd
order analysis by Scia Engineer shows that this is not
the case. As a result, different calculation methods will be used for C1 and
Cmy,0. In order to perform the verification using the same moment diagram, the
moment diagram from the reference was inputted in Scia Engineer through the
use of non-calculated internal forces.
- In Scia Engineer the C1 factor for LTB is calculated according to the formula for
end moment loading given in ENV 1993-1-1:1992. This formula results in a value
of 1,88 in case of a triangular moment diagram. The reference uses a similar
formula which results in a value of 1,77.
This slight difference in C1 results in a difference in Mcr.
Reference Mcr = 2488kNm Scia Engineer Mcr = 2645 kNm
- In the verification according to Method 1, the reference uses the ‘General Case’
for LTB. However, the reference also applies the reduction factor f to calculate
LT,mod in this case. In EN 1993-1-1 this reduction is only specified for the
‘Rolled sections and equivalent welded sections Case’ and not for the ‘General
Case’.
- To determine the interaction factors kij using alternative method 2 (Annex B) a
distinction is made between members not susceptible to torsional deformations
(Table B.1) and members susceptible to torsional deformations (Table B.2).
The reference concludes that the member is susceptible to torsional deformations
and uses Table B.2 leading to a kzy value of 0,936.
However, since LT = 1,00 the member is considered within Scia Engineer as
being non-susceptible to LT-buckling and thus Table B.1 is applied leading to a
kzy value of 0,6 kyy = 0,6 * 0,611 = 0,367
Benchmark 23: Access Steel Document SX002a-EN-EU
93
Benchmark 23: Access Steel Document SX002a-EN-EU
Project file: EN_Benchmark23.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX002a-EN-EU Buckling resistance of a
pinned column with intermediate restraints of Access Steel, http://www.access-
steel.com/, 2005.
This worked example concerns the procedure to determine the buckling resistance
of a pinned column with intermediate restraints.
Reference Results
The reference gives following results:
Member Buckling resistance in compression
Ncr,y 1964,5 kN Ncr,z 6206,0 kN
red ,y 1,019 red ,z 0,573
y 0,34 z 0,49
y 0,585 z 0,801
Nb,Rd 1193 kN
Benchmark 23: Access Steel Document SX002a-EN-EU
94
Scia Engineer Results
Comments
- The results correspond to the benchmark results.
- There are some small round-off differences between the cross-section properties.
In Scia Engineer the cross-section according to the Arbed catalogue has been
used.
Benchmark 24: Access Steel Document SX001a-EN-EU
95
Benchmark 24: Access Steel Document SX001a-EN-EU
Project file: EN_Benchmark24.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX001a-EN-EU Simply supported laterally
unrestrained beam of Access Steel, http://www.access-steel.com/, 2004.
This example gives the details for the verification of a simple non-composite
beam under uniform loading. The beam is laterally restrained at the supports only.
The loading is acting at the top flange (destabilizing). For Lateral Torsional
Buckling the ‘Rolled Sections or Equivalent Welded’ case is used.
Reference Results
The reference gives following results:
Classification
Flanges c/tf 5,07
Class 1 limit 9,0
Flanges Class 1
Web c/tw 36,1
Class 1 limit 72,00
Web Class 1
Bending resistance
Mc,y,Rd 189,01 kNm
Benchmark 24: Access Steel Document SX001a-EN-EU
96
Shear resistance
Av,z 3080 mm²
Vpl,z,Rd 417,9 kN
Shear buckling does not need to be considered
Member Buckling resistance in bending (Rolled)
C1 1,127
C2 0,454
Mcr 113,9 kNm
red LT 1,288
LT 0,49
LT 0,48
kc 0,94
f 0,984
LT,mod 0,488
Mb,Rd 92,24 kNm
Scia Engineer Results
Benchmark 24: Access Steel Document SX001a-EN-EU
97
Comments
The results correspond to the benchmark results.
Benchmark 25: Access Steel Document SX007a-EN-EU
98
Benchmark 25: Access Steel Document SX007a-EN-EU
Project file: EN_Benchmark25.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX007a-EN-EU Simply supported beam
with lateral restraint at load application point of Access Steel, http://www.access-
steel.com/, 2005.
This worked example deals with a simply supported beam with lateral restraints at
supports and at load application point.
For Lateral Torsional Buckling the ‘Rolled Sections or Equivalent Welded’ case
is used.
Reference Results
The reference gives following results:
Classification
Flanges c/tf 4,63
Class 1 limit 7,29
Flanges Class 1
Web c/tw 52,45
Class 1 limit 58,32
Web Class 1
Bending resistance
Mc,y,Rd 1115 kNm
Benchmark 25: Access Steel Document SX007a-EN-EU
99
Shear resistance
Av,z 7011,5 mm²
Vpl,z,Rd 1437 kN
Shear buckling does not need to be considered
Member Buckling resistance in bending (Rolled)
C1 1,77
Mcr 1590 kNm
red LT 0,837
LT 0,49
LT 0,74
kc 0,752
f 0,876
LT,mod 0,845
Mb,Rd 942,22 kNm
Scia Engineer Results
Benchmark 25: Access Steel Document SX007a-EN-EU
100
Comments
- The results correspond to the benchmark results.
- The reference assumes a linear bending moment diagram which is not the case
since the beam is loaded by both point loads and a line load. As a result, a
difference is obtained in the C1 and kc factors.
In Scia Engineer the actual moment diagram is used instead of a linear
approximation. This difference in C1 and kc results in a slight difference in
LT,mod. Reference LT,mod = 0,845 Scia Engineer LT,mod = 0,81
Benchmark 26: Access Steel Document SX030a-EN-EU
101
Benchmark 26: Access Steel Document SX030a-EN-EU
Project file: EN_Benchmark26.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX030a-EN-EU Elastic design of a single
bay portal frame made of fabricated profiles of Access Steel, http://www.access-
steel.com/, 2006.
A single bay portal frame made of welded profiles is designed according to EN
1993-1-1. This worked example includes the elastic analysis of the frame using
the 1st Order theory, and all the verifications of the members based on the
effective properties of the cross-sections (class4).
For Lateral Torsional Buckling the ‘General’ case is used. The interaction factors
kij for combined bending and compression are determined using alternative
method 1 (Annex A).
Reference Results
The reference gives following results:
Buckling amplification factor
cr 29,98
Benchmark 26: Access Steel Document SX030a-EN-EU
102
Sway imperfection
m 0,866
h 0,74
0,0032
Column Verification
Classification
Flanges c/tf 9,8
Class 3 limit 11,3
Flanges Class 3
Web c/tw 131,9
Class 3 limit 92,3
Web Class 4
Effective cross-section properties
Aeff 7586 mm²
Iy,eff 1215420000 mm4
Weff,y 2867400 mm³
Shear Buckling
Eta1 0,721
k 5,34
E 10,7 N/mm²
cr 57,14 N/mm²
red W 1,894
W 0,438
Vbw,Rd 430,9 kN
Eta 3 0,26
Benchmark 26: Access Steel Document SX030a-EN-EU
103
Member Buckling resistance in compression
Ncr,y 71920 kN Ncr,z 7199 kN
red ,y 0,1935 red ,z 0,6116
y 0,34 z 0,49
y 1,00 z 0,778
Nby,Rd 2693 kN Nbz,Rd 2095 kN
Member Buckling resistance in bending (General)
C1 1,31
Mcr 3873 kNm
red LT 0,5127
LT 0,76
LT 0,7705
Mb,Rd 784,3 kNm
Verification according to Method 1
y 1,0
z 0,995
Cmy,0 0,79
Mcr0 2957 kNm
red 0 0,587
aLT 1,00
Cmy 0,951
CmLT 1,00
kyy 0,953
kzy 0,948
eq. (6.61) 0,877
eq. (6.62) 0,890
Benchmark 26: Access Steel Document SX030a-EN-EU
104
Rafter Verification
Classification
Flanges c/tf 9,4
Class 3 limit 11,3
Flanges Class 3
Web c/tw 131,9
Class 3 limit 93,9
Web Class 4
Effective cross-section properties
Aeff 7346 mm²
Iy,eff 1175820000 mm4
Weff,y 2772100 mm³
Shear Buckling
Eta1 0,729
k 5,34
E 10,7 N/mm²
cr 57,14 N/mm²
red W 1,894
W 0,438
Vbw,Rd 430,9 kN
Eta 3 0,349
Determination of buckling length around yy-axis
cr 76,43
Ncr,y 9546 kN
Lcr,y 16180 mm
Benchmark 26: Access Steel Document SX030a-EN-EU
105
Member Buckling resistance in compression
Ncr,y 9546 kN Ncr,z 6370 kN
red ,y 0,5228 red ,z 0,6398
y 0,34 z 0,49
y 0,874 z 0,7619
Nby,Rd 2279 kN Nbz,Rd 1987 kN
Member Buckling resistance in bending (General)
C1 1,39
Mcr 3640 kNm
red LT 0,52
LT 0,76
LT 0,7653
Mb,Rd 753,1 kNm
Verification according to Method 1
y 0,9983
z 0,9953
Cmy,0 0,9927
Mcr0 2619 kNm
red 0 0,613
aLT 1,00
Cmy 0,9985
CmLT 1,014
kyy 1,024
kzy 1,021
eq. (6.61) 0,967
eq. (6.62) 0,972
Benchmark 26: Access Steel Document SX030a-EN-EU
106
Scia Engineer Results
Column Verification
Benchmark 26: Access Steel Document SX030a-EN-EU
107
Benchmark 26: Access Steel Document SX030a-EN-EU
108
Benchmark 26: Access Steel Document SX030a-EN-EU
109
Rafter Verification
Benchmark 26: Access Steel Document SX030a-EN-EU
110
Benchmark 26: Access Steel Document SX030a-EN-EU
111
Buckling shape for determination of Ncr,y:
Benchmark 26: Access Steel Document SX030a-EN-EU
112
Benchmark 26: Access Steel Document SX030a-EN-EU
113
Comments
- The results correspond to the benchmark results.
- There is a slight difference in the classification slenderness due to the weld throat
which is not accounted for in Scia Engineer.
- For calculating the in-plane buckling resistance of the rafter, the reference
assumes the frame to be restrained against horizontal displacement. Scia Engineer
takes into account the actual frame without this assumption.
- In the calculation of Cmy,0 the reference approximates the rafter as one straight
member of 30m. Scia Engineer uses the actual geometry of the rafter.
Benchmark 27: Access Steel Document SX029a-EN-EU
114
Benchmark 27: Access Steel Document SX029a-EN-EU
Project file: EN_Benchmark27.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX029a-EN-EU Elastic design of a single
bay portal frame of Access Steel, http://www.access-steel.com/, 2006.
A single bay portal frame made of rolled profiles is designed according to EN
1993-1-1. This worked example includes the elastic analysis of the frame using
the 1st Order theory, and all the verifications of the members under ULS
combinations.
For Lateral Torsional Buckling the ‘Rolled sections and equivalent welded
sections’ case is used. The interaction factors kij for combined bending and
compression are determined using alternative method 1 (Annex A).
Reference Results
The reference gives following results:
Buckling amplification factor
cr 14,57
Benchmark 27: Access Steel Document SX029a-EN-EU
115
Sway imperfection
m 0,866
h 0,74
0,0032
Column Verification
Classification
Flanges c/tf 4,21
Class 1 limit 8,28
Flanges Class 1
Web c/tw 42,83
Class 1 limit 59,49
Web Class 1
Compression resistance
Nc,Rd 4290 kN
Bending resistance
Mc,y,Rd 965,8 kNm
Shear resistance
Av,z 8380 mm²
Vpl,z,Rd 1330 kN
Shear buckling does not need to be considered
Benchmark 27: Access Steel Document SX029a-EN-EU
116
Member Buckling resistance in compression
Ncr,y 53190 kN Ncr,z 1956 kN
red ,y 0,284 red ,z 1,481
y 0,21 z 0,34
y 0,9813 z 0,3495
Member Buckling resistance in bending (Rolled)
C1 1,77
Mcr 1351 kNm
red LT 0,8455
LT 0,49
LT 0,7352
kc 0,7519
f 0,8765
LT,mod 0,8388
Verification according to Method 1
y 0,9999
z 0,9447
wy 1,144
wz 1,5
Mcr0 763,3 kNm
red 0 1,125
aLT 0,9982
Cmy,0 0,7896
Cmy 0,9641
CmLT 1,00
npl 0,03765
Cyy 0,9849
Czy 0,9318
Benchmark 27: Access Steel Document SX029a-EN-EU
117
kyy 0,9818
kzy 0,5138
eq. (6.61) 0,9534
eq. (6.62) 0,5867
Rafter Verification
Classification
Flanges c/tf 4,62
Class 1 limit 8,28
Flanges Class 1
Web c/tw 41,76
Class 1 limit 58,38
Web Class 1
Compression resistance
Nc,Rd 3176 kN
Bending resistance
Mc,y,Rd 603,4 kNm
Shear resistance
Av,z 5985 mm²
Vpl,z,Rd 950,3 kN
Shear buckling does not need to be considered
Benchmark 27: Access Steel Document SX029a-EN-EU
118
Member Buckling resistance in compression
Ncr,y 5082 kN Ncr,z 1233 kN
red ,y 0,7906 red ,z 1,605
y 0,21 z 0,34
y 0,8011 z 0,3063
Member Buckling resistance in bending (Rolled)
C1 2,75
Mcr 1159 kNm
red LT 0,7215
LT 0,49
LT 0,8125
kc 0,91
f 0,9556
LT,mod 0,8503
Verification according to Method 1
y 0,9946
z 0,9208
wy 1,138
wz 1,5
Mcr0 421,5 kNm
red 0 1,196
aLT 0,9981
Cmy,0 0,9803
Cmy 0,996
CmLT 1,072
npl 0,0428
Cyy 0,9774
Czy 0,9011
Benchmark 27: Access Steel Document SX029a-EN-EU
119
kyy 1,116
kzy 0,5859
eq. (6.61) 0,8131
eq. (6.62) 0,5385
Scia Engineer Results
Column Verification
Benchmark 27: Access Steel Document SX029a-EN-EU
120
Benchmark 27: Access Steel Document SX029a-EN-EU
121
Benchmark 27: Access Steel Document SX029a-EN-EU
122
Rafter Verification
Benchmark 27: Access Steel Document SX029a-EN-EU
123
Benchmark 27: Access Steel Document SX029a-EN-EU
124
Benchmark 27: Access Steel Document SX029a-EN-EU
125
Comments
- In the verification of the column, the reference gives a wrong value for c/tf in the
classification of the flanges. The value shown above has been corrected by a
manual calculation.
- In the verification of the column, in Scia Engineer the C1 factor for LTB is
calculated according to the formula for end moment loading given in ENV 1993-
1-1:1992.
This formula results in a value of 1,88 in case of a triangular moment diagram.
The reference uses a similar formula which results in a value of 1,77.
This slight difference in C1 results in a difference in Mcr.
Reference Mcr = 1351 kNm Scia Engineer Mcr = 1432 kNm
- In the verification of the rafter, the reference gives a wrong value for c/tf in the
classification of the flanges. The value shown above has been corrected by a
manual calculation.
- In the verification of the rafter, the reference uses an approximate graphic for
determining C1 for combined loading which gives 2,75. Scia Engineer uses the
method outlined in the Steel Code Check Theoretical Background which gives
2,47.
This slight difference in C1 results in a difference in Mcr.
Reference Mcr = 1159 kNm Scia Engineer Mcr = 1033 kNm
- In the verification of the rafter, the reference applies a fictitious restraint at the top
of the column to calculate the in-plane buckling length. Scia Engineer uses the
actual geometry of the structure. In order to execute the verification using the
same assumptions, the buckling length used by the reference was inputted in Scia
Engineer.
Benchmark 28: Access Steel Document SX021a-EN-EU
126
Benchmark 28: Access Steel Document SX021a-EN-EU
Project file: EN_Benchmark28.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX021a-EN-EU Simply supported IPE
profile purlin of Access Steel, http://www.access-steel.com/, 2006.
This example gives the details of the verification according to EN 1993-1-1 of a
simply supported purlin under a uniform loading. The purlin is an I-section rolled
profile which is laterally restrained by steel sheeting.
For Lateral Torsional Buckling the ‘Rolled sections and equivalent welded
sections’ case is used.
The purpose of this benchmark for Scia Engineer is to verify the calculation of the
LTB resistance for a member which is laterally restrained by sheeting at the
tension flange.
Reference Results
The reference gives following results:
Classification
Flanges c/tf 4,23
Class 1 limit 8,28
Flanges Class 1
Web c/tw 27,5
Class 1 limit 66,24
Web Class 1
Benchmark 28: Access Steel Document SX021a-EN-EU
127
Bending resistance
Mc,y,Rd 45,76 kNm
Shear resistance
Av,z 1120 mm²
Vpl,z,Rd 177,8 kN
Shear buckling does not need to be considered
Member Buckling resistance in bending (Rolled)
Mcr 27,20 kNm
red LT 1,297
LT 0,34
LT 0,525
Mb,Rd 24,02 kNm
Scia Engineer Results
Benchmark 28: Access Steel Document SX021a-EN-EU
128
Benchmark 28: Access Steel Document SX021a-EN-EU
129
Comments
- The results correspond to the benchmark results.
- The reference and Scia Engineer use a different method to calculate the shear
stiffness of the diaphragm. The reference gives insufficient data concerning the
K1 and K2 manufacturer factors (as specified in the Steel Code Check
Theoretical Background).
Therefore, the K1 factor has been inputted in Scia Engineer in such a way that the
same shear stiffness was obtained as in the reference. The reasoning behind this is
that purpose of this benchmark for Scia Engineer is to verify the calculation of the
LTB resistance for a member which is laterally restrained by sheeting at the
tension flange, not the actual calculation of the sheeting.
- The FriLo LTB solver was used to calculate Mcr through an eigenvalue analysis.
- For LTB the ‘Rolled sections and equivalent welded sections’ case is used.
According to EN 1993-1-1 in this case the reduction factor may be reduced by the
factor f. The reference does not apply this modification (however for this example
the modification has no effect).
Benchmark 29: Access Steel Document SX044a-EN-EU
130
Benchmarks EN 1993-1-2
Benchmark 29: Access Steel Document SX044a-EN-EU
Project file: EN_Benchmark29.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX044a-EN-EU Fire design of a protected
HEB section column exposed to the standard temperature time curve of Access
Steel, http://www.access-steel.com/, 2006.
This worked example illustrates the fire design of a column that is continuous
over two storeys. Heat transfer into the section is evaluated using the EN1993-1-2
calculation procedure. The resistance of the column is evaluated using the simple
calculation model for compression members given in EN1993-1-2.
The column, fabricated from a hot-rolled HEB section, supports two floors and is
fire protected with sprayed vermiculite cement. The required period of fire
resistance is R90.
Reference Results
The reference gives following results:
Fire Situation
Ap/V 159 m-1
g at 90 min 1006,0 °C
a,t at 90 min 553,8 °C
ky, 0,613
Benchmark 29: Access Steel Document SX044a-EN-EU
131
kE, 0,444
Classification
Flanges c/tf 5,05
Class 1 limit 6,22
Flanges Class 1
Web c/tw 14,35
Class 1 limit 22,80
Web Class 1
Buckling resistance
Lcr,z,fi 2,45 m
Ncr,z 4706 kN
red ,z 0,702
red ,z, 0,825
z,fi 0,581
Nb,fi, ,Rd 825,0 kN
Scia Engineer Results
Benchmark 29: Access Steel Document SX044a-EN-EU
132
Benchmark 29: Access Steel Document SX044a-EN-EU
133
Comments
- The results correspond to the benchmark results.
Benchmark 30: Access Steel Document SX046a-EN-EU
134
Benchmark 30: Access Steel Document SX046a-EN-EU
Project file: EN_Benchmark30.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX046a-EN-EU Fire design of an
unprotected IPE section beam exposed to the standard time temperature curve of
Access Steel, http://www.access-steel.com/, 2006.
The worked example illustrates the fire design of a simply supported non-
composite beam. The transfer of heat into the beam is evaluated using a step-by-
step calculation procedure. The structural resistance of the member at elevated
temperature is evaluated using the simple calculation model for members subject
to bending given in EN1993-1-2.
A beam made of hot-rolled IPE section is a part of the floor structure of an office
building. The beam is loaded uniformly and restrained against lateral torsional
buckling by a concrete slab. The beam is design to achieve a fire resistance rating
of R15.
Reference Results
The reference gives following results:
Fire Situation
Am/V 188 m-1
ksh 0,667
g at 15 min 738,6 °C
a,t at 15 min 613,8 °C
ky, 0,436
Benchmark 30: Access Steel Document SX046a-EN-EU
135
Classification
Flanges c/tf 5,3
Class 1 limit 7,07
Flanges Class 1
Web c/tw 35
Class 1 limit 56,6
Web Class 1
Shear resistance
Av,z 2568 mm²
Vfi,t,Rd 177,8 kN
Bending resistance
1 0,7
2 1,0
Mfi,t,Rd 107,6 kNm
Scia Engineer Results
Benchmark 30: Access Steel Document SX046a-EN-EU
136
Benchmark 30: Access Steel Document SX046a-EN-EU
137
Comments
- The results correspond to the benchmark results.
Benchmark 31: Access Steel Document SX047a-EN-EU
138
Benchmark 31: Access Steel Document SX047a-EN-EU
Project file: EN_Benchmark31.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX047a-EN-EU Fire design of protected
IPE section beam exposed to parametric fire curve of Access Steel,
http://www.access-steel.com/, 2006.
This worked illustrates the fire design of a simply supported non-composite beam.
Heat transfer into the section is calculated using the equation for protected
members given in EN1993-1-2, which is evaluated using an iterative calculation
procedure. The structural resistance is calculated using the simple calculation
model for members in bending, given in EN1993-1-2.
A steel beam forms part of a floor structure of an office building. The beam is
uniformly load and restrained against lateral torsional buckling by a concrete slab.
The beam is required to achieve 60 minutes fire resistance and will be fire
protected using sprayed vermiculite cement. The thermal actions will be
determined using the parametric temperature - time curve.
Reference Results
The reference gives following results:
Fire Situation
Ap/V 188 m-1
g at 42,5 min 562,1 °C
a,t at 42,5 min 582,5 °C
ky, 0,525
Benchmark 31: Access Steel Document SX047a-EN-EU
139
Classification
Flanges c/tf 5,3
Class 1 limit 7,07
Flanges Class 1
Web c/tw 35
Class 1 limit 56,6
Web Class 1
Shear resistance
Av,z 2568 mm²
Vfi,t,Rd 214,1 kN
Bending resistance
1 0,85
2 1,0
Mfi,t,Rd 106,7 kNm
Scia Engineer Results
Benchmark 31: Access Steel Document SX047a-EN-EU
140
Benchmark 31: Access Steel Document SX047a-EN-EU
141
Comments
- The results correspond to the benchmark results.
Benchmark 32: Access Steel Document SX048a-EN-EU
142
Benchmark 32: Access Steel Document SX048a-EN-EU
Project file: EN_Benchmark32.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX048a-EN-EU Fire design of protected
unrestrained HEA section beam exposed to the standard temperature time curve
of Access Steel, http://www.access-steel.com/, 2006.
This example illustrates the fire design of a simply supported beam with partial
lateral restraint. Transfer of heat into the section is calculated with the equation
given in EN1993-1-2, which is evaluated using an incremental calculation
procedure. The structural resistance is evaluated using the simple calculation
model for beams subject to LTB given in EN1993-1-2.
A hot-rolled HEA section has forming part of floor structure of an office building
supports a concentrated load. The beam is restrained at the ends and at the point
of load application. The beam is required to achieve R30 fire resistance and is to
be fire protected with sprayed vermiculite cement.
Reference Results
The reference gives following results:
Fire Situation
Ap/V 165 m-1
g at 30 min 841,8 °C
a,t at 30 min 396 °C
ky, 1,000
kE, 0,704
Benchmark 32: Access Steel Document SX048a-EN-EU
143
Classification
Flanges c/tf 8,6
Class 2 limit 8,5
Class 3 limit 11,9
Flanges Class 3
Web c/tw 24,5
Class 1 limit 61,2
Web Class 1
Shear resistance
Av,z 3174 mm²
Vfi,t,Rd 430,6 kN
Lateral Torsional Buckling
C1 1,77
Mcr 1362,7 kNm
red LT 0,438
red LT, 0,522
LT,fi 0,704
Mfi,t,Rd 167,6 kNm
Scia Engineer Results
Benchmark 32: Access Steel Document SX048a-EN-EU
144
Benchmark 32: Access Steel Document SX048a-EN-EU
145
Comments
- The results correspond to the benchmark results.
- Within Scia Engineer the C1 factor for LTB is calculated according to the
formula for end moment loading given in ENV 1993-1-1:1992.
This formula results in a value of 1,88 in case of a triangular moment diagram.
The reference uses a similar formula which results in a value of 1,77.
This slight difference in C1 results in a difference in Mcr.
Reference Mcr = 1362,7 kNm Scia Engineer Mcr = 1448,37 kNm
Benchmark 34: Temperature Domain
146
Benchmark 33: Access Steel Document SX043a-EN-EU
Project file: EN_Benchmark33.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX043a-EN-EU Fire design of unprotected
HEB section column exposed to the standard temperature time curve of Access
Steel, http://www.access-steel.com/, 2006.
This worked example illustrates the fire design of a column that is continuous
over two storeys. The resistance of the member at elevated temperature is
evaluated using the simple calculation model given in EN1993-1-2.
A column fabricated from a hot-rolled HEB section supports two floors. The
member is to be constructed without fire protection and its load bearing resistance
is to be checked for exposure to the standard temperature-time curve. The
required fire resistance is R15.
Reference Results
The reference gives following results:
Fire Situation
a,t at 15 min 565 °C
ky, 0,578
kE, 0,411
Buckling resistance
Lcr,z,fi 2,45 m
Benchmark 34: Temperature Domain
147
Ncr,z 4706 kN
red ,z 0,702
red ,z, 0,833
z,fi 0,577
Nb,fi, ,Rd 772,5 kN
Scia Engineer Results
Benchmark 34: Temperature Domain
148
Comments
- The results correspond to the benchmark results.
Benchmark 34: Temperature Domain
149
Benchmark 34: Temperature Domain
Project file: EN_Benchmark34.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark is based on an example worked out at the training for Fire Safety
Engineering as part of the TETRA project Brandveilig Constructief Ontwerp in
September 2009.
In this benchmark the fire resistance of an unprotected beam on two supports is
evaluated in case of flexure. The beam is part of an office building. At the top side
a concrete slab prohibits the occurrence of lateral torsional buckling. Due to this
slab the beam is exposed to fire at three sides.
The beam is exposed to the standard ISO 834 curve and is required to have a
resistance R15.
The verification is carried out in the Temperature domain using a manual
calculation. To determine the eventual fire resistance at the critical temperature a
monogram is used as published by Infosteel (http://www.infosteel.be/).
Reference Results
Length l = 7,4m
Properties
IPE 300
S 275
fy = 275 N/mm²
E = 210000 N/mm²
Density: ρa = 7850 kg/m³
Benchmark 34: Temperature Domain
150
Section properties
A = 5380 mm²;
Wpl,y = 628.4 10³mm³;
Iy = 8356 104mm4.
The section is taken as Class 1 in bending.
Loading:
Permanent:
gk = 4,8 kN/m
Variable:
qk = 7,8 kN/m
Accidental situation using ψ2,1 = 0,3 for office buildings.
reduction factor for the design load level for the fire situation:
dfifid EE ,
QkGk
kk
fiqg
qg 1,1 393,0
5,18,735,18,4
8,73,08,4fi
Section factor for an unprotected beam subjected to fire at three sides:
1188mV
Am
Box shape section factor for an unprotected beam subjected to fire at three sides:
1139²005380.0
3.0215.02m
m
mm
V
hb
V
A
bb
m
Correction factor for the shadow effect:
665,0188
1399.09.0
1
1
m
m
V
A
V
A
km
b
m
sh
Benchmark 34: Temperature Domain
151
The modified section factor thus becomes:
11 125188*665,0 mmV
AkP m
sh
Adaptation factors for non-uniform temperature distribution along the cross-
section and along the member:
- κ1 = 0,7 unprotected beam subjected to fire at three sides
- κ2 = 1,0 simply supported member
Degree of utilization at time t=0:
210 fi 275,00,17,0393,00
The critical temperature is calculated as:
48219674,0
1ln19.39
833,3
0
,cra
Ccra 6774821275,0*9674,0
1ln19.39
833,3,
Using the monogram this critical temperature corresponds to a fire resistance of
17 minutes.
The member thus meets the requirement of R15.
Benchmark 34: Temperature Domain
152
Benchmark 34: Temperature Domain
153
Scia Engineer Results
Comments
- The results correspond to the benchmark results.
Benchmark 35: Combined Compression and Bending
154
Benchmark 35: Combined Compression and Bending
Project file: EN_Benchmark35.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark is based on an example worked out at the training for Fire Safety
Engineering as part of the TETRA project Brandveilig Constructief Ontwerp in
September 2009.
In this example a beam of an office building is subjected to the combined loading
of bending and compression. At the top side a floor is resting on the beam
however the floor is not prohibiting lateral torsional buckling thus instability can
occur.
The beam is exposed to fire at three sides and protected by a hollow encasement
of gypsum.
The required fire resistance is R90.
The verification is carried out in the Resistance domain using a manual
calculation. To determine the critical temperature a monogram is used as
published by Infosteel (http://www.infosteel.be/).
This benchmark uses the correction to the interaction equations as published in
the EN 1993-1-2:2005/AC:2009 correction sheet.
Benchmark 35: Combined Compression and Bending
155
Reference Results
Length l = 10 m
Properties
Beam
HE 200 B
S 235
Section class 1
E = 210000 N/mm²
Aa = 7810 mm²
Iz = 2000 cm4
It = 59,3 cm4
Iw = 171100 cm6
Protection
Gypsium
dp = 20 mm (hollow encasement)
λp = 0,2 W/(m·K)
cp = 1700 J/(kg·K)
As a conservative measure the density of the protection is not accounted
for.
Loading:
Permanent:
Gk = 96,3 kN
gk = 1,5 kN/m
Variable:
qk = 1,5 kN/m
Accidental combination of actions in case of fire:
i,ki,,k,dkGAdA QQAGEE 2112
For office buildings ψ2,1 = 0.3.
Benchmark 35: Combined Compression and Bending
156
Design loading in the fire situation:
kNN dfi 3,963.960,1,
kNmM dfi 38,248
²105,13,05,10,1,
The steel temperature is calculated using the monogram published by Infosteel.
For a member subjected to fire at three sides and having hollow encasement
protection the following section factors are determined:
177007810
202022m
²m.
m.m.
A
bh
V
A
a
p
Km
W
m
mKWm
dV
A
p
pp
³770
020,0
2,077 1
This leads to the following critical temperature:
C540max,90a,
Benchmark 35: Combined Compression and Bending
157
First of all the combined effect of buckling and bending is checked:
1
,
,,
,,
,
,,
,
fiM
y
yypl
dfiyy
fiM
y
yfiy
dfi
fkW
Mk
fkA
N
The reduction factor χy,fi is used since it concerns single bending and thus in plane
effects need to be combined.
Relative slenderness at room temperature:
247,19.9354,8
1000
ay
cr
yi
L
10,29.9307,5
1000
az
crz
i
L
For a critical temperature of 540 °C the following reduction factors apply:
ky,θ = 0,656
kE,θ = 0,484
Relative slenderness in the fire situation:
,
,
,
E
y
yyk
k 452,1
484,0
656,0247,1,y
,
,
,
E
y
zzk
k 45,2
484,0
656,010,2,z
The reduction factor for flexural buckling can then be calculated:
65,0235
23565,0
23565,0
yf
²12
1,,, yyy 03,2²452,1452,165,01
2
1,y
²12
1,,, zzz 27,4²45,245,265,01
2
1,z
Benchmark 35: Combined Compression and Bending
158
²²
1
,,,
,
yyy
fiy 29,0²46,1²04,204,2
1, fiy
²²
1
,,,
,
zzz
fiz 13,0²45,2²27,427,4
1, fiz
Since the bending moment diagram is caused by a line load M,y = 1,3
8.029,0*44,0*5*2 ,,, yMyyMy with ,y limited to 1,1
(Using the EN 1993-1-2:2005/AC:2009 correction sheet)
8,0778,129,03,144,0)1,1;452,1min(53,12y
31
,
,,
,
fim
y
yafiy
dfiy
y fkA
Nk
349,1
1
²235656,0²781029,0
33,96778,11
mmNmm
Nek y
Check:
164,0²/235656,05,642
³1038,2449,1
²/235656,0²781029,0
³103,96
mmN
Nm
mmNmm
N
Benchmark 35: Combined Compression and Bending
159
Second the combined effect of compression and lateral torsional buckling is
checked
1
,
,,,
,,
,
,,
,
fiM
y
yyplfiLT
dfiyLT
fiM
y
yfiz
dfi
fkW
Mk
fkA
N
Relative slenderness at room temperature:
cr
yypl
LTM
fW , 0187,1
11,14549
5,235,642LT
With:
gg
z
t
z
w
w
zcr zCzC
IE
IGLk
I
I
k
k
Lk
IECM 22
2
1 ²)(²
)²(
)²(
²
2
2045,0²)
2
2045,0(
200021000²
3,598100)²10000,1(
2000
171100
0,1
0,1*
)²10000,1(
200021000²13,1
2
crM
With C1 and C2 determined according to ENV 1993-1-1 Annex F.
Relative slenderness in the fire situation:
,
,
,
E
y
LTLTk
k 19,1
484,0
656,00187,1,LT
The reduction factor for lateral torsional buckling can then be calculated:
²12
1,,, LTLTLT 59,1²19,119,165,01
2
1,LT
²²
1
,,,
,
LTLTLT
fiLT 38,0²19,1²59,159,1
1, fiLT
kNmMcr 49,145
Benchmark 35: Combined Compression and Bending
160
Since the bending moment diagram is caused by a line load M,LT = 1,3
9.015,015,0 ,, LTMzLT
9,0327,015,03,144,215,0LT
11
,
,,
,
fim
y
yfiz
dfiLT
LT fkA
Nk
1799,0
0,1
²235656,0²1,7813,0
3,96327,01
2
3
mmNmme
NekLT
Check:
1
,
,,,
,,
,
,,
,
fiM
y
yyplfiLT
dfiyLT
fiM
y
yfiz
dfi
fkW
Mk
fkA
N
113,1
0,1
²235656,0³5,64238,0
38,24799,0
0,1
²235656,0²1,7813,0
3,96
3
6
2
3
mmNmme
Nmme
mmNmme
Ne
The member thus does not meet the R90 requirement.
Benchmark 35: Combined Compression and Bending
161
Scia Engineer Results
Benchmark 35: Combined Compression and Bending
162
Benchmark 35: Combined Compression and Bending
163
Comments
The results correspond to the benchmark results. A slight difference is caused by
rounding errors in the manual calculation.
Benchmark 36: Designer’s Guide Ex. 13.1
164
Benchmarks EN 1993-1-3
Benchmark 36: Designer’s Guide Ex. 13.1
Project file: EN_Benchmark36.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 13.1: Calculation of section properties for
local buckling of Designer’s Guide to EN 1993-1-1 Eurocode 3, The Steel
Construction Institute, 2005.
The effective area and the horizontal shift in neutral axis due to local buckling is
calculated for a 200 x 65 x 1.6 lipped channel in zinc-coated steel with a nominal
yield strength of 280 N/mm^2 and a Young modulus of 210000 N/mm^2, and
subjected to pure compression. It is assumed that the zinc coating forms 0,04 mm
of the thickness of the section, and the contribution of the coating is ignored in the
calculations.
Reference Results
The reference gives following results:
Local Buckling calculation
Part [mm] k beff [mm]
Web 198,4 4,0 2,44 0,37 73,87
Flanges 63,4 4,0 0,78 0,92 58,31
Lips 14,2 0,43 0,53 1,00 14,2
Effective section properties
Aeff 341,5 mm^2
eNy 8,66 mm
Benchmark 36: Designer’s Guide Ex. 13.1
165
Scia Engineer Results
Results for CS1 – Actual C-section including rounded corners:
With cYLCS of the gross section 17,66 mm this gives:
eNy = 25,73 – 17,66 = 8,07 mm
Benchmark 36: Designer’s Guide Ex. 13.1
166
Results for CS2 – Idealized C-section without rounded corners:
With cYLCS of the gross section 16,46 mm this gives:
eNy = 25,12 – 16,46 = 8,66 mm
Comments
- The results correspond to the benchmark results.
- CS1 was inputted as an actual C-section including rounded corners. The notional
widths are thus calculated by Scia Engineer using the exact geometry. The
reference example however idealizes the cross-section to a section without
roundings. Within Scia Engineer this cross-section has been inputted as CS2. This
leads to an exact comparison with the benchmark results.
Benchmark 37: Designer’s Guide Ex. 13.2
167
Benchmark 37: Designer’s Guide Ex. 13.2
Project file: EN_Benchmark37.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns Example 13.2: Cross-section resistance to distortional
buckling of Designer’s Guide to EN 1993-1-1 Eurocode 3, The Steel Construction
Institute, 2005.
This example demonstrates the method set out in EN 1993-1-3 for the calculation
of cross-section resistance to (local and) distortional buckling. The example is
based on the same 200 x 65 x 1,6 mm lipped channel section of example 13.1,
where effective section properties for local buckling were determined.
Reference Results
The reference gives following results:
Local Buckling calculation
Part [mm] k beff [mm]
Web 198,4 4,0 2,44 0,37 73,87
Flanges 63,4 4,0 0,78 0,92 58,31
Lips 14,2 0,5 0,49 1,00 14,2
Distortional Buckling calculation - Lips
As 67,6 mm^2
Is 1132,4 mm^4
b1 53,6 mm
b2 53,6 mm
hw 198,4 mm
kf 1,0
K 0,22 N/mm^2
cr 212 N/mm^2
1,15
d 0,64
As,red 43,3 mm^2
Effective section properties
Aeff 292,8 mm^2
eNy 3,92 mm
Benchmark 37: Designer’s Guide Ex. 13.2
168
Scia Engineer Results
Results for CS1 – Actual C-section including rounded corners:
With cYLCS of the gross section 17,66 mm this gives:
eNy = 21,16 – 17,66 = 3,50 mm
Benchmark 37: Designer’s Guide Ex. 13.2
169
Results for CS2 – Idealized C-section without rounded corners:
With cYLCS of the gross section 16,46 mm this gives:
eNy = 20,38 – 16,46 = 3,92 mm
Comments
- The results correspond to the benchmark results.
- CS1 was inputted as an actual C-section including rounded corners. The notional
widths are thus calculated by Scia Engineer using the exact geometry. The
reference example however idealizes the cross-section to a section without
roundings. Within Scia Engineer this cross-section has been inputted as CS2. This
leads to an exact comparison with the benchmark results.
Benchmark 38: Access Steel Document SX022a-EN-EU
170
Benchmark 38: Access Steel Document SX022a-EN-EU
Project file: EN_Benchmark38.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX022a-EN-EU Calculation of effective
section properties for a cold-formed lipped channel section in bending of Access
Steel, http://www.access-steel.com/, 2005.
This example deals with the effective properties calculation of a cold formed
lipped channel section subjected to bending about its major axis.
Reference Results
The reference gives following results:
Local Buckling calculation
Part [mm] k beff
[mm]
be1
[mm]
be2
[mm]
Flange 72 4 0,789 0,914 65,8 32,9 32,9
Edge
fold
19,8 0,5 0,614 1,00 19,8
Benchmark 38: Access Steel Document SX022a-EN-EU
171
Distortional Buckling calculation – Iteration 1
As 103,3 mm^2
Is 3663 mm^4
b1 61,73 mm
hw 198 mm
kf 0
K 0,439 N/mm^2
cr 355,78 N/mm^2
0,992
d 0,753
Distortional Buckling calculation – Iteration n
be1 32,9 mm
be2,n 35,9 mm
ceff,n 19,8 mm
d,n 0,737
Local Buckling calculation
Part [mm] k beff [mm] be1 [mm] be2 [mm]
Web 198 22,58 0,914 0,959 97,5 39 58,5
Effective section properties
Aeff 689,2 mm^2
Ieff,y 4140000 mm^4
Weff,y,c 40460 mm^3
Weff,y,t 43260 mm^3
Scia Engineer Results
Result for the initial calculation i.e. without stiffener iterations:
Benchmark 38: Access Steel Document SX022a-EN-EU
172
Result using stiffener iterations:
Comments
- The results correspond to the benchmark results.
- The reference ignores the fact that the principal axis is not parallel to the flanges
(alfa = -1,47 deg). As a result, the top flange is not in uniform compression but
subject to a stress gradient. Scia Engineer accounts for the actual stress
distribution leading to small differences in the results.
- The reference does not detail the calculation of b2.
- The reference does not detail the different stiffener iteration steps.
Benchmark 39: Access Steel Document SX023a-EN-EU
173
Benchmark 39: Access Steel Document SX023a-EN-EU
Project file: EN_Benchmark39.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX023a-EN-EU Calculation of effective
section properties for a cold-formed lipped channel section in compression of
Access Steel, http://www.access-steel.com/, 2005.
This example deals with the effective properties calculation of a cold-formed
lipped channel section subjected to compression.
Reference Results
The reference gives following results:
Local Buckling calculation
Part [mm] k beff
[mm]
be1
[mm]
be2
[mm]
Upper
Flange
72 4 0,789 0,914 65,8 32,9 32,9
Lower
Flange
64 4 0,702 0,978 62,6 31,3 31,3
Upper Fold 19,8 0,5 0,614 1,00 19,8
Lower Fold 19,8 0,5 0,614 1,00 19,8
Web 198 4 2,171 0,414 82 41 41
Benchmark 39: Access Steel Document SX023a-EN-EU
174
Distortional Buckling calculation – Upper stiffener – Iteration 1
As 103,3 mm^2
Is 3663 mm^4
b1 61,73 mm
b2 54,41 mm
hw 198 mm
kf 0,97
K 0,331 N/mm^2
cr 309 N/mm^2
1,064
d 0,701
Distortional Buckling calculation – Lower stiffener – Iteration 1
As 100,2 mm^2
Is 3618 mm^4
K 0,406 N/mm^2
cr 350,7 N/mm^2
0,999
d 0,748
Distortional Buckling calculation – Upper stiffener – Iteration n
be1 32,9 mm
be2,n 36 mm
ceff,n 19,8 mm
d,n 0,683
Distortional Buckling calculation – Lower stiffener – Iteration n
be1 31,3 mm
be2,n 32 mm
ceff,n 19,8 mm
d,n 0,744
Effective section properties
Aeff 436,8 mm^2
Benchmark 39: Access Steel Document SX023a-EN-EU
175
Scia Engineer Results
Result for the initial calculation i.e. without stiffener iterations:
Result using stiffener iterations:
Benchmark 39: Access Steel Document SX023a-EN-EU
176
Comments
- The results correspond to the benchmark results.
- For the distortional buckling calculation (iteration 1) of the lower stiffener, the
reference uses a wrong value for kf. More specifically the reference uses kf 0,97
for both the upper and the lower stiffener however for the lower stiffener a value
of 1,031 should be used. Within Scia Engineer the correct kf value is used.
Benchmark 40: Access Steel Document SX024a-EN-EU
177
Benchmark 40: Access Steel Document SX024a-EN-EU
Project file: EN_Benchmark40.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX024a-EN-EU Design of a cold-formed
steel lipped channel wall stud in compression of Access Steel, http://www.access-
steel.com/, 2006.
This example deals with the design of a pinned wall stud subjected to
compression. The stud is composed of a cold-formed lipped channel section
where boards are attached to both flanges and they prevent buckling in the
weak direction and torsional buckling.
Reference Results
The reference gives following results:
Effective section properties
Aeff 118 mm^2
Weff,z,com 1274 mm^3
Weff,z,ten 2585 mm^3
Combined compression and bending
eNz 3,04 mm
Nc,Rd 41,3 kN
Mcz,Rd,com 0,45 kNm
Mz,Ed 0,077 kNm
UC 0,785
Benchmark 40: Access Steel Document SX024a-EN-EU
178
Scia Engineer Results
Comments
- The reference does not detail the calculation of the effective section properties.
The compression force causes a shift in neutral axis towards the edge folds. This
implies that the compression load, acting at the centroid of the gross section,
causes a weak axis moment which gives compression in the web and tension in
the edge folds.
The effective shape for this negative weak axis moment leads to only a reduction
of the web and causes the centroid to shift just to the left of the middle of the
flanges. As a result, the section modulus at the compression (web) side Weff,z,com is
slightly bigger than the section modulus at the tension (edge fold) side Weff,z,ten.
Benchmark 40: Access Steel Document SX024a-EN-EU
179
The reference however has the inverse i.e. a big modulus at the tension side
compared to a small modulus at the compression side.
This seems to correspond to a positive weak axis moment which causes tension in
the web and compression in the edge folds. For this effective shape there is
practically no reduction so the centroid nearly stays at its original location. This
causes a big section modulus at the tension (web) side Weff,z,ten and a small section
modulus at the compression (edge fold) side Weff,z,com.
The reference seems to be applying an incorrect sign/direction of the weak
axis bending moment, causing incorrect effective section moduli values.
Benchmark 41: Access Steel Document SX025a-EN-EU
180
Benchmark 41: Access Steel Document SX025a-EN-EU
Project file: EN_Benchmark41.esa
Scia Engineer Version 10.0.86
Introduction
This benchmark concerns the example SX024a-EN-EU Design of a cold-formed
steel lipped channel wall stud in compression of Access Steel, http://www.access-
steel.com/, 2006.
This example deals with the design of a pinned wall stud subjected to
tension. The stud is made of one thin-walled cold-formed lipped channel
section.
Reference Results
The reference gives following results:
Average Yield Strength
k 7
n 4
fya 359,1 N/mm^2
Axial Tension Check
Ag 198 mm^2
Nt,Rd 71,1 kN
UC 0,675
Benchmark 41: Access Steel Document SX025a-EN-EU
181
Scia Engineer Results
Comments
- The results correspond to the benchmark results.
- The reference does not check Fn,Rd while this is more limiting than Nt,Rd. Within
Scia Engineer, to account for this M2 has been set to 1,00 so Fn,Rd is not limiting.
Benchmark 42: Stiffened Cross-section
182
Benchmark 42: Stiffened Cross-section
Project file: EN_Benchmark42.esa
Scia Engineer Version 10.0.86
Introduction
In this benchmark the effective section calculation for a stiffened cross-section is
evaluated.
More specifically the effective area in compression for a Sadef Sigma Plus section
of type SADEFSP 420x2.00 is determined. This section contains both internal
stiffeners in the web and double edge folds at the flange tips.
The cross-section is made of S390GD+Z and has a metallic coating of 0,5 mm.
The results are verified by a manual calculation; therefore the optional stiffener
iterations are not applied.
Benchmark 42: Stiffened Cross-section
183
Reference Results
The results are checked by a manual calculation.
The following picture shows the part numbers for the different elements of the
cross-section:
Since the section is symmetric, the reductions are calculated for one half of the
section.
Benchmark 42: Stiffened Cross-section
184
From the Initial Shape
1: DEF w = 9,60 mm
3: I w= 22,25 mm
5: I w= 89 mm
7: I w= 70,69 mm
9: RI w= 44,70 mm
11: I w= 218,36 mm
rm = 4 + 1,50 / 2 = 4,75 mm
From Profile library shape the depression angle is determined as 21,252 degrees.
Notional widths
1: DEF bp = 9,60 + 4,75 * sin ( (90 - 20,05) / 2) = 12,323 mm
3: I bp = 22,25 + 4,75 * sin ( (90 - 20,05) / 2) + 4,75 * sin (90 / 2) = 28,33 mm
5: I bp = 89 + 4,75 * sin (90 / 2) + 4,75 * sin (90 / 2) = 95,718 mm
7: I bp = 70,69 + 4,75 * sin (90 / 2) + 4,75 * sin ( (90 - 21,252) / 2) = 76,75
mm
9: RI bp = 44,70 + 4,75 * sin ( (90 - 21,252) / 2) + 4,75 * sin ( (90 - 21,252) / 2)
= 50,06 mm
11: I bp = 218,36 + 4,75 * sin ( (90 - 21,252) / 2) + 4,75 * sin ( (90 - 21,252) /
2) = 223,724 mm
Epsilon = sqrt ( 235 / 390) = 0,77625
Slenderness Limit for internal compression elements in case psi = 1,00: 0,5 + sqrt
( 0,085 - 0,055 * 1,00) = 0,673205
Slenderness Limit for outstand compression elements: 0,748
Benchmark 42: Stiffened Cross-section
185
Centerline Lengths of web elements
7: I lc = 76,75 + 4,75 * [tan (90 / 2) - sin (90 / 2)] + 4,75 * [tan ( (90 -
21,252) / 2) - sin ( (90 - 21,252) / 2)] = 78,708656 mm
9: RI lc = 50,06 + 4,75 * [tan ( (90 - 21,252) / 2) - sin ( (90 - 21,252) / 2)] +
4,75 * [tan ( (90 - 21,252) / 2) - sin ( (90 - 21,252) / 2)] = 51,1948267
mm
11: I lc = 223,724 + 4,75 * [tan ( (90 - 21,252) / 2) - sin ( (90 - 21,252) / 2)] +
4,75 * [tan ( (90 - 21,252) / 2) - sin ( (90 - 21,252) / 2)] = 224,8588267
mm
13: RI lc = 50,06 + 4,75 * [tan ( (90 - 21,252) / 2) - sin ( (90 - 21,252) / 2)] +
4,75 * [tan ( (90 - 21,252) / 2) - sin ( (90 - 21,252) / 2)] = 51,1948267
mm
15: I lc = 76,75 + 4,75 * [tan (90 / 2) - sin (90 / 2)] + 4,75 * [tan ( (90 -
21,252) / 2) - sin ( (90 - 21,252) / 2)] = 78,708656 mm
Local buckling
1: DEF k = 0,43 Lambda,p = (12,323 / 1,50) / (28,4 * 0,77625 * sqrt(0,43) ) =
0,568 => Rho = 1,00 => beff = 1,00 * 12,323 = 12,323 mm
3: I k = 4 Lambda,p = (28,33 / 1,50) / (28,4 * 0,77625 * sqrt(4) ) = 0,4284
=> Rho = 1,00 => beff = 1,00 * 28,33 = 28,33 mm => be1 =
be2 = 0,5 * 28,33 = 14,165 mm
5: I k = 4 Lambda,p = (95,718 / 1,50) / (28,4 * 0,77625 * sqrt(4) ) =
1,4473 => Rho = 0,5859 => beff = 0,5859 * 95,718 = 56,081 mm
=> be1 = be2 = 0,5 * 56,081 = 28,04 mm
7: I k = 4 Lambda,p = (76,75 / 1,50) / (28,4 * 0,77625 * sqrt(4) ) = 1,160
=> Rho = 0,69857 => beff = 0,69857 * 76,75 = 53,615 mm => be1 =
be2 = 0,5 * 53,615 = 26,8076 mm
9: RI No reduction for local buckling
11: I k = 4 Lambda,p = (223,724 / 1,50) / (28,4 * 0,77625 * sqrt(4) ) =
3,383 => Rho = 0,27637 => beff = 0,27637 * 223,724 = 61,83 mm
=> be1 = be2 = 0,5 * 61,83 = 30,915 mm
Benchmark 42: Stiffened Cross-section
186
Distortional buckling Double Edge Fold 1-2-3-4-5
1: Fully effective => w = 9,60 mm
2: Rounding with angle (90 - 20,05)° => w = 2 * pi * 4,75 * ((90 -
20,05)/360) = 5,80 mm
3: Fully effective => w = 22,25 mm
4: Rounding with angle 90° => w = 2 * pi * 4,75 * (90/360) = 7,4613 mm
5: be2 = 28,04 mm => be2,w = 28,04 - 4,75 * sin (90 / 2) = 24,681 mm
=> As = [ 9,60 + 5,80 + 22,25 + 7,4613 + 24,681 ] * 1,50 = 104,69 mm^2
This section is inputted as a general cross-section to calculate the section
properties:
Is = IYLCS = 17426,81 mm^4
Benchmark 42: Stiffened Cross-section
187
b1 = 100 - (1,5 / 2) - 1,5 - 4 - 24,681 + cYLCS = 91,33 mm
b2 = 91,33 mm ( symmetrical section)
kf = 1,00 (symmetrical section in compression)
hw = sum of the centerline lengths of all elements in the web (7, 9, 11, 13, 15) =
78,708656 + 51,1948267 + 224,8588267 + 51,1948267 + 78,708656 = 484,67
mm
E = 210000 N/mm^2
mu = 0,3
=> K = [ 210000 * (1,5)^3 ] / [ 4 * (1 - (0,3)^2)] * [1 / [ 91,33^2 * 484,67 +
91,33^3 + 0,5 * 91,33 * 91,33 * 484,67 * 1,00 ] ] = 0,02852567 N/mm^2
=> Sigma,cr,s = [ 2 * sqrt ( 0,02852567 * 210000 * 17426,81 ) ] / 104,69 =
195,192 N/mm^2
=> Lambda,d = sqrt ( 390 / 195,192 ) = 1,4135 >= 1,38
=> Chi,d = 0,66 / 1,4135 = 0,4669198
=> As,red = 0,4669198 * 104,69 = 48,8818 mm^2
Distortional buckling Intermediate stiffener 7-8-9-10-11
7: be2 = 26,8076 mm => be2,w = 26,8076 - 4,75 * sin ( (90 - 21,252) / 2) =
24,1258 mm
8: Rounding with angle (90 - 21,252)° => w = 2 * pi * 4,75 * ((90 -
21,252)/360) = 5,70 mm
9: Fully effective => w = 44,70 mm
10: Rounding with angle (90 - 21,252)° => w = 2 * pi * 4,75 * ((90 -
21,252)/360) = 5,70 mm
11: be2 = 30,915 mm => be2,w = 30,915 - 4,75 * sin ( (90 - 21,252) / 2) =
28,2332 mm
=> As = [ 24,1258 + 5,70 + 44,70 + 5,70 + 28,2332 ] * 1,50 = 162,6885 mm^2
Benchmark 42: Stiffened Cross-section
188
This section is inputted as a general cross-section to calculate the section
properties:
Is = IZLCS = 64167,8190 mm^4
b1 = (1,5 / 2) + 4 + 70,69 - 24,1258 + cZLCS = 89,6672 mm
centerline length element 11: 218,36 + 4,75 * tan ( (90 - 21,252) / 2) + 4,75 * tan (
(90 - 21,252) / 2) = 224,8585 mm
b2 = 224,8585 - 4,75 * tan ( (90 - 21,252) / 2) - 28,2332 + (77,868 - CZLCS) =
232,891 mm
E = 210000 N/mm^2
mu = 0,3
=> K = [ 0,25 * (89,6672 + 232,891) * 210000 * 1,5^3 ] / [ (1 - 0,3^2) * 89,6672
* 89,6672 * 232,891 * 232,891 ] = 0,14402 N/mm^2
=> Sigma,cr,s = [ 2 * sqrt ( 0,14402 * 210000 * 64167,8190 ) ] / 162,6885 =
541,571 N/mm^2
=> Lambda,d = sqrt ( 390 / 541,571) = 0,8486 => between 0,65 and 1,38
Benchmark 42: Stiffened Cross-section
189
=> Chi,d = 1,47 - 0,723 * 0,8486 = 0,8565
=> As,red = 0,8565 * 162,6885 = 139,34 mm^2
Effective Area
Aeff = 1132,8549 - 2 * (1 - 0,5859) * 95,718 * 1,5 - 2 * (1 - 0,69857) * 76,75 * 1,5
- (1 - 0,27637) * 223,724 * 1,5 - 2 * (104,69 - 48,8818) - 2 * (162,6885 - 139,34)
= 543,387 mm^2
Scia Engineer Results
Benchmark 42: Stiffened Cross-section
190
Comments
- The results correspond to the benchmark results.
- A slight difference is due to rounding errors.
Benchmark 43: Purlin Design in Uplift
191
Benchmark 43: Purlin Design in Uplift
Project file: EN_Benchmark43.esa
Scia Engineer Version 10.0.86
Introduction
In this benchmark the uplift purlin design according to EN 1993-1-3 chapter 10 is
evaluated.
The member has a KU80/40x3.0 cross-section, a length of 3m and is fabricated of
S390GD+Z material.
At the top flange the member is connected to a diaphragm of type E96/1.50. The
bolts are positioned in the bottom flange of the diaphragm and each rib is
connected.
The extremities of the purlin are simply supported.
The purlin is loaded in uplift by two loads: a permanent point load of 5 kN in the
middle of the member and a variable line load of 2 kN/m. Both loads are
combined according to a ULS Set B combination.
Both the resistance of the cross-section according to article 10.1.4.1 as well as the
buckling resistance of the free flange according to 10.1.4.2 are checked.
The results are verified by a manual calculation in the middle of the member, at
1,5m. Due to the fact that a point load is applied at this position, also the
resistance to local transverse forces is evaluated.
Benchmark 43: Purlin Design in Uplift
192
Reference Results
The results are checked by a manual calculation.
In a first step the shear stiffness of the diaphragm is determined using MathCad
and compared to the required stiffness as given in article 10.1.1(6). In the same
calculation the rotational stiffness of the diaphragm is determined.
Benchmark 43: Purlin Design in Uplift
193
Benchmark 43: Purlin Design in Uplift
194
Benchmark 43: Purlin Design in Uplift
195
Since the shear stiffness is higher than the required stiffness the purlin may be
considered as being laterally restrained in the plane of the sheeting and thus the
provisions of chapter 10 may be applied.
Benchmark 43: Purlin Design in Uplift
196
A) Cross-section Resistance of the free flange
Equivalent Lateral Load
The combination ULS returns in the mid section a bending moment of -8,44 kNm
=> qEd = 8 * M / L^2 = 8 * 8,44 / 3^2 = 7,5022222 kN/m (Printed positive due to
uplift)
Since Iyz = 0 for this section this implies that kh0 = 0
The loading concerns Uplift loading. For Uplift the loading is assumed to act in
the middle of the flange.
=> kh = kh0 - f / h with h = 80 mm and f = 24,05 - 11,38 + 20 = 32,67 mm
=> kh = 0 - 32,67 / 80 = -0,408375
The minus sign indicates that the loading is acting in the opposite sense as
indicated in the code.
Benchmark 43: Purlin Design in Uplift
197
=> qh,Ed = -0,408375 * 7,5022222 kN/m = -3,06372 kN/m
The code indicates that the loading is acting from the web to the tip of the flange.
However, due to the minus sign of kh the loading works in inverse direction, thus
from the tip of the flange to the web (i.e. causing compression in the tip and
tension in the web)
Free Flange Geometry
For a cold formed channel section the height of the free flange is taken as 1/5 h
=> 1/5 * 80 mm = 16 mm
This length is measured including the length of the rounding. The rounding has
length (Pi/2) * (3 + 3/2) = 7,0686 mm
=> The length of the web part is: 16 - 7,0686 = 8,9314165294 mm
Benchmark 43: Purlin Design in Uplift
198
Af = 149,97 mm^2
Ifz = IZLCS = 24074 mm^4
Distance from centroid to web: 16,34 mm
=> Wfz,web = 24074 / 16,34 = 1473,32 mm^3
Distance from centroid to flange tip: 40 - 16,34 = 23,66 mm
=> Wfz,flange tip = 24074 / 23,66 = 1017,50 mm^3
Lateral Spring Stiffness
Since no anti-sag bars have been defined the length La = 3m
The connected flange with b = 40 mm
The fastener distance a = 0,5 b = 20 mm
Since this concerns a simple U-section the developed height of the web hd is
taken as the full height h => hd = h = 80 mm
Benchmark 43: Purlin Design in Uplift
199
The determination of qh,Ed indicated that the loading is pointing from the tip to
the web due to the minus sign of kh
Therefore, qh is bringing the purlin into contact with the sheeting at the purlin
web
=> bmod = a = 20 mm
The rotational spring stiffness of the diaphragm is calculated as CD = cvorh =
0,4064 kNm/m (see MathCad calculation above).
=> (1 / K ) = [[4 * (1 - 0,3 * 0,3) * 80 * 80 * (80 + 20)] / [210000 * 3 * 3 * 3] +
[80 * 80] / [ 0,4064 * 1000] = 16,158896 mm^2/N
=> K = 0,061885 N/mm^2 = 61,8854 kN/m^2
=> R = [ 0,061885 * 3000^4 ] / [ pi^4 * 210000 * 24074 ] = 10,179
Lateral Bending Moment
Since it concerns a single span member the boundary conditions are taken as
Hinged - Hinged.
Since the member is loaded by uplift the free flange is in compression.
Using the analytical solution for Hinged-Hinged boundary conditions the Mfz,Ed
value is determined in each section using MathCad:
Benchmark 43: Purlin Design in Uplift
200
Properties for the final check
Since there is no axial force, Aeff is taken as Ag from the initial shape:
Aeff = 450,36 mm^2
The cross-section has a cZLCS coordinate of 40 mm. Using the Run Analysis
tool, the effective shape for negative y-y bending is determined for a stress of 390
N/mm^2
This effective shape has an inertia Iy,eff = 4,2557 * 10^5 mm^4 and a cZLCS
coordinate of 40,79 mm (Using iterations)
=> shift in neutral axis: 40,79 - 40 = 0,79 mm upward
Weff,restrained flange (top) = Iy,eff / (80 - 40,79) = 10853,61 mm^3
Weff, free flange (bottom) = Iy,eff / (40,79) = 10433,19 mm^3
Since Weff,y is different from Wel,y the safety factor Gamma M is taken as
Gamma M1 = 1,00
Wfz = Wfz,flange tip = 1017,50 mm^3 since the lateral load causes compression
in the flange tip.
Unity Check
(10.3a) : - [(8,44 * 10^6) / 10853,61] / [390 / 1,00] + [ 0 / 450,36] / [390 / 1,00 ]
= - 1,99 + 0 = 1,99 (using absolute values)
(10.3b) : [(8,44 * 10^6) / 10433,19] / [390 / 1,00] + [ 0 / 450,36] / [390 / 1,00 ] +
[0,222 * 10^6 / 1017,50] / [390 / 1,00 ] = 2,07 + 0 + 0,56 = 2,64
Benchmark 43: Purlin Design in Uplift
201
B) Buckling resistance of the free flange
Test to see if the free flange is in tension or compression: (tension is negative,
compression is positive)
[(8,44 * 10^6) / 10433,19] + [ 0 / 450,36] = 809 => compression
=> The buckling resistance needs to be checked
Free Flange Buckling Length
La = 3000 mm
R = 10,179
Situation: Since the member has only one part for system length Ly it is seen as
Simple span.
For uplift table 10.2b is used:
There are no anti-sag bars present on the member
Eta1 = 0.694
Eta2 = 5.45
Eta3 = 1.27
Eta4 = -0.168
=> lfz = 0.694 * 3000 * ( 1 + 5.45 * 10,179 ^1.27 ) ^ -0.168 = 952,933 mm
Reduction factor for flexural buckling of the free flange
ifz = sqrt ( Ifz / Af ) = sqrt (24074 / 149,97) = 12,67 mm
Lambda1 = pi * [210000 / 390] ^ 0,5 = 72,90
Lambda,fz = ( 952,933 / 12,67 ) / 72,90 = 1,0317
Lambda,0,LT = 0,4
LTB curve b => Alpha,LT = 0,34
Fi,LT = 0,5 * [ 1 + 0,34 * (1,0317 - 0,4 ) + 0,75 * 1,0317 * 1,0317 ] = 1,006552
Chi,LT = 1 / [ 1,006552 + sqrt ( 1,006552 * 1,006552 - 0,75 * 1,0317 * 1,0317) ]
= 0,68025
Benchmark 43: Purlin Design in Uplift
202
Unity Check
(10.7) : (1 / 0,68025 ) * [ [(8,44 * 10^6) / 10433,19] / [390 / 1,00] + [ 0 / 450,36]
/ [390 / 1,00 ] ] + [0,222 * 10^6 / 1017,50] / [390 / 1,00 ] = 3,61
C) Resistance to Local Transverse Forces
Resistance to local transverse force alone
The cross-section has a single unstiffened web. The resistance is determined
according to article 6.1.7.2.
The transverse load of 6,75 kN is applied at 1,5m in the middle of the beam.
With a default bearing length Ss of 10 mm the distance of the edge of the load to a
member end becomes c = 1500 – 10/2 = 1495 mm.
hw = 80 – 3/2 – 3/2 = 77 mm
c > 1,5 hw which implies the loading is categorized as Internal Loading.
With t = 3 mm Ss/t = 10 / 3 = 3,33 < 60 which implies (6.15d) needs to be
used
k = 390 / 228 = 1,71
k3 = 0,7 + 0,3 * (90 / 90)^2 = 1,00
k4 = 1,22 – 0,22 * 1,71 = 0,84368
k5 = 1,06 – 0,06 * (3 / 3) = 1,00
Rw,Rd = = 1,00 * 0,84368 * 1,00 * [14,7 - (77 / 3) / 49,5 ] * [1 + 0,007 * 10 / 3]
* 3 * 3 * 390 / 1,00 = 42,976 kN
Unity check: 6,75 / 42,976 = 0,16
Bending resistance
The section modulus of the gross section is Wel,y = 10978,33 mm^3
The effective section modulus under uplift loading is Weff,y = 10647,88 mm^3
Since the effective section modulus is smaller than the gropss section modulus the
bending resistance is determined according to article 6.1.4.1 formula (6.4)
Benchmark 43: Purlin Design in Uplift
203
Mc,Rd = 10647,88 * 390 / 1,00 = 4,1527 kNm
With MEd = 8,44 kNm this gives:
Unity check: 8,44 / 4,1527 = 2,03
Combined bending and local transverse force
MEd = 8,44 kNm
Mc,Rd = 4,1527 kNm
FEd = 6,75 kN
Rw,Rd = 42,976 kN
Unity Check: [ (8,44 / 4,1527) + (6,75 / 42,976) ] / 1,25 = 1,75
Benchmark 43: Purlin Design in Uplift
204
Scia Engineer Results
Benchmark 43: Purlin Design in Uplift
205
Benchmark 43: Purlin Design in Uplift
206
Benchmark 43: Purlin Design in Uplift
207
Comments
The results correspond to the benchmark results.