en1993 benchmarks

1

Benchmarks Steel Code Check EN 1993

EN 1993-1-1 EN 1993-1-2 EN 1993-1-3 EN 1993-1-5


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


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


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


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


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


Part 1 - 1 : General rules and rules for buildings

EN 1993-1-1:2005/AC:2009 Corrigendum

Eurocode 3


Part 1 - 2 : General rules - Structural fire design

EN 1993-1-2:2005

Eurocode 3


Part 1 - 2 : General rules - Structural fire design


Eurocode 3


Part 1-3: General rules

Supplementary rules for cold-formed members and sheeting

EN 1993-1-3:2006


2

Eurocode 3


Part 1-3: General rules

Supplementary rules for cold-formed members and sheeting


Eurocode 3


Part 1.5 : Plated structural elements

EN 1993-1-5 : 2006

Eurocode 3


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/





http://www.access-steel.com/



3


Benchmarks 36 to 37 concern examples of Designer’s Guide to EN 1993-1-1 Eurocode 3,

The Steel Construction Institute, 2005.




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



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.


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




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


7

Reference Results


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.


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.


Comments


Benchmark 3: Material Yield Strength

9




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


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


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


Results for CS1 - HE1000X393 – S235


11

Results for CS2 - RHSCF300/100/12.5 – S275

Comments


Benchmark 4: Effective Cross-Section Area

12




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


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


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


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


15


Comments


Benchmark 5: Designer’s Guide Ex. 5.1

16




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


17


Classification under pure compression

Classification under combined loading

Comments



18




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


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


19


Comments



20




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


Shear resistance

Av 2092 mm²

Vpl,Rd 332 kN

Shear buckling does not need to be considered


Comments



21




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


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



22

Resistance to combined bending and shear

My,V,Rd 386,8 kNm


Comments



23




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


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


Npl,Rd 2937,5 kN

Bending resistance

Mpl,y,Rd 524,5 kNm


24

Resistance to combined bending and axial force

MN,y,Rd 342,2 kNm


Comments



25




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


Classification

Tube d/t 24,5

Class 1 limit 42,7

Tube Class 1


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


26


Comments



27




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


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


28

Bending resistance

Mc,y,Rd 1704 kNm

Shear resistance

Av 9813 mm²

Vpl,Rd 1959 kN


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


29



30

LTB for segment BC:

LTB for segment CD:


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.


32




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


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.


The interaction factors kij for combined bending and compression are determined

using alternative method 1 (Annex A).


33

Reference Results


Classification (under pure compression)

Web c/tw 9,50

Class 1 limit 26,85

Web Class 1


Nc,Rd 2946,5 kN

Shear resistance

Av 5533,3 mm²

Vpl,Rd 1134 kN


Bending resistance

Mc,y,Rd 174,3 kNm

Resistance to combined bending, shear and axial force

My,NV,Rd 174,3 kNm


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


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


35



36


37

Comments


- 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.


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.


39




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


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.



using alternative method 2 (Annex B).

Reference Results


Classification

Flanges c/tf 3,51

Class 1 limit 8,32

Flanges Class 1


40

Web c/tw 10,73

Class 1 limit 30,51

Web Class 1


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



My,NV,Rd 773,8 kNm

Mz,NV,Rd 503,9 kNm

2

2,04



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


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


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



42


43


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.




0,26 < 0,40 => LT = 1,00


- 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


45




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


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


Effective section properties

Aeff 341,5 mm²

eN 8,66 mm


46


Comments



47




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


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


48


Comments


- 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




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


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


50


Comments


Benchmark 17: ECCS N°119 Worked Example 1

51


Project file: EN_Benchmark17_1.esa & EN_Benchmark17_2.esa


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.


using both alternative method 1 (Annex A) and alternative method 2 (Annex B).

Reference 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


52


Nc,Rd 669 kN

Bending resistance

Mc,y,Rd 51,8 kNm

Shear resistance

Av,z 1400 mm²

Vpl,z,Rd 190 kN



My,NV,Rd 44,7 kNm


Ncr,y 3287 kN

red ,y 0,451

y 0,21

y 0,939


Cmy,0 0,782

bLT 0

Cmy 0,782

y 0,996

wy 1,135

Cyy 1,061

eq. (6.61) 0,985


53


Cmy 0,6

kyy 0,65

eq. (6.61) 0,874



54



55


Comments


- 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.


56




Introduction



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.



Reference 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


57


Nc,Rd 2714 kN

Bending resistance

Mc,y,Rd 516 kNm

Shear resistance

Av,z 5990 mm²

Vpl,z,Rd 819 kN



My,NV,Rd 468 kNm



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


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


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


59


Cmy 0,495

Cm,LT 0,495

kyy 0,492

kzy 0,847

eq. (6.61) 0,628

eq. (6.62) 1,006



60



61


62


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.


63




350 / 2310,41 = 0,1515 < 0,16 => LT = 1,00


- 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






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


64


Project file: EN_ Benchmark19_1.esa & EN_Benchmark19_2.esa


Introduction



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.



Reference Results


Classification

Web c/tw 14,0

Class 1 limit 33,00

Web Class 1


Nc,Rd 1316 kN


65

Bending resistance

Mc,y,Rd 82,7 kNm

Shear resistance

Av,z 3600 mm²

Vpl,z,Rd 488 kN




red ,y 0,605 red ,z 1,065

y 0,21 z 0,21

y 0,888 z 0,62


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


66


Cmy 0,95

kyy 1,213

kzy 0,728

eq. (6.61) 0,904

eq. (6.62) 1,112



67



68


Comments


- 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.


69




Introduction



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.



Reference Results


Classification

Web c/tw 14,0

Class 1 limit 33,00

Web Class 1


Nc,Rd 1316 kN


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



My,NV,Rd 82,7 kNm

Mz,NV,Rd 44,9 kNm

1,763

1,763



red ,y 0,605 red ,z 1,065

y 0,21 z 0,21

y 0,888 z 0,620


y 0,990

wy 1,266

z 0,883

wz 1,184

Cmy,0 0,998

Cmy 0,998


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


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



72


73



74


Comments


- 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.


75




Introduction



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.



Reference 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


76


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



My,NV,Rd 516 kNm

Mz,NV,Rd 78,9 kNm

2

1



red ,y 0,195 red ,z 0,927

y 0,21 z 0,34

y 1,00 z 0,644


77


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


red LT 0,697

LT 0,49

LT 0,827

kc 0,907

f 0,954

LT,mod 0,867


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


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


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


79



80



81


82



83

Comments


- 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’




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




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.



Reference Results


Sway imperfection

m 0,791

h 2/3

0,00264


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


Nc,Rd 2782 kN

Bending resistance

Mc,y,Rd 302 kNm

Shear resistance

Av,z 3755 mm²

Vpl,z,Rd 509,5 kN



My,NV,Rd 232,5 kNm


red ,y 0,258 red ,z 0,566

y 0,34 z 0,49

y 0,979 z 0,805


86


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


red LT 0,339

LT 0,34

LT 1,00

kc 0,752

f 0,929

LT,mod 1,00


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


87

dLT 0

Cyy 1,037

Czy 0,998

eq. (6.61) 0,588

eq. (6.62) 0,534


Cmy 0,6

CmLT 0,6

kyy 0,612

kzy 0,936

eq. (6.61) 0,508

eq. (6.62) 0,674


88



89



90


91



92

Comments


- 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’




Case’.






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




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



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




94


Comments


- 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.


95




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


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



96

Shear resistance

Av,z 3080 mm²

Vpl,z,Rd 417,9 kN



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



97

Comments



98




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


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




99

Shear resistance

Av,z 7011,5 mm²

Vpl,z,Rd 1437 kN



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



100

Comments


- 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


101




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


Buckling amplification factor

cr 29,98




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


103



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


C1 1,31

Mcr 3873 kNm

red LT 0,5127

LT 0,76

LT 0,7705

Mb,Rd 784,3 kNm


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


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


105



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


C1 1,39

Mcr 3640 kNm

red LT 0,52

LT 0,76

LT 0,7653

Mb,Rd 753,1 kNm


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


106


Column Verification


107


108


109

Rafter Verification


110


111

Buckling shape for determination of Ncr,y:


112


113

Comments


- 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.


114




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


Buckling amplification factor

cr 14,57



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


Nc,Rd 4290 kN

Bending resistance

Mc,y,Rd 965,8 kNm

Shear resistance

Av,z 8380 mm²

Vpl,z,Rd 1330 kN



116



red ,y 0,284 red ,z 1,481

y 0,21 z 0,34

y 0,9813 z 0,3495


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


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


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


Nc,Rd 3176 kN

Bending resistance

Mc,y,Rd 603,4 kNm

Shear resistance

Av,z 5985 mm²

Vpl,z,Rd 950,3 kN



118



red ,y 0,7906 red ,z 1,605

y 0,21 z 0,34

y 0,8011 z 0,3063


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


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


119

kyy 1,116

kzy 0,5859

eq. (6.61) 0,8131

eq. (6.62) 0,5385


Column Verification


120


121


122

Rafter Verification


123


124


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.


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.


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.


126




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


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



127

Bending resistance

Mc,y,Rd 45,76 kNm

Shear resistance

Av,z 1120 mm²

Vpl,z,Rd 177,8 kN



Mcr 27,20 kNm

red LT 1,297

LT 0,34

LT 0,525

Mb,Rd 24,02 kNm



128


129

Comments


- 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).


130





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


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



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



132


133

Comments



134




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


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



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



136


137

Comments



138




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


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



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



140


141

Comments



142




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


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



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



144


145

Comments


- 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.


Reference Mcr = 1362,7 kNm Scia Engineer Mcr = 1448,37 kNm

Benchmark 34: Temperature Domain

146




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


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


Fire Situation

a,t at 15 min 565 °C

ky, 0,578

kE, 0,411

Buckling resistance

Lcr,z,fi 2,45 m



147

Ncr,z 4706 kN

red ,z 0,702

red ,z, 0,833

z,fi 0,577

Nb,fi, ,Rd 772,5 kN



148

Comments



149




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


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


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.


152


153


Comments


Benchmark 35: Combined Compression and Bending

154




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.

http://www.infosteel.be/


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.


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,


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


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


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


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.


161



162


163

Comments

The results correspond to the benchmark results. A slight difference is caused by

rounding errors in the manual calculation.


164





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


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


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


Aeff 341,5 mm^2

eNy 8,66 mm


165


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


166

Results for CS2 – Idealized C-section without rounded corners:


eNy = 25,12 – 16,46 = 8,66 mm

Comments


- 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.


167




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



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


Aeff 292,8 mm^2

eNy 3,92 mm


168


Results for CS1 – Actual C-section including rounded corners:


eNy = 21,16 – 17,66 = 3,50 mm


169

Results for CS2 – Idealized C-section without rounded corners:


eNy = 20,38 – 16,46 = 3,92 mm

Comments


- 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.


170




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


This example deals with the effective properties calculation of a cold formed

lipped channel section subjected to bending about its major axis.

Reference Results



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



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


Part [mm] k beff [mm] be1 [mm] be2 [mm]

Web 198 22,58 0,914 0,959 97,5 39 58,5


Aeff 689,2 mm^2

Ieff,y 4140000 mm^4

Weff,y,c 40460 mm^3

Weff,y,t 43260 mm^3


Result for the initial calculation i.e. without stiffener iterations:


172

Result using stiffener iterations:

Comments


- 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.


173




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



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



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


Aeff 436,8 mm^2


175


Result for the initial calculation i.e. without stiffener iterations:

Result using stiffener iterations:


176

Comments


- 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.


177




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



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




178


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.


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.


180




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


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




181


Comments


- 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




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.


183

Reference Results


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.


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


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


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


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


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


21,252)/360) = 5,70 mm



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


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


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



190

Comments


- A slight difference is due to rounding errors.

Benchmark 43: Purlin Design in Uplift

191




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.


192

Reference Results


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.


193


194


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.


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.


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


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


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:


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


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


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)


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


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Comments


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