alexandria.pdf

Reprinted October 2000

Supervisors: Prof. ir H.H. Snijder Dr. ir. J.C.D. Hoenderkamp

T9 Report: Experimental research

Lateral torsional buckling of channel shaped sections

TUE BCO 99.06

Dagowin la Poutré April 1999

i

Preface This project originated as a result of unanswered questions during my graduation project. The experiments fill some gaps in the understanding of the structural behaviour of channel sections. Since the first proposal for this project many people have reacted enthusiastically. I am very indebted to all those people and institutions that have supported me to make it a success. Special thanks go to Salzgitter AG for donating channel sections and to Delta Staal BV for delivering them free of charge. Many thanks to Prof. If. H.H. Snijder and Dr. if. J.C.D. Hoenderkamp for supervising this project. I wish to record my appreciation of TNO Building and Construction Research for its financial support for this research project. Thanks are due ir. H.M.G.M. Steenbergen and ir. F.S.K. Bijlaard for the useful remarks. Thanks to the people of the laboratory for helping me executing the experiments. In particular I wish to express my gratitude to Martin Ceelen for setting up the measurement system and patiently explaining how it works.

Dagowin la Poutré

April 1999 Note with reprinted version: This report has been reprinted to meet demands and to be able to publish it in the online electronic library of the Eindhoven University of Technology (http://www.tue.nl/bib/). For this version some minor errors in typing have been corrected as well as some incorrect numbers. Specifically, the row starting with ‘ωkip’ in table 3 has been corrected. These numbers were typed incorrectly and do not effect the rest of the table. Non of these changes effect either the experimental results or the conclusions.

Dagowin la Poutré October 2000

iii

Summary In literature very little data on the structural behaviour of channel sections is found. To obtain data on the structural behaviour of channel sections a serie of experiments have been set-up. Channel sections with a span of 2.8 meters are loaded by two concentrated loads acting at a quarter span from each support, see figure A.

s.c.

c.g.

e

e+φy

xz2.8 m

figure A: section for testing To test the effect of the point of load application load has been applied at the shear centre, at the top flange, in the middle of the web and at the bottom flange. If the load is applied at the shear centre the section buckles suddenly when the ultimate load is reached. The failure load is also highest when the load is applied at the shear centre. Sudden buckling does not occur when the load is applied onto the web. Furthermore the failure load is lower than if the load is applied at the shear centre, see figure B.

Fu;2A = 37.5 = 75.8% Fu;1B

Fu;1B= 49.4 [kN]

e

Fu;2B = 43.7 = 88% Fu;1B

Fu;2C = 47.3 = 96% Fu;1B

Experiments 2XX

0

10

20

30

40

50

0 10 20 30

w 1 [mm]Vertical deflection at midspan

aver

age

load

F [k

N] 1B3

2A12B2

2C11 st order

figure B: failure load in relation to load at shear centre and load-displacement graphs of experiments The failure loads have been predicted using the Merchant-Rankine postulate. The predictions were quite conservative and need further improvement.

1

Table of contents INTRODUCTION...............................................................................................................................3

OBJECTIVES .......................................................................................................................................4 RESTRICTIONS ...................................................................................................................................4 BASIC ASSUMPTIONS .........................................................................................................................4

1 CHOICE OF EXPERIMENT....................................................................................................5 1.1 SELECTED SECTION................................................................................................................5 1.2 LOAD CASES ..........................................................................................................................6 1.3 TYPE OF LOADS......................................................................................................................8 1.4 POINT OF LOAD APPLICATION................................................................................................9 1.5 NUMBERING OF EXPERIMENTS ............................................................................................10

2 MATERIAL PROPERTIES ....................................................................................................11

3 ANALYTICAL CALCULATIONS.........................................................................................13 3.1 EXPERIMENT 1.....................................................................................................................13 3.2 EXPERIMENT 2.....................................................................................................................15

4 TEST RIG ..................................................................................................................................17 4.1 INFLUENCE OF PINNED-END JACK........................................................................................18 4.2 INFLUENCE OF PRESSURE AND TENSION JACKS ...................................................................18 4.3 APPLYING LOAD ONTO THE SECTION...................................................................................19 4.4 INFLUENCE OF SUPPORTS.....................................................................................................20

5 MEASUREMENTS...................................................................................................................29 5.1 OFFSET.................................................................................................................................29 5.2 DEFLECTION IN Z .................................................................................................................30 5.3 DEFLECTION IN Y.................................................................................................................32 5.4 ROTATION ABOUT X.............................................................................................................33 5.5 STRAIN MEASUREMENTS .....................................................................................................34

6 RESULTS...................................................................................................................................36 6.1 EXPERIMENT 1.....................................................................................................................36 6.2 EXPERIMENT 2.....................................................................................................................38

7 DISCUSSION ............................................................................................................................42 7.1 EXPERIMENT 1.....................................................................................................................42 7.2 EXPERIMENT 2.....................................................................................................................44 7.3 GENERAL CONSIDERATIONS ................................................................................................46

8 CONCLUSIONS........................................................................................................................47 8.1 EXPERIMENT 1.....................................................................................................................47 8.2 EXPERIMENT 2.....................................................................................................................47

9 RECOMMENDATIONS ..........................................................................................................48

REFERENCES..................................................................................................................................50

APPENDICES .....................................................................................................................................1

A. SECTIONS...............................................................................................................................3 UNP SECTIONS ..................................................................................................................................6

2

UPE SECTIONS...................................................................................................................................7 UAP SECTIONS ..................................................................................................................................8 PFC SECTIONS ...................................................................................................................................9

B. TENSILE TESTS ..................................................................................................................11 SPEED OF INCREMENTS....................................................................................................................12 YIELD RANGE...................................................................................................................................13 ELASTIC RANGE (STRAIN MEASUREMENTS) ....................................................................................15 RESULTS ..........................................................................................................................................15

C. LATERAL TORSIONAL BUCKLING..............................................................................16 THEORETICAL ELASTIC LATERAL TORSIONAL BUCKLING LOAD......................................................16 ULTIMATE ELASTIC-PLASTIC LATERAL TORSIONAL BUCKLING LOAD .............................................16

D. BENDING AND TORSION .................................................................................................18 BENDING..........................................................................................................................................18 TORSION DUE TO SHEAR FORCE.......................................................................................................22

E. INTRODUCING LOAD ON SECTION .............................................................................29

F. TEST RIG ..................................................................................................................................32 PLANS ..............................................................................................................................................32 PHOTOGRAPHS.................................................................................................................................36 LOAD BEARING YOKES ....................................................................................................................38 DETAILS...........................................................................................................................................40 FITTINGS TO SECTIONS ....................................................................................................................43

G. SUPPORTS............................................................................................................................47

H. CALIBRATION ....................................................................................................................49 USED EQUIPMENT ............................................................................................................................49

I. DIMENSIONS ...........................................................................................................................58 RESTRICTIONS ON DIMENSIONS.......................................................................................................58

3

Introduction This project is part of the curriculum of the study of building and construction of the Eindhoven University of Technology. At the ninth trimester students have to perform an exercise in either designing and calculating a building, doing numerical research or experimental research. This project is mainly comprised of experiments, however analytical and numerical calculations were to be carried out as well for the execution of the experiments. Chosen is to perform experiments on channel sections because they have quite a different structural behaviour from other beam elements, especially where stability is concerned. Since very little has been published about this subject the experiments are performed to obtain a better understanding of the structural behaviour. The experimental data can also be used for comparison with finite element analyses. To predict the load at which failure occurs it is proposed to use the Merchant – Rankine postulate. This postulate is used to predict the failure of frames by determining the theoretical buckling load and the plastic capacity of the frame. For channel sections it is possible to determine both the plastic section capacity and the theoretical buckling load. With these two dimensions the critical load, at which failure would occur, is predicted. In figure 1 it is shown how the failure load is predicted for a channel section. In the same picture it is shown how failure is simulated by finite element method (FEM) simulation. The experimental data will prove if this is a productive approach. More on the Merchant – Rankine postulate can be found in Over Spannend Staal [7] and other literature.

Load -deflection

F-buckling

FEM

1st order

F-plastic

F-critical

05

101520

2530

3540

0 10 20 30 40 50deflection [mm]

load

F [k

N]

failure

figure 1: Merchant Rankine postulate

This report follows largely the chronology of the project, which is illustrated in figure 2. This time frame depicts the progress as it took place, not as it was planned out.

June July August September October November December January February commencement

proposal for the project

1998 1999 conclusion

ordering sections

delivery sections

tensile tests, design test rig

changes in design test rig

experiments

writing of report

figure 2: time frame

4

Objectives • Studying the structural behaviour of channel sections, specifically stability In literature very little data on the structural behaviour of channel sections is found. The experiments will yield load – displacement curves revealing the specific structural behaviour of channel sections. • Gathering data concerning strength, stiffness and stability With analytical models first order approximations can be made to the structural behaviour of channel sections. With finite element analyses the geometrically and physically non linear behaviour can be simulated. Both methods need to be validated by experimental data. • Repetitive testing of each kind of experiment to obtain mean and variation on failure load Performing one test of each kind of experiment would yield data on structural behaviour but would leave out certainty about the failure load, i.e. the one experiment might be, due to many reasons, an exception to normal behaviour. • To check the proposed verification method It is proposed to use the Merchant – Rankine postulate to predict failure loads. These experiments will either confirm or exclude the possibility of using this verification method.

Restrictions • Members subjected to shear force, bending and torsion The effect of a lateral load, causing bending and torsion, is analysed only. No axial load is applied. • Load must be applied in such manner that no lateral support is given or failure is precipitated The stability of the member is studied. On one had the member must be free to rotate and deflect laterally otherwise it may be fail at a higher load. On the other hand the line of work of the load must remain vertical as not the precipitate failure. • The section must retain its shape • Local buckling, such as web or flange buckling, may not occur. In beam theory sectional properties do not alter under loading. This theory is used to analyse the behaviour of the sections and therefor the sections must retain it shape and no local buckling may occur.

Basic assumptions • The span will be chosen in the range of ten to thirty times the height of the section It is common engineering practise to use spans in this range.

5

1 Choice of experiment

1.1 Selected section There are two types of channel sections: hot-rolled sections and cold-formed sections. Hot-rolled sections can almost always be assigned to class 1 or class 2 cross sections according to NEN 6770 [1]. Cold-formed sections usually have thinner webs and flanges than hot-rolled sections and therefore belong to class 3 or 4. The restrictions of this project mention that the sections must retain their shape and local buckling may not occur to be able to use beam theory. This means that a section of class 1 or 2 must be chosen. Thus hot-rolled sections are chosen for the experiments.

figure 3: connecting UNP sections (obtained from [8])

Hot-rolled sections are available in different series: with parallel flanges (UPE-, UAP- and PFC- series) and with tapered flanges (UNP- serie). The UNP sections have a disadvantage in making connections, such as gables, due to the tapered flanges, see figure 3. However they are widely available. UPE, UAP and PFC sections can make connections easier due to the parallel flanges, see figure 5. They are more expensive than UNP sections and are usually not in stock.

figure 5: connecting a parallel flange channel (from [8])

These four series of channel sections are compared in figure 6 for sections having a height of 100 millimetres and it can be noticed that the difference is very little except for the UNP section. In table 1 the programmes of available heights are shown: they are different for all series. The UPE-programme has heights that correspond to IPE-sections, see figure 4.

figure 4: comparison between UPE and IPE sections (from [8])

6

table 1: heights of hot-rolled channel sections (heights with a * are available in two widths) height [mm]

UNP 80 100 120 140 160 180 200 220 240 260 280 300 320 350 350 380 400UPE 80 100 120 140 160 180 200 220 240 270 300 330 360 400PFC 100 125 150* 180* 200* 230* 260* 300* 380* 430UAP 80 100 130 150 175 200 220 250 300 The UPE section is chosen because it’s dimensions are associated with the widely used IPE-sections and because sections with parallel flanges will probably be used more often in the future.

PFC UAP UPE UNP

figure 6: comparison of the different series of hot-rolled channel sections

1.2 Load cases In structural engineering two types of loading dominate: concentrated loads and distributed loads. Concentrated loads are usually the reaction force of other structural elements such as roof purlins or columns, see figure 7. Distributed loads are reaction forces of other structural elements such as sheeting or walls, see figure 8.

A

A’

section A-A’

B

B’

section B-B’

C

C’

section C-C’ figure 7: the origin of concentrated loads

A

A’

section A-A’

sheeting wall

figure 8: the origin of distributed loads

The objective of the experiments is to test the stability of channel sections. Thus the way in which load is applied may not facilitate lateral support. In the laboratory it is virtually impossible to generate a distributed load that can move freely with the deformations of the section. Therefore concentrated loads must be used.

7

In order to determine the number of concentrated loads the mechanical consequences should be considered. The more loads the more complicated this will be and therefore only one and two concentrated load configurations will be considered. Single load: The maximum bending moment coincides with the maximum shear force. If torsion is

present in the member the sum of warping restraint torsion and Saint Venant-torsion has a distribution which is identical to the shear force distribution. However, warping restraint torsion dominates at midspan whereas Saint Venant-torsion is largest at the supports, see figure 9.

0 700 1400 2100 2800 -2

0

2 X [mm]

Mx [kNm]

Saint Venant-torsion warping restraint-torsion summation of torsion

Mx warping free to occur, axial rotation prevented

��

��

F

0,5 L 0,5 L

shear force distribution

moment distribution

figure 9: distribution of forces through member with single concentrated load

Double loads: The maximum bending moment coincides with the maximum shear force at the points

of load application but between those point shear force is absent. If torsion is present the distribution is similar to that of the shear force distribution, which means that no torsion is present between the points of load application. However, the graph shows that the Saint Venant-torsion and warping torsion are not equal to zero. They are equal but opposite in sign and thus they do not resist the externally applied torsion, see figure 10.

0 700 1400 2100 2800 -2

0

2 X [mm]

Mx [kNm]

Saint Venant-torsion warping restraint-torsion summation of torsion

Mx Mx ��

��F F

0,5 L 0,25 L

shear force distribution

moment distribution

figure 10: distribution of forces through member with two concentrated loads

From these two load configurations the second one is selected. The constant value of the bending moment between the points of load application makes it best suitable for testing the stability phenomena. An extreme value might cause large deformations due to yielding before the section buckles.

8

1.3 Type of loads The load can be applied in two ways: one using dead weight and the other using jacks. These two methods shall be discussed briefly. Dead weight: By using dead weight one can be certain that the direction of the load doesn’t change,

see figure 11. However, applying loads gently and balancing them evenly over the two points of application might be difficult. Another difficulty is the volume of weights needed to apply relatively large loads. Figure 12 shows the design of a test rig for experiments using dead weight. The load bearing yokes can move freely with de deformation of the section, hinging about the point of load application. The load is comprised of concrete blocks that are laid down on the two yokes. A possible imbalance between the two yokes will be registered by the load cells.

Jacks: Using jacks the problem emerges that one side of the jack is pinned. When lateral displacements occurs the direction of the load changes, see figure 11. A distinction can be made in tension and pressure jacks; tension jacks pull the section back by a horizontal force and thus supporting the section against buckling, whereas pressure jacks pushes the section outwards and thus precipitating buckling. The magnitude of this effect depends on the length of the rod that connects the jack with the section as well as the amount of lateral deflections. The advantage of jacks is that loads can be applied smoothly, balancing can be done quit accurately and relatively large loads can be generated easily.

F FF

Fhorz

Fvert

1 2 1 2

F

Fhorz

Fvert

pinned end

1 = commencement of loading

2 = deformed state1 2

dead weight pressure jack tension jack

jack

figure 11: different ways of applying load

The initial choice was to use dead weight due to the apparent advantages over jacks. The maximum amount of load that could be applied was about 6 tons. In order to obtain failure of the section three criteria had to be met: the span couldn’t be too short, the rigidity of the section must not be too large and the yield stress low. In the proposal for this project UPE sections with a height between 100 and 200 mm and a span between 10 and 30 times their heights have been considered for use. From these considerations the section with a height of 160 millimetres was chosen. Beforehand, uncertainty remains about one parameter, the yield stress. The quality indication, such as S235JR for example, given by the manufacturer of the product is a guaranteed lowest estimated value. The actual value is usually much higher. To by-pass this uncertainty members of 12 meters length were ordered and were not cut until the yield stress had been determined and a decision was made about the span of each experiment. The tensile tests to determine the yield stress will be discussed later on in this report.

9

front view side view

concrete weight

supports are stilleligible for change

line of work

load bearing yoke

load cells

figure 12: test rig for experiments using dead weight

Designing of the test rig, figure 12, and a number of special fittings to apply load onto the section had commenced. However during this process objections were raised against using dead weight and it was decided to go-ahead with jacks instead. An entirely different rig had to be designed and a well-founded decision to use either pressure or tension jacks had to be made. This is reported in chapter 4 Test rig.

1.4 Point of load application To decide on the point of load application one must consider the specific properties of channel sections. The sections are symmetric about the y-axis and asymmetric about the z-axis. This causes a shift in the location of the shear centre and centre of gravity. The shear centre remains on the y-axis, due to symmetry, but is removed from the centre of gravity to the other side of the web. If the section is loaded at the centre of gravity then this is considered an eccentric load, causing bending as well as torsion. If the load is applied outside of the section at the shear centre then this is considered centric loading and yields bending only, see figure 13.

F

z

y

F C S

eccentric loading

⇒ w φ

F

C S ⇒

w

centric loading

F S = shear centre C = centre of gravity w = bending deformation φ = angle of twist

figure 13: deflection configuration

In figure 7 it can be seen that nearly all connections with channel sections are eccentric. It makes a difference whether the load is applied at the top flange, y-axis or bottom flange, which will be discussed later. It has been decided to apply loads at all three points of the cross section.

10

Even though centric loading is uncommon in practice, in an experiment it yields valuable information about the stability phenomena. When the section is loaded at the shear centre it will bent and at a certain load it suddenly buckles by torsion and lateral flexure. The point at which this happens is called the bifurcation point. At this point an unstable equilibrium path of continued in plane bending (bending about the y-axis, without torsion or lateral deflection) is also possible, see figure 14.

figure 14: bifurcation point

If the section is loaded eccentrically the bifurcation point will not appear in the load-deformation graph. Therefore it was decided to include an experiment with centric loading to compare the structural behaviour and failure loads to those of an eccentrically loaded section.

1.5 Numbering of experiments The load can be applied at different locations in the cross section, see figure 16. The number stands for the location with regard to the y-axis; if the force is applied at the location of the shear centre (at y = ys) then it is referred to as ‘1’, if it is applied at the centre line of the web (y = yc) than it is referred to as ‘2’. The character stands for the location with regard to the z-axis; if the load is applied at the top flange (z = -h/2) then it is referred to as ‘A’, if it is applied at the line of symmetry (z = 0) than it is referred to as ‘B’ and if it is applied at the bottom flange (z = h/2) it is referred to as ‘C’. All tests will be done twice or more and to indicate the individual specimen for each test a number is added after the character. In figure 15 it is shown how the specimen are cut from the three original members and how they are numbered.

1B

2A

2B

2C

z-axis

y-axis

1A

1C

centre of gravityshear centre

figure 16: points of load application

tensile test 2B1 2B3member 1:

member 2:

member 3:

2B2

1B1 1B2 1B41B3

2A1 2A2 2C22C1

12 [m]3 [m]

figure 15: order in which specimens were cut from the members

11

2 Material properties The value of the yield stress is needed in the analytical calculations and must be determined first. The material is of quality S235JR as described by NEN-EN 10025 [4]. To determine the yield stress specimens must be taken from the channel sections. The proportions of the dimensions of the specimens are obtained from EN 10 002-1 [3] and are applied on the UPE 160 section, resulting in the dimensions given in figure 17.

Top view flange Cross section

R12

50 12 12 50

44

70

20

markingsTolerance +/- 0.33 [mm]

Lo = 79.9 +/- 0.8

Lc = 110

Lt = 234

b =2/3 b = 47.3

b/3 = 22.3

figure 17: dimensions of specimen for tensile test

Four tests were carried out from which the average yield stress was determined. In figure 18 (a) a plot of the second test is shown. The yield range is enlarged to make the fluctuations better visible.

(a)

Tensile test 2σ max

= 446.9

0

100

200

300

400

500

0 0.1 0.2strain [%]

Stre

ss [N

/mm

^2]

enlarged area

(b)

Tensile test 2, Yield range

Min. = 295,9

ReH = 302,4

Max = 305,9

ReL = 298.5

295

300

305

0 0.008 0.016strain [%]

Stre

ss [N

/mm

^2]

figure 18: tensile test and yield range

Code EN 10 002-1 makes a distinction between the upper yield stress, ReH, and the lower yield stress, ReL. The upper yield stress is described as ‘the stress at which the first clear descent of stress is observed’ and the lower yield stress is ‘the lowest stress that occurs in the yield range, disregarding brief irregularities.’ The upper yield stress can be determined in a straight forward manner but the lower yield stress is a bit more ambiguous to determine. The crosses in the graph of figure 18 (b) resemble points that have been measured by an evenly spaced time interval. Thus the number of crosses per equal sized line segment gives the speed of yielding, and from the density of crosses the speed can be read. In this test the lower yield stress is fitted as shown but could have been fitted through the two lowest curves since they do have a reasonable number of crosses in comparison.

12

table 2: results tensile tests

Test fy;max fy;min ReH ReL Rm 1 298.2 290.8 298.2 290.8 443.8 2 305.9 295.9 302.4 298.5 446.9 3 310.1 300.5 310.1 302.6 452.6 4 301.6 291 298.2 291.1 447.1

average 304 294.5 302.2 295.7 447.6 deviation 5.17 4.63 5.62 5.8 3.65 The results of all four tests are given in table 2 and it is noted that there is only a small difference between the lower yield stress ReL and the lowest yield stress fy;min. The lower yield stress ReL will be used in the calculations in the next chapter. Problems occurred by measuring strain and therefore the values of strain should be disregarded.

13

3 Analytical calculations

3.1 Experiment 1 The objective of this experiment is to determine the lateral torsional buckling moment. In order to do so at least three tests must be carried out to determine a mean value and deviation. If experiments 1A, 1B and 1C are performed, nine tests have to be carried out, which is too many. Therefore only experiment 1B, with the load applied at the shear centre, will be carried out. This is the purest instability problem and therefore most valuable. The experimental value on the lateral torsional buckling moment will be compared to the value obtained from formula 12.2-3 of NEN 6771 [2]. The concentrated loads F1 and F2 should be equal. During loading some differences may occur due to imbalance of the load. To be able to make a distinction between the two, different notations are used, see figure 19. In order to apply the load in the shear centre, plates are welded perpendicular onto the web.

S

F1

F2

CL/4

L/4

L/2

y

xz

(a)

S

z-axis

y-axis

1B

C

weld

plate

(b) figure 19: loading scheme of experiment 1

3.1.1 Lateral torsional buckling load and section capacity Since the load acts in the shear centre the theoretical lateral torsional moment can be determined analogous to an I-section. This is shown in Appendix C, § Theoretical elastic lateral torsional buckling load. However, this bending moment is a theoretical value only and buckling will occur at a smaller bending moment. Therefore the value of this moment needs to be reduced, see figure 20. Again, for I-sections a reduction method exists. This method is applied on channel sections but the validity is uncertain. In this method the plastic section capacity must be known and this is determined in Appendix D, § Plastic section capacity. For short spans the elastic section capacity is reached before buckling occurs. In order to investigate at which length the buckling moment governs the elastic section capacity must be determined. This is determined in Appendix D, § Elastic section capacity, by using the Von Mises criterion.

3.1.2 Proposed verification method Another approach to predict the ultimate elastic-plastic load carrying capacity is using the method of Merchant-Rankine. This method is given by the postulate of equation 1.

Mke

Mpl

M

10

ke

plrel M

M=λ

theoretical buckling

actual buckling

figure 20: difference between actual and

theoretical buckling

14

1 1 1F F Fc pl ke

= + Eq. 1

In which: Fc = ultimate elastic-plastic load carrying capacity Fpl = plastic load carrying capacity Fke = theoretical elastic lateral torsional buckling load

3.1.3 Results The results of calculations of Appendix C and Appendix D, § Bending are given in table 3 for spans from 2.8 up to 4.8 meters. The elastic-plastic lateral torsional buckling load, Fmax;s;d from NEN 6771, is considered to be the most accurate prediction for the failure load. The critical load Fc, as has been determined by using the Merchant – Rankine method, is lower and is assumed to be less accurate as prediction of the failure load for this load case. At an early stage the load was restricted to 25 kN and thus the span had to be 4 meters or over. However the idea of using dead weight was abandoned and any span could be used. Now the choice was to select a span in the range where the relative slenderness λrel was smaller or equal to one because the failure load is furthest removed from the theoretical failure load, see figure 20. Thus the selection was narrowed down to a span of 3.2 meters or shorter. Since the three available sections were 12 meters each, they could be cut in four parts of three meters. Taking into account some length for the supports the actual span came down to 2.8 meters. The predicted failure load at that span is 41.7 kN. Experiment 1B will be performed four times to determine the mean and standard variation of the actual failure load. table 3: results of calculations from Appendix D, § Bending

17.5 · h 20 · h 22.5 · h 25 · h 27.5 · h 30 · h Span 2.8 [m] 3.2 3.6 4.0 4.4 4.8 Fke 69.2 [kN] 51.3 39.6 31.5 25.7 21.4 F el 50.6 44.34 39.46 35.54 32.33 29.65 F pl 57.3 50.6 45.3 40.9 37.3 34.3 λrel 0.91 [−] 0.993 1.069 1.14 1.205 1.267 ωkip 0.727 [−] 0.67 0.617 0.569 0.527 0.489 Fmax;s;d 41.7 33.9 27.9 23.3 19.7 16.8 Fc 31.3 25.5 21.1 17.8 15.2 13.2

Experiment 1B

0

10

20

30

40

50

60

70

2.8 3.2 3.6 4 4.4 4.8Span L [m]

Forc

e [k

N]

F-keF-elF-plF-max;s;d - [2]F-c - M.R.

figure 21: results of table 3 in graph

15

3.2 Experiment 2 At this experiment bifurcational buckling does not occur due to torsion in the member. Therefore the first yield stress is taken as an indication for failure. The objective of experiments 2A, 2B and 2C is to determine the structural behaviour and failure load in comparison to experiment 1B.

s.c.F1

F2 c.g.F1

F2

experiment 2

e

e+φy

xz

(a)

2A

2B

2C

z-axiz

y-axis

e

(b) figure 22: scheme of experiment 2

To analyse this load case bending and torsion are separated. To analyse the bending effects the load is shifted to the shear centre and the same analysis as for experiment 1 is used. However, due to the eccentricity of the load a torsional moment is present in the member. The effects of this torsional moment is analysed in Appendix D, § Torsion due to shear force. The elastic section capacity is determined using the Von Mises criterion. However, determining the plastic section capacity for this load case is much more difficult than for experiment 1. To approximate the plastic section capacity the elastic section capacity is multiplied by the section shape factor.

3.2.1 Proposed verification method To determine the ultimate load the Merchant – Rankine postulate is used. The analysis in Appendix D only considers equilibrium from the undeformed state. Whether the load acts at the top flange, the web or bottom flanges makes no difference for the torsional moment. In figure 23 it can be seen that once the section is twisted the eccentricity of the load at the top flanges will increase and that of the load at the bottom flange decrease. The load at the top flanges has a negative influence on the ultimate load and the one at the bottom flange a positive one. To incorporate this effect in the prediction for the ultimate load the load, Fke is determined for a load applied at the top flange in experiment 2A, a load in at the shear centre in experiment 2B and a load applied at the bottom flange in experiment 2C. Thus the ultimate loads decrease from experiment 2C up to 2A.

3.2.2 Results For these experiments the same span is chosen as for experiment 1 to be able to compare the failure loads to experiment 1. In experiment 1 the elastic-plastic lateral torsional buckling load Fmax;s;d was considered to be the failure load. For the eccentrically loaded experiments such a failure load can not be obtained and thus the critical load obtained form the Merchant – Rankine postulate is considered to be the failure load. table 4 through table 6 give the results of the analysis of Appendix D, § Torsion due to shear force. These results are shown graphically in figure 24 through figure 26.

z-axis

2A

2C

>e

<e

-φ ��S C

��

+φ

z

y

figure 23: influence of load point location

16

table 4: experiment 2A

Span 2.8 [m] 3.2 3.6 4.0 4.4 4.8 Fke;top 55.8 [kN] 42.2 33.1 26.8 22.1 18.6 F el;torsion 30.1 27.9 25.9 24.3 22.8 21.6 Fpl;torsion 35.3 32.6 30.4 28.4 26.7 25.2 Fc;top 21.6 18.4 15.8 13.8 12.1 10.7

Experiment 2A

0

10

20

30

40

50

60

2.8 3.2 3.6 4 4.4 4.8Span L [m]

Forc

e [k

N

F-keF-elF-plF-c - M.R.

figure 24: results for experiment 2A in graph

table 5: experiment 2B

Span 2.8 [m] 3.2 3.6 4.0 4.4 4.8 Fke;S 69.2 [kN] 51.3 39.6 31.5 25.7 21.4 Fc;S 23.4 19.9 17.2 14.9 13.1 11.6

Experiment 2B

0

10

20

30

40

50

60

70

2.8 3.2 3.6 4 4.4 4.8Span L [m]

Forc

e [k

N]


figure 25: results for experiment 2B

table 6: experiment 2C

Span 2.8 [m] 3.2 3.6 4.0 4.4 4.8 Fke;bottom 85.8 [kN] 62.3 47.3 37.1 29.9 24.6 Fc;bottom 25 21.4 18.5 16.1 14.1 12.4

Experiment 2C

0102030405060708090

2.8 3.2 3.6 4 4.4 4.8Span L [m]

Forc

e [k

N]


figure 26: results for experiment 2C

17

4 Test rig A decision was made to use jacks to apply the load onto the sections. To minimise the effect of the pinned end of the jacks, the jacks had to be placed as high above the section as possible. The available height to build a test rig was restricted to 5.1 meters. For these experiments the available test rig was originally build to test heavy concrete members. This test rig was composed of a lot more beams than strictly would be needed for these experiments, resulting in a tight space for manoeuvring the sections in and out of the rig. However, with a few adjustments this test rig could be used for the experiments. In figure 27 two sections of the test rig are depicted, showing how the load is applied and how the jacks are connected to the load bearing yokes. In Appendix F more plans as well as photographs of the test rig are given. In figure 28 the mechanical working of the test rig is shown.

scale

z - axis

x - a

xis

hinge

2a Section C-C'

jack

6000 300 900

z - axisload cell

prestressing

anker

y - axis

ankerhinge

jack

3a Section A-A'

hinge

support section

load bearing yoke

cable

figure 27: test rig, in two sections

Fhorz

Fvert

pinned-end

h

v = lateral displacement

α

jack

��

��

�� pressure of bothjacks is coupled

support

(b) front view(a) cross section

F F

figure 28: loading scheme

18

4.1 Influence of pinned-end jack In § 1.3 Type of loads the advantages and disadvantages have been discussed. In this paragraph the influence of the pinned end of the jacks is investigated. The influence of a point fixed in location depends on the length of the connecting rod and the lateral displacement. The effect of these two parameters on the horizontal load in relation to the absolute load is given by equation 2. Fhorz has been plotted for 2 < h < 4 metre and v < 30 millimetre in figure 29.

( ) Fhvarctansinh,vFhorz ⋅

= Eq. 2

It is clear that when the length of the connecting rod increases the horizontal load decreases. However this length is bound to a maximum due to the available height in the laboratory. Within the available height the test rig is designed such that the maximum possible length for the connecting rod can be used. In case of experiment 2A this is about 4.3 meters, see figure 27, thus the influence is as small as possible.

4.2 Influence of pressure and tension jacks To investigate the difference in using tension or pressure jacks, a simulation has been carried out using a finite element method (FEM) analysis. The cross section of the channel section (UPE 160) has been modelled by twelve elements, as is shown in figure 30 (b) The elements are 100 millimetres long, thus making a total of 336 elements. This model of the section omits the contribution of the root radii to the torsional and bending stiffnesses. The results are therefore taken only as an indication for the difference in failure load, not as an attempt to closely simulate the experiments. The simulated load case is the one of experiment 2A. The load has been applied at the end of a connection element, having the same length as the connecting rod from figure 27. Thus, as the section moves laterally, the load will make an angle with the original direction thereby pushing the section outward or pulling it back.

tf = 10

tw = 6.5 h0 = 150

b0 = 66.75

F2F1

y,v

z,w

x,u

v = 0

v,u = 0

v,w = 0

u,v,w = 0

v = 0

h = 4260 mm

(a) (b)

F

load acts at corner of flange and web

connecting element

figure 30: finite element model

figure 29: relation between length of connecting rod and lateral load

19

The simulation has been carried out for both a pressure and a tension load and the effect on the load and displacement at failure are compared. The results are given in figure 31, the dots resemble the steps that are taken in the calculation. The load deflection curves are almost identical. A closer look at the numbers shows that the failure load for tension is 1.2 % larger than for pressure whereas the lateral displacement at failure is 0.8 % smaller.

Finite Element Method simulation

0

10000

20000

30000

40000

0 10 20 30Lateral displacement [mm]

Load

[N] pressure Jack

tension jack

figure 31: finite element simulation of the difference between tension- and pressure jacks

Since the difference in the effect of using tension or pressure is very small it has been decided to use tension jacks. They make the experiments easier, i.e. all points at which the connection rod can hinge, see figure 27, will align when using tension.

4.3 Applying load onto the section To apply load at the determined points in the cross section special fittings had to be designed. For experiment 2A it was not possible to apply load exactly on the top flange and for 2C exactly on the bottom flange. In NEN 6771 [2] art. 12.2.1 b) it is stated that a load may act at a maximum of one tenth the height of a section above the top flange to analysing lateral torsional buckling with the given method. For experiments 2A and 2C a similar restriction is imposed: the load must act at a maximum of one tenth of the height of the section above or below it. It is shown in figure 32 that the fittings comply with this restriction.

F

FF

h =

< h

/10

= 16

h /2 = < h

/10

1B 2A 2B2C

figure 32: solutions for applying load at the different points in the cross section

One of the demands for the design of the fittings was that the section must be able to rotate freely about a longitudinal axis so that twisting is not restrained. Because the section bends as well, the fittings must also allow a small angle of rotation about the y-axis of the section (perpendicular to the longitudinal axis). In Appendix F plans and photographs of the specially designed fittings are given.

20

1B: In the fitting of experiment 1B a bolt is used to carry the load across from the fitting to the weld-on plate of the section. This bolt fits exact into the fitting, but to allow rotation about the y-axis it has been notched at where it sticks through the plate.

2A: For experiment 2A a ball and socket joint is used, ensuring free rotation about all axes. 2B: In experiment 2B a solid pin is used to apply the load onto the web at the line of symmetry. In

this pin a straight groove has been cut to allow the section to rotate about the longitudinal axis, see figure F11. Because the groove is straight it does not allow rotation about the y-axis, but it is assumed that the section will come to rest on one side of it and thus rotation is not prevented.

2C: For experiment 2C a shackle is used to apply the load to the section. On the section a plate is welded in which the shackle is hooked. In this plate a rounded notch, see figure F18, makes sure it stays in position, while it can rotate freely about the longitudinal- and y-axis.

4.4 Influence of supports At the supports ball and socket joints are used to allow free rotation about the y- and z-axis. Two metal rods prevent the section from rotation about its longitudinal axis, while not preventing the section to warp, see Appendix G, figures G1 and G2. The above description of the performance of the supports is only theoretical, in practice friction and a lack of a perfect geometry will influence the performance. These influences are discussed point by point hereafter.

4.4.1 Friction (rotation about z) The ball has a surface of hardened chrome and the socket is covered with woven PTFE, which ensures low friction. In figure G3, ‘tabel 8’, maximum and minimum values of the friction coefficients are given and it is shown how the frictional moment can be calculated. The frictional moment arises due to the normal force in the joint and can thus be written as a function of the applied load. The frictional-moment is bound between a maximum and minimum value and expressions for maximum and minimum values can be obtained from the equation in Appendix G. Here they are given by equations 3 and 4. In these equations dk is the size of the diameter of the joint and is in this case equal to 114 millimetres.

( ) 0005.011415.0F0005.0dFFM kmaxmax;friction ⋅⋅⋅=⋅⋅µ⋅= Eq. 3

( ) 0005.011403.0F0005.0dFFM kminmin;friction ⋅⋅⋅=⋅⋅µ⋅= Eq. 4

α L = 700 [mm]

Mfriction

Ffriction = Mfriction / α L

y-axis

x

z-axis pointingupwards

ball and socketjoint web point of vertical

load application

top view of section:

figure 33: effect of friction in joint

If the section deflects laterally it will have to overcome the friction in the joint; the friction moment is not an active force acting on the section. To get an idea of the amount of support it is assumed that a horizontal load, Ffriction, works on the section at the position where the vertical load is applied, see figure 33. • In experiment 1B no lateral deflection will occur until buckling occurs. This means that the

friction might support the section in an unstable path of equilibrium, see figure 14, and buckling could occur at a higher level of loading then would be the case if the joint was without friction.

21

• In case of experiments 2A, B and C the section deflects from commencement of loading on, so

friction must be overcome during the whole process. Since pinned-end jacks are used a horizontal component of the applied load will arise, as has been discussed in § 4.1. The effect of this component will be countered by the friction in the joint. In figure 34 (a) the maximum and minimum force due to friction (equations 3 and 4) are plotted as well as the horizontal component of the vertical load. To plot the horizontal component as a function of the applied load and lateral displacement, equation 2 has been rewritten, in which the relation between ‘v’ and Fvert from figure 34 (b) is used.

( ) F4000

varctansinFhvarctansinF,vFhorz ⋅

=⋅

= Eq. 5

��

0 12.5 25 37.5 500

0.2

0.4

0.6

Influence of suport and jack

applied load [kN]

horizontalforce[kN]

Ffriction;min

Ffriction;max

Fhorz

(a)

area in which Ffriction

must be

v [mm]

Fvert[kN]

30

50

assumed relation

actual relation

(b) figure 34: influence of supports and jacks

The horizontal component of the applied load, Fhorz , is largely within the upper and lower bounds of the friction load, see the shaded area of figure 34 (a). Therefore it can be concluded that for experiments 2A, B and C much of the effect of friction is cancelled out by the fact that pinned-end jacks are used. For experiment 1B the effect can not be specified in the same way as for experiment 2 and thus uncertainty remains.

4.4.2 Centre point of rotation (rotation about y) At the supports rotation should be free about the y- and z-axes and the centre point of rotation should be at the centre of gravity of the section. Figure G1, from Appendix G, shows that for rotation about the z-axis this is nearly so, and the effects coming from it will be disregarded. However, for rotation about the y-axis this is explicitly not the case and has been investigated further

4.4.2.1 Support fixed in location First it will be assumed that the joint is solidly fixed in location and that no slip between the section, the support-member and the joint can occur.

22

−θx-axis

point of rotation of joint

z-axis

neutral axis

+θ

socketball

figure 35: position of joint and point of rotation

First order effects: If the socket is fixed in place the section is forced to rotate about the centre point of the ball. At the neutral axis a displacement ‘u’ will occur, which is related to the angle of rotation θ. Τhis is illustrated by figure 35. The relation between ‘u’ and θ is given by equation 6, in which θ is the first derivative of equation D3, multiplied by minus one and with x = 0 entered in.

( ) ( ) ( )( ) ( )( )

⋅α−⋅α⋅−

⋅β−⋅β⋅−⋅=

=−⋅=θ⋅=LEI6

LLLFLEI6

LLLFsin56dx

0xdwEI

1sin56sin56uy

22

y

22

y

( )( ) ( )( )( )2222

yLLLLLL

LEI6F56u α−⋅α−β−⋅β−⋅

⋅⋅≈ Eq. 6

A displacement ‘u’ of the neutral axis must be due to a normal force in the member. The normal force is the horizontal reaction force from the joint and acts at the top flange, thus adding a bending moment to the section as well, see figure 36.

FreactMreact

=

u

reactie force from joint equivalente forces deformation

Freact

Mreact = Freact·h/2

h/256

+θx-axis

z-axis

figure 36: reaction force from fixed joint

To determine the reaction force the displacement at the centre point of rotation of the joint must be equal to the displacement from equation 6. An expression is derived for this displacement due to the reaction force:

⋅⋅

⋅+⋅

⋅=

y

reactreact

EI4LhF

sin56EA2

LFu Eq. 7

Equation 7 cannot be solved for Freact. Since the angle of rotation at the support is very small (about 1º) the argument is nearly equal to the sine of the argument and thus equation 7 can be approximated by just the argument. The reaction force Freact can now be factored from it:

y

react

EI4Lh56

EA2L

uF

⋅⋅⋅+

⋅

= Eq. 8

23

Entering equation 6 for ‘u’ in equation 8 yields an expression for the reaction force. The reaction force can be written as function of the applied load, see figure 39.

( )( )( ) ( )( )[ ]

y

2222

yreact

EI4Lh56

EA2L

LLLLLLLEI6

F56

FF

⋅⋅⋅+

⋅

α−⋅α−β−⋅β⋅⋅

⋅

= Eq. 9

Second order effects: The used ball and socket joints do not allow horizontal displacements. Due to second order effects of the deflection, the supports want to come closer to each other, see figure 37. Because this displacement is restraint horizontal reaction forces are induced at the supports.

Freact;2nd

u

Freact;2ndx-axis

z-axis

x = 0

u

x = L

supports move closer to each other …

deflection w(x)

… resulting in horizontal reaction forces

u u

figure 37: second order effect of deflection

The displacement of both supports can be found by solving equation 10 for the value of ‘t’ and then subtracting it from L. The integral in equation 10 is the definition of the length of a function. The function is the bending displacement of the member, which is given by equations D3 through D5 in Appendix D..

∫ =

+

t

0

2

Ldxdx

)x(dw1 Eq. 10

Since the deflection in the experiments is small compared to the overall length of the member, the displacement of the supports can be approximated by equation 11.

∫ −

+=⋅

L

0

2

Ldxdx

)x(dw1u2 Eq. 11

The supports do not allow the horizontal displacement ‘u’ and so the second order reaction force is induced. This force can only be introduced at the top flange where the joint is resting on the section. Thus not only a normal force is induced but a bending moment as well, see figure 38.

Freact;2nd

Mreact;2nd

=

u

second order reactie force from joint equivalente forces deformation

Freact;2nd

Mreact;2nd = Freact;2nd·h/2

h/2

−θx-axis

z-axis

figure 38: effect of supports moving closer

24

The reaction force due to second order effects is equal to the force that is necessary to resist the displacement ‘u’ from figure 38. An expression for this load has been derived in equation 12, in which the first part of the denominator describes the effect of the supports moving closer and the second part the effect of the rotation θ of the section.

y

2nd2;react

EI8Lh

EA2L

)F(u)F(F

⋅+

⋅

= Eq. 12

This reaction force works in opposite direction of the first order reaction force (equation 9) and can therefore be subtracted from it. Plotting equation 9 gives the graph of figure 39.

0 1.25 104 2.5 104 3.75 104 5 1040

6 104

1.2 105

1.8 105Horizontal reaction force

applied load [N]

reactionforce [N]

first order reaction force

reaction force with secondorder effects

figure 39: horizontal reaction force at one support

The reaction force, as calculated for supports solidly fixed in place, is nearly four times as great as the applied load, as is illustrated by figure 39. The second order effects lowers the reaction force only slightly and will therefore be disregarded. It is obvious that such a force would causes deformations is the supports. If the supports deform the reaction force declines. Therefore the stiffness of the support-member will be investigated.

25

4.4.2.2 No slip at support The ball and socket joint is placed between the support-member and the section. The joint can slide over both surfaces, but will be held in place by the normal force acting at the support.

section for testing cross section at support

cross section at test rig

support-member

figure 40: top view of test rig

Freact

Freact

part of the supportmember that carriesthe reaction force

joint

supportmembers

(a) cross section at support

clamped

(c) cross section at test rig

roughsurface

smoothsurfaces

gap while loading

(b) conveyance of reaction force

x-axis

z-axis

figure 41: attachment of support to test rig

If it is assumed that the joint can not slip, then the support-member carries the full reaction force. The bottom flange of the support member is clamped onto the test rig, see figure 41 (c). This connection is modelled as a hinge, see figure 42 (b), but probably can slide as well due to the gap that emerges during loading, see figure 41 (c). Further more it will be assumed that only the bottom flange will carry the horizontal reaction force.

26

x-axis

z-axis

k

(a) support-member works as a spring

test rig

��

��

��

Freact

0.9 m

0.9 m

(b) stiffness of support-membertest rig

150

z

3react

EI48LFw

⋅=

⋅=

⋅⋅⋅⋅=

=⋅

=

−

Nmm106.2

103.2101.2481800k

EI48Lk

465

3z

3

8

[ ]463

z mm103.2121508I ⋅=⋅=

zx

moment of inertia:

bottom flange of support - member

stiffness of support-member:

figure 42: stiffness of support-member

The stiffness of the support-member is determined in figure 42 (b) and expressed as coefficient ‘k’. In § 4.4.2.1 the displacement ‘u’ of the support has been determined. If the sections is loaded, the support-member will be pressed sideways, resulting in a reaction force. This force is equal to the imposed displacement divided by the stiffness of the support-member, or Freact = u / k.

0 1 104 2 104 3 104 4 104 5 1040

1000

2000

3000

4000Horizontal reaction force

applied load [N]

Rea

ctio

n fo

rce

[N]

figure 44: reaction force from support-member

In figure 44 it is shown that the reaction force that the support-member is able to give is much lower than the reaction force determined in case of a solidly fixed joint, see figure 39.

4.4.2.3 Measurements In the previous paragraphs it was assumed that the joint could not slide over both the section and the support member. In figure 41 (b) it is pointed out that the joint has smooth steel surfaces resting on other steel surfaces. In figure 49 (a) it is shown that to obtain such a surface on the section the rolling skin has been removed. The smooth surface ensures low friction so that the displacement ‘u’ can also be facilitated by sliding of the joint, besides bending of the support-member. It is difficult to specify the amount of friction and therefore difficult to determine how much of the displacement happens due to slip. To learn what actually happens at the joint, both the sliding displacement of the ball in relation to the channel section and the bending displacement of the support-member, see figure 45, have been measured in four tests.

0 1 104 2 104 3 104 4 104 5 1040

0.25

0.5

0.75

1Horizontal displacement

applied load [N]

disp

lace

men

t u [m

m]

figure 43: calculated displacement at support (Eq. 6)

27

z-axis

x-axis u

89 6

7��

��

fixed to section

fixed to floor

L

channel 6

section

socket

ball

figure 45: measurements of displacements of supports

(a)

Experiment 2C2

0

10

20

30

40

50

-0.75 -0.5 -0.25 0 0.25 0.5 0.75

displacement of joint [mm]

load

F [k

N]

channel 8

channel 7

(b)

Experiment 2C2

0

10

20

30

40

50

-0.15 -0.1 -0.05 0 0.05 0.1 0.15

slip of joint over section [mm]

load

F [k

N]

channel 6

channel 9

figure 46: measured displacement of joint

In figure 46 the measured displacements are given; the sign of the displacements corresponds with the sign of the axes, as they are given in figure 45. The total displacement at the supports is the sum of the absolute values of the displacements from figure 46 (a) and (b). For example: at F = 40 kN the measured displacement is 0.5 millimetre (in figure 46 a) plus 0.075(in figure 46 b) is together approximately 0.6 millimetre, whereas the calculated displacement is about 0.8 millimetre, see figure 43. The discrepancy between these two values may be due to erroneous assumptions in the above presented analysis.

4.4.2.4 conclusion If the calculated reaction force is taken from figure 44 one can see that at a load of 50 kN the reaction force is 4 kN. In figure 47 the effect of the reaction force on the bending moment distribution is analysed, causing a reduction of the bending moment of 0.32/35 · 100 = 0.64%. From this it can be concluded that the shift of the point of rotation at the ball and socket joint has an effect on the overall experiment that can be neglected.

F F = 50 [kN]

Freact Freact = 4 [kN]

h/2

L/αMsupport = Freact·h/2 = 0.32 [kNm]

M= F·α·L -Msupport = 35 –0.32M ≈ 35 [kNm]

section for testing moment distribution figure 47: effect of reaction force on the bending moment distribution

This conclusion is further supported by experiment 1B1. In figure 48 the load displacement graph as it has been measured is displayed. It is clearly visible that at about 50 kN the trend of the graph changes, which indicates yielding of the section. In table 3 the load at which first yield occurs, Fel, is presented to be 50.6 [kN]. This is in good agreement with the measured value. If the supports would have caused a change in the moment distribution the measured point of yielding would be different from the calculated value.

28

Experiment 1B1

0

10

20

30

40

50

60

0 10 20 30

vertical displacement [mm]

aver

age

load

[kN

]

figure 48: load displacement of experiment 1B1

4.4.3 Torsional restraint (rotation about x) The support must prevent rotation about the longitudinal axis while leaving the section free to warp. This has been achieved by placing two rod on both sides of the section, preventing it from twisting. In appendix G the support is depicted in figures G1 and G2. Because the section bends, about the y- and z-axes friction might arise between the section and the rods. Therefore a double layer of Teflon foil has placed between them. If one looks closely at figure G1, section A-A’, it can be seen that the tip of the top flange just sticks out above the rod. Due to the direction of loading this flanges is pressed against the rod and after testing an imprint is visible. In figure 49 (a) it is shown how the supportive rod leaves marks on the tip of the flanges and in the enlarged area of the top flange, shown in figure 49 (b), the imprint is clearly visible.

contact withsupportive rod

enlarged

area strippedof rolling skin

(a) (b) figure 49: end of section with imprint

This notch might have caused the section to hook on to the support and thus impede free rotation about y- and z-axes as well as warping. This effect is difficult to quantify thus we stick to simple reasoning: if the section was to hook on to the support a reaction force in x-direction will occur. Since the support-member is very slack in bending about z, which is discussed extensively in § 4.4.2.2, this reaction force will be small and thus the influence on the overall experiment, similar to the cases discussed in §4.4.2.4, will be small.

29

5 Measurements In construction most loads are gravity induced and cause downwards deflections. The co-ordinate axes are chosen such that these deflections are considered to be positive, see figure 19 and figure 22. Due to practical reasons the sections are loaded upwards. By choosing the z-axis upwards the deflections are positive. In figure 27 it is shown how the co-ordinate system is placed. The measurements consists of displacements in y- and z-directions, rotation about the x-axis, the applied load and in experiments 1B4, 2A2, 2B3 and 2C2 of strain. In Appendix H all used equipment is listed and the calibration of the equipment is reproduced. As for the load cells: they registered different loads whereas they should have been identical. This could not have been a mechanical phenomena since the two jacks were interconnected and thus the load on both jacks must be equal. After all experiments had been carried out the load cell were calibrated once more and it turned out that the calibration factor of load cell 91 had changed considerably. After long deliberation it has been decided to use the latest calibration factors as well as the average of both registered loads in the load – displacement curves. Most measured values could not be used right away because they must first be corrected, for example, for rigid body displacements. To make a difference between the actual measured quantity and the corrected quantity the following convention is used: wm;2 = deflection measured by channel 2 in z-direction

w2 = corrected displacement at the location of channel 2

5.1 Offset At commencement of loading the section rests on blocks and the load bearing yokes rest on the section. The registration of data is started and the first number acquired is taken as zero load. Once the jacks start to go up the load bearing yoke will start to carry the section and the load cells register an increases in load. Before contact with the supports the section moves up, but no load increase is registered. This is visible from the horizontal part of the graph in figure 50 (b). Once the section is pressed up against the supports the load will start to increase again. At that point the curve has a clearly visible change in direction. The value at that point is set back to zero and is subtracted from all registered values thereafter.

(a)

Experiment 1B3

0

20

40

0 10 20 30 40vertical displacement [mm]

Load

[kN]

Wm;1 vs lc91

Wm;2 vs lc90

enlarged area

(b)

Experiment 1B3

0

0.5

1

1.5

2

2.5

0 1 2 3 4 5vertical displacement [mm]

Load

[kN

]

Wm;1 vs lc91Wm;2 vs lc90

value set to zero

(c)

Experiment 1B3

0

20

40

0 10 20 30 40vertical displacement [mm]

Load

[kN

]

Wm;1 vs lc91Wm;2 vs lc90

figure 50: removing offset

In figure 50 the deflection near one load point, wm;2, and at midspan, wm;1, are plotted against the two load cells, lc90 and lc91. The two load cells clearly give different values in figure 50 (b) while they should be equal. This might be due to the fact that the cells are not completely free of any load at commencement. The load bearing yoke, the fitting, one half of the section and one support together weigh 134 kg which is equal to a load of 1.3 kN. If we look at the graph we see that the horizontal line is at about 1.3 kN. Thus it is quite certain the right offset is subtracted from the graph of figure 50 (a), resulting in graph (c).

30

5.2 Deflection in z The displacements near the points of load application and at mid span are measured at the shear centre so no corrections need to be made for rotation of the section. Figure F10 in Appendix F shows that the wires for measuring displacements (both horizontal and vertical) are hooked onto a bolt that sticks out from the web to the location of the shear centre. In the same figure it can also be seen that it is not physically possible to measure right at the point of load application. Therefore the displacements are measured 150 millimetres removed from it, towards midspan at x = 850 and x = 1950, see figure 51.

��

700 700100 150 550 550 150 100

z-axis

w

lc90 lc91

wm;71 wm;72wm;0 wm;1 wm;2

x-axis

mid span,x = 1400

point of load applicationsupport

y-axis

��

shear centre

x = 850x = 1950

front view cross section figure 51: points at which the deflection in z-direction is measured

deflection rigid body displacement

position and shape of member atcommencement of loading

wm;71wm;72

deformed shape

figure 52: rigid body displacement and bending deformation

At the span displacements are measured from a fixed point so no distinction can be made between the rigid body displacement of the section and the deflection of the section, see figure 52. To obtain the sole bending displacement the rigid body displacements at the supports must be subtracted.

Experiment 2B1

05

1015202530354045

0 10 20 30 40

vertical displacement [mm]

Load

F [k

N]

Wm;0

Wm;1

Wm;2

figure 54: measured displacements at span

At the supports displacements up to about 3 to 5 millimetres are measured. The difference between the displacements at both supports is at most one millimetre, so the section will rotate slightly.

Experiment 2B1

05

1015202530354045

0 1 2 3 4 5

vertical displacement at supports [mm]

Load

F [k

N]

Wm; 71 vs lc90

Wm;72 vs lc91

figure 53: vertical displacements at supports

31

Therefore the displacements measured at the load points could be corrected as is shown in equations 13 and 14.

( )

−+−= 71;m72;m71;m0;m0 ww

2800850www Eq. 13

( )

−+−= 71;m72;m71;m2;m2 ww

28001950www Eq. 14

At mid span the measured displacement is corrected by the average displacements of the supports, see equation 15

+−=

2ww

ww 72;m71;m1;m1 Eq. 15

Now, if we look at the numbers of figure 53 we see that if the applied load is 40 [kN] the displacements wm;71 ≈ 2.5 and wm;72 ≈ 3.5 millimetres. By entering these values in equations 13 and 14 we find:

8.2w2800850www 0;m71;m0;m0 −=

+−= and 2.3w

28001950www 2;m71;m2;m2 −=

+−=

The average displacement of the supports is 3 millimetres. If we look at the measured displacement of channel 0 and 2 (these are the displacements that have been measured near the points of load application) in figure 54 we see that at a load of 40 kN these are approximately 16 millimetres. Now to compare the method of correcting the measured displacements of equations 13 and 14 with correcting the displacements with just the average displacements of the supports we find the

following error at channel 0: %25.110016

38.2 −=⋅− and at channel 2: %25.110016

32.3 =⋅− . These

errors are quite small and it has been decided to correct the vertical displacements with just the average displacements of the supports.

32

5.3 Deflection in y The lateral deflection at the span is measured at the same points as the vertical deflection is measured. At the supports there is no room to measure so the readings of the displacements are taken right besides the supports, see figure 55 and figure 58.

��

610 610100 150 550 550 150 100

y-axis

v

lc90 lc91

vm;70 vm;63vm;60 vm;61 vm;62

x-axis

mid spanpoint of load applicationsupport

vm;73vm;64

z-axis

9090

shear centre

top view of section

support

figure 55: points at which the deflection in y-direction is measured

At the top flange the measured displacement is less than at the bottom flange, due to the difference in stiffness of the support, see figure 56. The lateral displacement of the shear centre at the supports is considered to be the average of the two measured displacements. At both supports the displacements are about the same size therefore the average displacement of the two supports, see equation 16, is subtracted from the deflection measured at the span (vm;60, vm;61 and vm;62).

22

vv2

vv

v

64;m63;m73;m70;m

average

++

+

= Eq. 16

reading ofhorizontaldisplacements

reading ofverticaldisplacement

d

number ofchannel

figure 58: measuring of displacements at support

Experiment 2B1

051015202530354045

-6 -4 -2 0 2

Lateral displacement at support [mm]

Load

F [k

N]

Vm;73

Vm;70

figure 56: measured displacements at support

��

��

d

vm;73

vm;70

top flange average displacement

figure 57: determining average lateral displacement at support

33

5.4 Rotation about x The rotation is measured near the points of load application and at midspan by special rotation recorders. In figure 61 a recorder is depicted. A metal bar is suspended from a turning point, thus remaining upright. The angle that the bar makes with the recorder housing is registered. The housing has been clamped to the top flange but is able to hinge about the y-axis. Due to the stiffness of the supports small rotations can occur and thus the rotations at the span need to be corrected. Since the rotation at both supports is about the same size the average rotation of both supports is subtracted from the rotation measured at the span.

��

610 610100 150 550 550 150 100

y-axis

v

lc90 lc91

vm;70 vm;63

φm;5 φm;4 φm;3

x-axis

mid spanpoint of load applicationsupport

vm;73vm;64

z-axis

9090

top view figure 59: points at which rotation of the member is measured

In figure 60 and in equation 17 it is shown how the rotation at the supports can be determined from the lateral displacement.

[ ]radd

vv 73;m70;msupport

−=φ Eq. 17

hinge

turning point

weight

figure 61: measuring of rotation

��

��

d

vm;73

vm;70−φ

vm;73

vm;73

vm;70 - vm;73

figure 60: determining rotation at supports

34

5.5 Strain measurements Only in one test of the different series of experiments (1B, 2A, 2B and 2C) strain measurements have been performed. In experiment 1B the sections bends only until lateral torsional buckling occurs. In the other experiments the sections are subjected to torsion as well and the interpretation of the results is slightly different.

5.5.1 Experiment 1 In this experiment strain has been measured at two positions along the member, at midspan and at x = 850 millimetres, see figure 63. At these points displacements are measured as well. At these positions two strain gauges were placed on the top flange, see figure 62, and two on the bottom flange, symmetrically to the ones on top. The measured strain should be equal in both strain gauges on top and equal but opposite in sign to the ones on bottom. If not it indicates that torsion and/or lateral bending is present as well. Furthermore equation D22 shows how stress can be calculated at the extreme fibres if the bending moment is known. By measuring the strain and using the relation σ = E · ε the bending moment can be determined as a check, see equation 18.

ε=⇒=ε⋅=σz

EIMz

IM

E yy

y

yMy

Eq. 18

(a)

midspan

x = 850

2425

2021

2627

2223 (b)

figure 63: numbering of stain gauges and position along the member

strain gauge

strain distribution

figure 62: position of strain gauges on section

35

5.5.2 Experiments 2 In these experiments the strain is measured identically to experiment 1B with the exception that at x = 850 [mm] a rosette is added between the two strain gauges on the top flange. Axial strain: This rosette is placed such that the axial strain is measured at the point where the warping function is equal to zero on the top surface. Due to a miscalculation they are all placed just next to that point (30 millimetres from the web instead of 33), see figure 64. By measuring the strain at this point a distinction can be made between strain due to bending and strain due to the bi-moment. If the strain measured by the axial component of the rosette is entered in equation 18 the contribution of bending is found. By subtracting the strain measured in the rosette from the strain measured in the two strain gauges and then using equation 19 (derived from equation D23) the bi-moment can be determined.

( )( ) ε

ω=⇒

ω=ε⋅=σ

sEI

BI

sBE wx

w

xBx

Eq. 19

Shear strain: Between the points of load application no shear stress is present due to bending. However, due to Saint-Venant torsion and warping restraint torsion there are shear stresses, but these are opposite in sign and should cancel each other out, which is shown in figure 10. By measuring the shear strain at x = 850 this can be checked.

midspan

x = 850

2425

2021

2627

2223

26

27

43

4445

top view

(a) (b) figure 65: numbering of strain gauges and rosette, and position along the member

strain gauge

rosette

warping function equal tozero on top surface

ω = -1622

ω = -213

ω = 1900

figure 64: position of strain gauges and rosette on section

36

6 Results The corrections on the measured quantities are implemented on the recorded data. This has resulted in a veriety of load-displacement curves, which are reproduced hereafter.

6.1 Experiment 1 In figure 66 the corrected displacements and rotation of the section are given at three points along the longitudinal axis: x = 850, midspan and x = 1950. In table 7 the failure loads are given. table 7: failure loads

Experiment Failure load Fu [kN] 1B1 55.64 ⇐ omission of this value will be discussed in § 7.1.1 1B2 49.8 49.8 1B3 50.02 50.02 1B4 48.37 48.37

average 50.96 49.4 standard deviation 3.204 0.895

Exp e r im e nts 1BX

0

10

20

30

40

50

60

0 10 20 30 40

w 0 [m m ]Ve rtica l de fle c tion a t x = 19 50

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

0 10 20 30 40

w 1 [m m ]Ve rtica l de fle c tion a t m idspa n

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

0 10 20 30 40

w 2 [m m ]Ve rtica l de fle c tion a t x = 85 0

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

-10 -5 0 5 10

fi 3 [ º ]rota tion a t x = 1 950

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

-10 -5 0 5 10

fi 4 [ º ]rota tion a t m idspan

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

-10 -5 0 5 10

fi 5 [ º ]rota tion a t x = 8 50

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

-15 -10 -5 0 5 10 15

v 60 [m m ]la tera l de fle c tion a t x = 195 0

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

-15 -10 -5 0 5 10 15

v 61 [m m ]la tera l de fle c tion a t m ids pa n

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4


0

10

20

30

40

50

60

-15 -10 -5 0 5 10 15

v 62 [m m ]la tera l de fle c tion a t x = 850

aver

age

load

F [k

N]

1B1

1B2

1B3

1B4

figure 66: load-displacement graphs of experiments 1BX

37

Strain measurements

Ex pe rim e nt 1B4stra in a t m idspa n

0

10

20

30

40

50

-2 -1 0 1 2stra in [‰]

aver

age

load

F [k

N]

s train gauge 20

s train gauge 21

s train gauge 24

s train gauge 25

Ex pe rim e nt 1B4stra in a t x = 850

0

10

20

30

40

50

-2 -1 0 1 2

stra in [‰]

aver

age

load

F [k

N]

s train gauge 22

s train gauge 23

s train gauge 26

s train gauge 27

figure 67: strain measured in experiment 1B4

table 8: bending moment at failure obtained from strain

at x = 850 at midspan strain gauge

My [kNm]

strain gauge

My [kNm]

26 40.41 24 35.33 27 34.53 25 32.48 22 35.49 20 34.98 23 36.27 21 36.77

average 36.67 average 34.89 The applied bending moment at failure is: My = Fu;1B4 · L/4 = 48.37 · 0.7 = 33.86 [kNm]

38

6.2 Experiment 2 Table 9 : failure loads of experiments 2xx

Experiment Failure load [kN] Experiment Failure load [kN] Experiment Failure load [kN] 2A1 37.68 2B1 43.13 2C1 48.01 2A2 37.22 2B2 43.84 2C2 46.67

2B3 44.17 average 37.45 average 43.71 average 47.34 standard deviation 0.323 standard deviation 0.529 standard deviation 0.946

Exp e r im e n ts 2XX

0

10

20

30

40

50

0 10 20 30 w 0 [m m ]

Ve rtica l de fle ction a t x = 1950

aver

age

load

F [k

N]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

0 10 20 30 w 1 [m m ]

Ve rtica l de fle ction a t m idspa n

aver

age

load

F [k

N]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

0 10 20 30 w 2 [m m ]Ve rtica l de fle ction a t x = 850

aver

age

load

F [k

N]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

-15 -10 -5 0 fi 3 [ º ]

rota tion a t x = 1950

aver

age

load

F [k

N]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

-15 -10 -5 0 fi 4 [ º ]

rota tion a t m idspa nav

erag

e lo

ad F

[kN

]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

-15 -10 -5 0 fi 5 [ º ]

rota tion a t x = 850

aver

age

load

F [k

N]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

-30 -20 -10 0v 60 [m m ]

la te ra l de fle ction a t x = 1950

aver

age

load

F [k

N]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

-30 -20 -10 0v 61 [m m ]

la te ra l de fle ction a t m idspa n

aver

age

load

F [k

N]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2


0

10

20

30

40

50

-30 -20 -10 0v 62 [m m ]

la te ra l de fle c tion a t x = 850av

erag

e lo

ad F

[kN

]

2A 1

2A 2

2B1

2B2

2B3

2C1

2C2

figure 68 : load-displacement graphs of experiments 2xx

39

Strain measurements:

Ex pe rim e nt 2A2stra in a t x = 850

0

5

10

15

20

25

30

35

40

-3 -2 -1 0 1 2Strain [‰ ]

aver

age

load

F [k

N]

s train gauge 22

s train gauge 23

s train gauge 26

s train gauge 27

Ex pe rim e nt 2A2stra in in rose tte

0

5

10

15

20

25

30

35

40

-0.5 0 0.5 1 1.5 2Strain [‰ ]

aver

age

load

F [k

N]

s train gauge 43

s train gauge 44

s train gauge 45

Ex pe rim e nt 2A2stra in a t m idspa n

0

5

10

15

20

25

30

35

40

-3 -2 -1 0 1 2Strain [‰ ]

aver

age

load

F [k

N]

s train gauge 21

s train gauge 20

s train gauge 24

s train gauge 25

Ex pe rim e nt 2A2stra in a t top fla nge a t x = 850

0

5

10

15

20

25

30

35

40

0 0.5 1 1.5 2Strain [‰ ]

aver

age

load

F [k

N]

s train gauge 26

s train gauge 27

s train gauge 43

figure 69: strain measured in experiment 2A2

Bending moment at failure: - determined from strain gauge 43: My = 40.2 [kNm] - determined from failure load: My = Fu;2A2 · L/4 = 37.22 · 0.7 = 26.1 [kNm] Bi-moment at failure: - determined from strain gauges 26: Bx = -0.17 [kNm2] - determined from strain gauges 27: Bx = -0.07 [kNm2] - determined from failure load: Bx = -0.145 [kNm2]

40

Ex pe rim e nt 2B3stra in a t m idspa n

0

5

10

15

20

25

30

35

40

45

-3 -2 -1 0 1 2Strain [‰ ]

aver

age

load

F [k

N]

s train gauge 21

s train gauge 20

s train gauge 24

s train gauge 25

Ex pe rim e nt 2B3stra in in rose tte

0

5

10

15

20

25

30

35

40

45

-0.5 0 0.5 1 1.5 2Strain [‰ ]

Forc

e (c

hann

el 9

0) [k

N]

s train gauge 44

s train gauge 43

s train gauge 45

Ex pe rim e nt 2B3stra in a t x = 850

0

5

10

15

20

25

30

35

40

45

-3 -2 -1 0 1 2Strain [‰ ]

aver

age

load

F [k

N]

s train gauge 22

s train gauge 23

s train gauge 26

s train gauge 27

Ex pe rim e nt 2B3stra in a t top fla nge a t x = 850

0

5

10

15

20

25

30

35

40

45

0 0.5 1 1.5 2

Str ain [‰ ]

ave

rag

e lo

ad F

[k

N]

s train gauge 26

s train gauge 27

s train gauge 43

figure 70: strain measured in experiment 2B3

Bending moment at failure: - determined from strain gauge 43: My = 37.64 [kNm] - determined from failure load: My = Fu;2B3 · L/4 = 44.17 · 0.7 = 30.9 [kNm] Bi-moment at failure: - determined from strain gauges 26: Bx = -0.07 [kNm2] - determined from strain gauges 27: Bx = -0.08 [kNm2] - determined from failure load: Bx = -0.172 [kNm2]

41

Ex pe rim e nt 2C2stra in a t m idspa n

0

5

10

15

20

25

30

35

40

45

50

-3 -2 -1 0 1 2Strain [‰ ]

aver

age

load

F [k

N]

strain gauge 20

strain gauge 21

strain gauge 24

strain gauge 25

Ex pe rim e nt 2C1stra in in rose tte

0

5

10

15

20

25

30

35

40

45

50

-0.5 0 0.5 1 1.5Strain [‰ ]

aver

age

load

F [k

N]

strain gauge 43

strain gauge 44

strain gauge 45

Ex pe rim e nt 2C2stra in a t x = 850

0

5

10

15

20

25

30

35

40

45

50

-3 -2 -1 0 1 2

Str ain [‰ ]

ave

rag

e lo

ad F

[k

N]

strain gauge 22

strain gauge 23

strain gauge 26

strain gauge 27

Ex pe rim e nt 2C1 a nd 2C2stra in a t top fla nge a t x = 850

0

5

10

15

20

25

30

35

40

45

50

0 0.5 1 1.5 2

Str ain [‰ ]

ave

rag

e lo

ad F

[k

N]

strain gauge 26

strain gauge 27

strain gauge 43 (2C1)

figure 71: strain measured in experiment 2C2 (measurements with rosette are from experiment 2C1)

Bending moment at failure: - determined from strain gauge 43: My = 35.68 [kNm] - determined from failure load: My = Fu;2C1 · L/4 = 48 · 0.7 = 33.6 [kNm]

42

7 Discussion In general it can be said that the objectives of this study have been achieved and that the experiments have been performed according to the restrictions. A few points are articulated: • The wanted experimental data on strength, stiffness and stability has been acquired • This data can be compared to analytical models and numerical calculations, as will be hereafter. • By repetitive testing mean and variation on failure loads have been determined. The variation

turned out to be very low, indicating conscientious execution of the experiments. • The Merchant – Rankine postulate has been checked against the experimental data. The results

are not satisfying yet but leave room for improvement. • According to the restrictions the sections must be loaded such that no support is given or failure

is precipitated. In chapter 4 a long investigation has been carried out to quantify the effects of the test rig on the course of the experiments. It can be concluded that most effects are minimal. Only the effect on the lateral torsional buckling load was not possible to quantify.

7.1 Experiment 1

7.1.1 Failure The objective of experiment 1 is to determine the lateral torsional buckling load. In figure 14 theoretical paths of displacements and rotation are shown. The results, as presented in figure 66, show that the experiments follow the theoretical path closely, i.e. the section hardly rotates nor deflects laterally until failure. Thus the highest registered value of the applied load is taken as the lateral torsional buckling load. The fact that the sections did not rotate until failure is remarkable since there is a small eccentricity in applying load, due to the welding on of the perpendicular plates. This eccentricity ranges form 2 to 4 millimetres, see Appendix I. Experiment 1B1 has a failure load that is about 10 % higher than that of the other experiments. The section also buckled in the other direction. At this experiment no teflon was used at the supports. Whether or not this has contributed to the higher failure load is unclear. Nevertheless, this experiment will be disregarded in determining the mean and standard deviation of the lateral torsional buckling load. In table 7 the final results of the failure loads are shown. The average failure load of the experiments Fu = 49.4 is quite a bit larger than the predicted failure load of Fmax;s;d = 41.7 [kN] (see table 3), i.e. Fmax;s;d = 84 % Fu. This might be a reason to reconsider using the approach of code NEN 6771 [2], using the reduction curves, to predict lateral torsional buckling failure of channel sections. Perhaps a special reduction curve can be obtained for channel sections. In figure 72 it is shown how the theoretical lateral torsional buckling moment Mke is reduced with the a-curve of NEN 6771 [2]. At λrel = 0.91 (slenderness of experiment 1B, see table 3) Mke is reduced with ωkip = 0.727. If we look at the actual buckling moment Mu the reduction would be Mu / Mke = 34.6 / 48.4 = 0.7. By performing experiments on sections with other spans a new reduction curve can be constructed. It might be possible to construct a reduction curve based on finite element analyses as well. A simulation of this experiment using the finite element method will show whether or not the results are in good agreement with the experimental data.

43

Mke = 48.4 Mpl = 40

M [kNm]

1 0

ke

plrel M

M=λ

reduction of Mke with a-curve [2] Mmax;s;d = 29.2 (ωkip = 0.727)

actual buckling at Mu = 34.6 ⇒ reduction ω = 0.86

0.91

figure 72: predicted load for lateral torsional buckling and actual load

The predicted failure load Fmax;s;d is plotted in figure 73 to compare with the actual failure loads of experiments 1B2, 1B3 and 1B4.

7.1.2 Stiffness In Appendix D it is shown how bending and torsion can be analysed. In this experiment only bending occurs and thus the first order displacements at any point along the longitudinal axis can be found by using equations D3 through D5. With equation 15, obtained form equations D4, the bending displacement at midspan (x = 1400) can be analysed. w( )F ...F a ( )L 1400

...6 E Iy LL2 a2 ( )L 1400 2 ...F a 1400

...6 E Iy LL2 a2 14002

Eq. 20 Experiments 1BX

0

10

20

30

40

50

0 5 10 15 20 25 30

w2 [mm]Vertical deflection at x = 850

aver

age

load

F [k

N]

1B2

1B3

1B3

[Eq. 20]

NEN 6771

figure 73: bending displacement at midspan

The results of the experiments and of equation 15 are plotted in figure 73 and it is clear that there is a difference in stiffness. This is remarkable because one would expect an exact fit since only bending is present in the member. However equation 15 does neglect shear deformations. For the members of this experiment, with a slenderness of L/h = 17.5, this is expected to be very little.

7.1.3 strain In experiment 1B4 strain has been measured. In figure 67 it is shown that the strain at midspan is nearly the same as at x = 850. This is expected since the bending moment is of constant value between the points of load application. Strain gauges 20 and 21 show a small difference, indicating a small bending moment in the bottom flange. To determine the bending moment from the strain seems inaccurate: a small deviation in strain means a large deviation in bending moment. However, the values are in the order of the bending moment determined from the applied load, see table 8.

44

7.2 Experiment 2

7.2.1 Failure The difference between experiments 1B and 2B is that the load is applied eccentrically to the shear centre, causing torsion. In failure load this difference becomes apparent: Fu;1B = 49.4 versus Fu;2B = 43.7 [kN], or Fu;2B = 88 % Fu;1B. If the load is applied at the top or bottom of the section then this further influences the failure load. In figure 74 (a) the failure loads of experiment 2 are compared to the one of experiment 1. In figure 74 (b) the failure loads of experiments 2 are compared amongst themselves. Fu;2A = 37.5 = 75.8% Fu;1B

Fu;1B= 49.4 [kN]

e

Fu;2B = 43.7 = 88% Fu;1B

Fu;2C = 47.3 = 96% Fu;1B

(a) failure in relation to load at shear centre

79% Fu;2C

92% Fu;2B

Fu;2C = 47.3 [kN]

(b) failure in relation to load at bottom flange figure 74 : influence of load point location on failure load determined by experiments

Fc;2A = 21.6 =52% Fmax;s;d

Fmax;s;d = 41.7 [kN]

e

Fc;2B = 23.4 = 56% Fmax;s;d

Fc;2C = 25 = 60% Fmax;s;d

prediction relative to load at shear centre

86% Fc;2C

93.6% Fc;2C

Fc;2C = 25 [kN]

prediction relative to load at bottom flange figure 75: predicted failure loads and influence of load point location on failure

The predicted failure loads, from table 4, table 5 and table 6, are shown in figure 75 (a) in a similar fashion as the experimental failure loads. It can be concluded that they are far off from the actual failure loads, leaving all but one conclusion: the Merchant-Rankine postulate can not accurately predict the failure load. The Merchant-Rankine postulate could possibly be adapted such that it can more accurately predict failure loads. Some considerations: • In the expression for the Merchant – Rankine postulate (equation 1) the plastic section capacity

Fpl needs to be entered. In case of experiment 2 this capacity had been approximated at Fpl = 35 [kN]. If we look at the failure loads of experiments 2 we see that these are all higher, i.e. the plastic section capacity is greater than the approximated value. If the value of Fpl is increased in equation 1, the value of the critical load Fc will increase as well. Thus, by further studying how to determine the plastic section capacity the critical load might be predicted more accurately.

45

• In figure 75 (b) the predicted failure loads are given relative to the load applied at the bottom flange. If this figure is compared to figure 74 (b) we see that the relative prediction is in good agreement with the failure loads in relation to the load applied at the bottom flange.

7.2.2 Stiffness Similar to § 7.1.2 the first order deformations of the member can be analysed. In this case torsion is present in the member and the first order rotation at midspan can be analysed by using equation D18 and entering x = 1400 millimetres in it (which is midspan).

φ ( )F ..sinh( )..β λ Ltanh ( ).λ L

cosh ( )..β λ L sinh( ).1400 λ.λ L

.sinh( )..α λ Ltanh ( ).λ L

sinh( ).1400 λ .sinh( )..λ L α cosh ( ).1400 λ

.λ Lα

..F e L.G It Eq. 16

(a)

Experiments 2XX

0

10

20

30

40

50

0 10 20 30


aver

age

load

F [k

N]

2A1

2B2

2C1

1st orderdisplacem ent

(b)

Experiments 2XX

0

10

20

30

40

50

-15 -10 -5 0

fi 4 [ º ]rotation at midspan

aver

age

load

F [k

N]

2A1

2B2

2C1

1st orderrotation

figure 76: bending displacement and rotation of section at midspan

Contrary to the bending displacement the first order rotation overestimates the actual twist of the section, see figure 76 (b). The cause of this is not clear.

7.2.3 Strain In figure 69, figure 70 and figure 72 the results of strain measurements of experiment 2 is given. Due to bending and bi-moment one would expect antimetric graphs. However, strain gauges 21 and 23, which are placed on the tip of the bottom flange (in compression), switch near failure from compression to tension. Strain gauges 25 and 27, placed on the tip of the top flange (under tension) do not shown this behaviour. This is due to the fact that, besides the bending moment and the bi-moment, the sections also bents about the minor axis. In § 5.5.2 it has been expected that the flanges would be free of shear strain. The shear strain is measured by the rosette, more particularly by strain gauge 44. From the results of the measurements with the rosette it appears that this expectation holds true only for experiment 2B3. Determining the bending moment and bi-moment from the measured strain turns out to be very inaccurate. It can be concluded that, in order to get an accurate view on the strain distribution more strain gauges need to be used in one cross section. Besides gauges on the flanges, there should also be gauges on the web and on the inside of the section.

46

7.3 General considerations It can be stated that it is difficult to estimate the lateral torsional buckling load of channel sections as well as to determine the bending deflections. However, this load case is quite rare in engineering practice because one would have to adjust the section to apply load at the shear centre. In common practice if load is applied directly onto the section, as is illustrated in figure 7, it is even more difficult to predict the failure load. As for bending deformations, both centric and eccentric loading have about the same stiffness in these experiments, as is shown in figure 77. If the span is increased, it will be expected that the bending stiffness of eccentric loading in comparison to centric loading will decrease due to the twisting of the section, i.e. the section will bent about both the major and minor axes.

Exp e r im e n ts 2A

0

10

20

30

40

-20 -15 -10 -5 0

f i 4 [ º ]r o tatio n at m id s p an

ave

rag

e lo

ad F

[kN

]

2A 12A 2

1s t orderFEM


0

10

20

30

40

0 10 20 30

w 1 [m m ]V e rtica l de fle ction a t m idspa n

aver

age

load

F [k

N]

2A 12A 21 s t orderFEM


0

10

20

30

40

-30 -20 -10 0

v 61 [m m ]late r al de f le ctio n at m id s p an

ave

rag

e lo

ad F

[kN

]

2A 1

2A 2

FEM

figure 78: comparison of experimental data to analytical and numerical data

In § 4.2 Influence of pressure and tension jacks, experiment 2A has been simulated. The model still needs further fine tuning to completely comply with the properties of the actual section. Nevertheless, the results are compared to the test results. The failure load is very close to the test results: Fu;FEM = 36.7 versus Fu;2A = 37.5 [kN]. The displacements at failure is much larger indicating that the stiffness of the FEM model is less than that of the actual section, see figure 78. Further analysis using the FEM will yield accurate data on failure loads. By varying the span and the height of the section tables can be constructed on the stability of channel sections.

Experiments 2XX

0

10

20

30

40

50

0 10 20 30


aver

age

load

F [k

N] 1B3

2A12B2

2C11 st order

figure 77: bending stiffness

47

8 Conclusions

8.1 Experiment 1 In experiment 1 the lateral torsional buckling load has been determined, by applying the load in the shear centre. • In three test the section buckled in the same direction. For these tests the average failure load has

been determined at Fu;1B = 49.4 [kN] with a standard deviation of 0.895 [kN]. • In one test the section buckled in the other direction. The failure load appeared to be 10 % higher

than those of the other three experiments and was recorded at Fu;1B1 = 55.6 [kN]. • For all test hardly any rotation nor lateral bending occurred until buckling. When the ultimate

load was reached the section buckled suddenly. This is what is expected in theory. • To predict the failure load using the method of NEN 6771 [2] yields conservative values. • The bending stiffness of the section appeared to be less than calculated by first order theory, see

figure 73.

8.2 Experiment 2 In experiment 2 load has been applied onto the section at the top flange, the web and the bottom flange. The failure loads decreases when the point of load application is moved from the bottom flange towards the top flange, see figure 74 (b). This is in agreement with the predicted failure loads, see figure 75(b). However, the experimental failure loads were much higher than the predicted ones. The bending stiffness of the section appeared to be less than calculated whereas the rotational stiffness was higher than calculated by first order theory, see figure 76.

48

9 Recommendations In § 5.3 Deflection if y and in § 5.4 Rotation about x, it is shown that one has to go through a great deal of trouble to correct the measured lateral displacement and rotation of the section due to the lack of stiffness of the support. In follow-up experiments the supports might be designed as suggested in figure 79 to increase stiffness. Than the displacements at the supports must be measured at the first test. If they are small enough they can be omitted in the following test, simplifying processing the acquired data.

current situation

improved situationsection

support-members

test rig

support of bottomflange lacks stiffness

figure 79: current and improved situation at support

In figure 49 it is shown how torsion at the support causes an imprint at the tip of the flange. What the effect of the imprint is can not be quantified exactly so it is better to avoid this to happen. This can simply be done by placing a metal strip between the section and the joint, see figure 80. The disadvantage of this is that the centre point of rotation, extensively discussed in § 4.4.2, is shifted further away from the neutral axis of the section. By placing teflon foil between the metal strip, the section and the joint, most of the negative effect are countered by slip between the strip on one hand and the section and the joint on the other hand.

metal strip

joint

rod

flange doesnot stick outabove rod

figure 80: adjusting support

49

In § 4.4.1 the effect of friction at the supports is investigated. It has been stated that for experiment 1 the influence can not be specified. The joints that were used were quite old and had some damage in the PTFE-covering which might greatly increase the amount of friction. By ordering new joints and retaining them very carefully (one grain of sand will increases the friction) one can be certain that the friction at the supports is as low as possible. Measuring rotation of the section by the special rotation recorders caused some unexplained problems. In figure 81 a load rotation curve is given. The curve should be nearly equal to the tangent, which has been drawn in, but sudden changes in directions of the curve are encountered. To avoid these problems the lateral displacement of the section can be measured at the top and bottom flange. The rotation can then be obtained from the difference in lateral deflection divided by the height of the section. Nota bene: this will not increase the numbers of measuring channels since the channels of the rotation recorders become available. In Appendix I it has been investigated if the sections was within the tolerance on shape and dimension according to Ontw. NEN EN 10279 [5]. As for determining the deviation on dimensions one could confine to taking random checks. If the results of these checks are satisfying it is useless to systematically check dimensions on all sections, as has been done in these experiments. As for checking the deviation in shape, visual inspections seems to be adequate. If one section seems to be crooked one could still check the shape deviations on a flat table. Just in case of experiment 1, the sections should be checked because the sections will bend due to the welding of the perpendicular plates onto the section.

Experiments 2C2

0

10

20

30

40

50

-6 -4 -2 0

fi 5 [ º ]rotation at x = 850

aver

age

load

F [k

N]

tangent to curve

irregularity

figure 81: problems at measuring rotation

50

References [1] STAALCONSTRUCTIES TGB 1990,

Basiseisen en basisrekenregels voor overwegend statisch belaste constructies NEN 6770, NNI, Delft, 1991, 175 p.

[2] STAALCONSTRUCTIES TGB 1990, Stabiliteit

NEN 6771, NNI, Delft, 1991, 98 p. [3] Metalen –Trekproef

Deel 1: Beproevingsmethode (bij omgevingstemperatuur) EN 10 002-1, CEN, Brussel, 1990, 15 p.

[4] Warmgewalste produkten van ongelegeerd constructiestaal.

Technische leveringsvoorwaarden (bevat wijzigingsblad A1: 1993) NEN-EN 10025, NNI, Delft, 1993, 12 p.

[5] Warmgewalste U-profielen van staal.

Toleranties op vorm en afmetingen Ontwerp NEN-EN 10279, NNI, Delft, 8 p.

[6] Terrington, J.S.

COMBINED BENDING AND TORSION OF BEAMS AND GIRDERS (part I & II) London: The British Constructional Steelwork Association ltd. 1968, 103 p.

[7] Rolloos, A., et al

OVER SPANNEND STAAL, DEEL 3: CONTRUEREN B Rotterdam, Stichting Kennisoverdracht SG, 1996, 542 p.

[8] UPE, An intelligent alternitive Peine: Salzgitter AG

Appendices

Appendix Page 3

A. Sections In this appendix copies of product information on the different hot-rolled channel sections are given. Each manufacturer uses different notations for dimensions. To avoid confusion the dimensions of the chosen section, a UPE 160 section, are given in the notations of this report below and on the next page. Table A1: properties of UPE160 section quantity dimension unit height h = 160 [mm] width b = 70 [mm] h - tf h0 = 150 [mm] b - tw/2 b0 = 66.75 [mm] thickness of flange tf = 10 [mm] thickness of web tw = 6.5 [mm] thickness at root tr = 16.9 [mm] root radius r = 12 [mm] distance from centre of gravity to E

yc = 18.9 [mm]

distance from c.g. to shear centre ys = 44.13 [mm] Eccentricity of centroid of web E e = 25.23 [mm] plastic lever arm sz = 118.9 [mm] area A = 2372 [mm2] moment of inertia about major axis

Iy = 9.65 •106 [mm4]

moment of inertia about minor axis

Iz = 1.14 •106 [mm4]

torsion moment of inertia It = 6.5 •104 [mm4] warping constant Iw = 4.54 •109 [mm6] Wy;el 120670 [mm3] Wy;pl 141004 [mm3]

tf

r

tw

tr

( ) 2w

2fr tt12rt ++−=

figure A2: thickness at the root of the section

hh0

ys

e yc

bb0

r

s.c. c.g.

tf

tw

A*

A

B*B

D

C

E

z-axis

y-axis

figure A1: dimensions of UPE160 section

Table A2: values of the warping function position in

cross section

z-ordinate

[mm]

warping function [mm2]

A* - 80 ωA* = - 2604 A - 75 ωA = - 3036

A** - 70 ωA* = - 3469 B* - 80 ωB* = 2326 B - 75 ωB = 1966 E 0 ωE = 0 C 75 ωC = - ωB

C* 80 ωC* = - ωB* D 75 ωD = - ωA

D* 80 ωD* = - ωA*

ωA**

ωA

ωA*ωB*

ωB

y-axis

z-axis

x-axis,warping function ω

centre line

figure A3: warping function of a UPE 160 section

Appendix Page 5

figure A4: specifications of the ordered sections

UNP sections Data obtained from: Bridé, K.J.

Tabellen voor bouw- en waterbouwkunde Leiden: Spruyt, van Mantgem & de Does B.V., 1991

Appendix Page 7

UPE sections Data obtained from: R. Kindmann

Die Neuen UPE, 80-400, Konstruktion und Bemessung Peine Germany: Preussag Stahl AG, 55 p.

UAP sections Data obtained from: Sales programme, Structural Shapes Profil Arbed, Luxembourg, 64 p.

Appendix Page 9

PFC sections Data obtained from: Parallel Flange Channels, Section Properties and Member Capacities Ascot UK: The Steel Construction Institute, 1996, SCI-P-210

Appendix Page 11

B. Tensile Tests Determining the yield stress has been carried out as closely to codes EN 10 002-1 [3] and NEN-EN 10025 [4]. In this appendix a report is given of all facts concerning the execution of the tensile tests. First the location of the samples had to be determined. Parallel flanged channels are not mentioned in [4] so the designated location of sloped flanged channels was chosen, see figure B1. It is assumed that it will be the same for parallel flanges. The location of the specimen is at one third of the width from the tip of the flange.

figure B1: location of samples [4]

R12

50 12 110 12 50

44

79.9 +/- 1 %

20

Tolerance +/- 0.33 [mm]

234

Lo =

Lc =

Lt =

b

cros

s sec

tion

figure B2: dimensions of specimen Index: a = thickness of specimen b = width of specimen So = area of original cross section Lo = original measuring length Lc = length of the parallel section Le = measuring length of the strain gauge, see figure B3 Lt = length of the entire specimen The specimen are called proportional specimen because the original measuring length Lo is related to the original cross section by: oo SkL = with k = 5.65 In figure B2 the tolerance of the width of the specimen is given and in table B1 the area of the original cross section is determined. The width of specimen 3 has been cut too small but is used anyway.

Table B1: dimensions of cross section

Test 1 Test 2 Test 3 Test 4 a B A a b A a b A a b A

9.8 20 196 9.8 19.9 195 9.75 18.85 183.79 9.8 19.85 194.53 9.75 19.9 194.03 9.65 19.9 191.6 9.65 18.9 182.39 9.7 19.8 192.06 9.75 19.9 194.03 9.7 19.9 193 9.7 18.85 182.85 9.85 19.75 194.54 9.8 19.95 195.51

9.75 19.9 194.03 So = 194.72 So = 193.2 So = 183.01 So = 193.71

Le = 50 mmparallel section

clamped on strain gauge

figure B3: specimen in testing bench

Speed of increments

1 elastic range

2 yield range

3 plastic range

strain

stress

figure B4: different parts of the test In figure B4 the different parts of the test are indicated For each part a stress or strain increment speed for the parallel section is determined by [3]. 1 elastic range stress increment 6 – 30 [N/mm2 s-1] 2 yield range strain increment 0.00025/ s-1 - 0.0025/ s-1 3 plastic range strain increment ≤ 0.008/ s-1

Appendix Page 13

1 elastic range: The speed on the testing bench was set to 0.6 [mm/min]. From this the stress incremental speed is

calculated as follows:

⋅=

⋅=∆⋅=σ⇒

=∆=

⋅⋅

smmN0.19

11060

6.0210000

LLE

minmm6.0L

AELF

2t

t

Which falls in the allowed range. 2 yield range: During yielding the speed can not be changed, but since the yield stress was unknown before the first test the speed of the test bench has not been changed. To determine the settings for the test bench the given limits can be changed as follows:

=⋅⋅<<=⋅⋅

minmm5.1660L0025.0speedtest 65.160L00025.0 tt

The test speed of 0.6 mm/min was too slow. This had not been noticed until the last test, test number 3, at which the speed was erroneously incremented up to 1.2 mm/min which is also too slow. 3 plastic range: In this range no lower value for the strain increment is given, the upper value is considerably higher:

=⋅⋅<

minmm8.5260L008.0speedtest t

In this range only a maximum is set down and at no time speeds were reached that exceeded this maximum. In table B2 the speed at which individual experiments were carried out are given.

Yield range The specimen were tested in the 100 kN testing bench and the data was acquired by computer. The time interval of measuring was 1 second. In table B2 the date and duration of the experiment is shown. table B2: duration of experiments

number of speed [mm/min] test date duration observations elastic- yield- plastic- range 1 September 7, 1998 73 min 4380 0.6 0.6 0.6 2 September 8, 1998 65 3900 0.6 0.6 0.6 3 September 9, 1998 34 2040 0.6 1.2 3.2 4 September 8, 1998 55 3300 0.6 0.6 1.9

Te nsile Te st 1Yie ld ra nge

Min. = 290,8

Max. = 298 ReH = 298

ReL = 290,8

290

295

300

0 0.005 0.01 0.015 0.02

s tr ain [%]

Stre

ss [N

/mm

^2]

Te nsi le te st 2Yie ld ra nge

Min. = 295,9

ReH = 302,4

Max = 305,9

ReL = 298.5

295

300

305

0 0.005 0.01 0.015 0.02

s tr ain [%]

Stre

ss [N

/mm

^2]

Te nsi le Te st 3Yie ld ra nge

Min. = 295,5

Max./ ReH = 310,1

ReL = 302,6

300

304

308

312

0 0.005 0.01 0.015 0.02

s tr ain [%]

Stre

ss [N

/mm

^2]

Te nsi le te st 4Yie ld ra nge

Min. = 291

ReH = 298,2

Max. = 301,6

ReL = 291

290

295

300

305

0 0.005 0.01 0.015 0.02

s tr ain [%]

Stre

ss [N

/mm

^2]

figure B5: yield range of the four tensile tests

Appendix Page 15

Elastic range (strain measurements) In figure B3 it is shown how strain is measured at the parallel section of the test specimen. The strain plotted against the stress yields figure B4. If we just look at the elastic range we expect to find a straight line with a tangent equal to the elasticity modulus, E = 210,000 [N/mm2]. However, the linear fit shows a tangent of only approximately 140,000 [N/mm2]. It is understood that the enormous discrepancy between the measured value and the theoretical value is due to slip between the clamped-on strain gauge and the specimen. Subsequently all values of strain must be disregarded.

Tensile Test 2, elastic range

linear fit: y = 142480x - 1.3284

R2 = 0.9966

0

50

100

150

200

250

300

0 0.0005 0.001 0.0015 0.002 0.0025strain [%]

Stre

ss [N

/mm

^2]

commencement of yielding

measured data

figure B6: strain measurements at the elastic range

Results The results of the tensile tests are given in table B3. The average value or the lower yield stress, ReL, will be used as value for the yields stress in further calculations. All values of measured strain will be disregarded. table B3: results of tensile tests

Test fy;max fy;min ReH ReL Rm 1 298.2 290.8 298.2 290.8 443.8 2 305.9 295.9 302.4 298.5 446.9 3 310.1 300.5 310.1 302.6 452.6 4 301.6 291 298.2 291.1 447.1

average 304 294.5 302.2 295.7 447.6 deviation 5.17 4.63 5.62 5.8 3.65

C. Lateral torsional buckling

Theoretical elastic lateral torsional buckling load The theoretical elastic lateral torsional buckling moment, Mke, is determined using formulae from code NEN 6771 [2]. The notations of [2] code are used in this appendix. Table C1 gives the coefficients that apply to the chosen load case. table C1: load case coefficients according to [2] Load case

Load configuration C1 C2

4 F F

��

��

0,25 L 0,5 L 0,25 L

1.04

0.42

A few parameters need to be determined to find the coefficient C:

S h E IG I

d z

d t=

2 NEN 6771: (12.2-11)

S = 539.5 [mm] Coefficients C1 and C2 are from table C1; C2 = 0 when the load is applied at the shear centre:

( )CC ll

Sl

C C Sl

g= + + +

π π π1

1

2 2

12 2

2 2

11 1 NEN 6771: (12.2-10)

C = 3.5 [-] The coefficient kred is set down for double symmetric sections in [2] but not for channels loaded in the shear centre. It is assumed that this coefficient will be equal to unity. The theoretical elastic lateral torsional buckling moment can now be determined using the coefficients kred and C.

M k Cl

E I G Ike redg

d z d t= NEN 6771: (12.2-10)

This moment has a linear relation with the elastic lateral torsional buckling load. From the moment distribution it can be derived that Fke. = α ⋅ Mke / L, with α = 0.25 in this case.

Ultimate elastic-plastic lateral torsional buckling load The theoretical elastic lateral torsional buckling load has been determined. However, the material is not fully elastic, buckling may occur at a smaller load than has been determined. To determine the ultimate elastic-plastic lateral torsional buckling load, denoted by My;max;s;d, equation 12.2-3 from NEN 6771 is used.

Appendix Page 17

MM

y s;d

kip y u d

;max;

; ;ω≤ 1 NEN 6771: (12.2-3)

Thus M My; s d kip y;u dmax; ; ;≤ ω and for class 1 and 2 cross sections the following applies: My;u;d = My;pl;d NEN 6770: (11.2-6) The plastic section capacity, My;pl , is determined in Appendix D. The relative slenderness, λ, can be determined from Mke and My;u;d :

λ =MM

y;u d

ke

; NEN 6771: (12.2-4)

Table 23 of [1] assigns instability curves for different structural shapes. For channel sections curve c is assigned for instability about any axis. However this table seems to apply to buckling due to normal force. Therefore the analogy with lateral torsional buckling of an I-section is used in determining the instability curve. The section is bent about the y-y axis and both demands h/b = 160/70 = 2.1 > 1.2 and tf = 10 < 40 [mm] are met. Thus the ‘a’ curve is used. Using equation 12.1-14 of NEN 6770 ωkip can be determined and entered in: M My; s d kip y;u dmax; ; ;≤ ω The ultimate elastic-plastic lateral torsional buckling load, Fmax;s;d , can be determined from the moment distribution.

D. Bending and Torsion

Bending The chosen load configuration is one with two concentrated loads at a quarter span of each support. The distribution of forces, moments and deformations is given in figure D1 qualitatively.

x

z

F1 F2

α·L β·L L

Vz

My

w

θ

shear force

bending moment

angular rotation about the y-axis

displacement in z-direction

α = 0.25

β = 0.75

Vz = F

Vz = -F

My = α·L·F

( ) ( )dx

xdwx −=θ

( ) ( )dx

xdEIxM yθ=

( ) ( )dx

xdMxV =

w(x)

figure D1: distribution of shear force, bending moment and deformations. In order to find the exact values at each point along the beam axis an expression must be found for the bending displacement w. By differentiating expressions are found subsequently for the angular rotation θ, the bending moment My and the shear force Vz. For a single load an expression for the bending displacement w can be found. The load F1 is placed at a = α·L from one support and b = β·L is the complementary distance from the other support, see figure D2 (a).

F1

L

a = α·L b = β·L

(a)

F1

L

a b -aF2

a

(b) figure D2: position of concentrated load The expression is given in two parts:

0 ≤ x < a: ( )222

yxbL

LEI6xbF)x(w −−⋅⋅

⋅⋅= Eq. D1

a < x ≤ L: ( ) ( )[ ]222

yxLaL

LEI6xLaF)x(w −−−

⋅⋅−⋅= Eq. D2

Appendix Page 19

If a second load F2 is placed symmetrically from midspan to F1 the expression can be found by simply swapping the parameters ‘a’ and ‘b’ in equations D1 and D2. If both loads are applied simultaneously the expression for the bending displacement is found by superposing the expressions for the individual loads, yielding equations D3 through D5.

0 ≤ x < a: ( ) ( )222

y

222

yxaL

LEI6xaFxbL

LEI6xbF)x(w −−

⋅⋅⋅⋅+−−

⋅⋅⋅⋅= Eq. D3

a < x ≤ b: ( ) ( ) ( )[ ]222

y

222

yxLaL

LEI6xLaFxaL

LEI6xaF)x(w −−−

⋅⋅−⋅+−−

⋅⋅⋅⋅= Eq. D4

b < x ≤ L: ( ) ( )[ ] ( ) ( )[ ]222

y

222

yxLbL

LEI6xLbFxLaL

LEI6xLaF)x(w −−−

⋅⋅−⋅+−−−

⋅⋅−⋅= Eq. D5

In figure D3 the bending displacements due to F1 and F2 are given with α = 0.25 and subsequently β = 0.75. The chosen load is F1 = F2 = 50 [kN] and the span is 2.8 [m]. If the two graphs for the individual loads F1 and F2 are added up the bending displacement for experiment 1 is found. The angular rotation θ is found by differentiation equations D3 through D5, the moment distribution by differentiating twice and the shear force by differentiating three times. This relation is also shown in figure D1 The moment distribution and the shear force can be found quite easily as is shown in figure D1.

0 700 1400 2100 28000

5

10

15

super posed displacement

displacement due to F1

displacement due to F2

Bending displacement

w[mm]

x [mm] figure D3: bending displacement due to F1, F2 and superposed displacement

Elastic section capacity The load at which first yield occurs is called the elastic section capacity. First yield will occur at the largest combination of axial stress, due to bending, and shear stress, due to the shear force. The axial stress is largest at the top- and bottom flange at z = ± h/2, see figure D4 c). If the shear stress distribution along the centre lines is considered the largest stress is at point B and C (see figure A1 for the location of point B and C) in the cross section, see figure D4 a). However, at B and C the thickness tr is larger than the thickness tf of just the flange. This is due to the root radius, see figure A1.

a)

C

��

��

��

τmax

shear stress at centre lines b)

C

h0/2

��sB**

C**elastic section modulus Sy

s = 0

s = b0

s

c)

C

��

��

��

��

σx,max -

+axial stress distribution

h/2

figure D4: stress distribution To investigate at which point the shear stress is largest two points are considered, point B and point B**. The shear stress is given by equation D6, in which Sy(s) is the elastic section modulus which is a function of ‘s’ and t(s) is the thickness at ‘s.’

( )( )stI

sSV

y

yzxs =τ Eq. D6

Section modulus at: B Sy = b0 ⋅ tf ⋅ h0/2 = 50062.5 [mm3]

B** Sy = (b-tw-r) ⋅ tf ⋅ h0/2 = 38625 [mm3] Shear stress at: B τxs = Vz /Iy ⋅ Sy /tr = Vz /Iy ⋅ 50062.5/16.9 = 2962.3 Vz /Iy [N/mm2] B** τxs = Vz /Iy ⋅ Sy /tf = Vz /Iy ⋅ 38625/10 = 3862.5 Vz /Iy [N/mm2] This shows that the largest combination occurs at point B** and C**. The Von Mises criteria is used to determine the elastic section capacity at these points. This criteria is defined by equation D7.

y2tot

2tot f3 ≤τ+σ Eq. D7

with:

3

f

f

yVtot

yMtot

z

y

≤τ=τ

≤σ=σ

Eq. D8

in which σMy stands for axial stress due to bending and τVz shear stress due to the shear force. If the expressions for the axial stress and shear stress are entered in equation D7 one finds:

( )( )

( )M zI

V S sI t s

L FI

F t b t rI t

fy

y

z y

y

h

y

f wh

y fy

+ ⋅

=⋅ ⋅

+ ⋅⋅ − −

≤2 2

2

2

2

2

3 30α

Eq. D9

Appendix Page 21

It must be know at which load the elastic section capacity is reached. Writing equation D9 explicit for the load F yields:

( )F

f

LI

b t rI

ely

h

y

wh

y

=⋅ ⋅

+ ⋅− −

α 2

2

2

2

30

Eq. D10

With equation D10 the load at which the elastic section capacity is reached, denoted by Fel, can be analysed for any span.

Plastic section capacity If the load is increased after the elastic section capacity is reached an increasing area of the section will yield. The plastic capacity is reached when the entire section yields, see figure D6 for the stress distribution. The top part is in compression and the bottom part in tension. Of both halves of the section the centre of gravity is determined and represented by two axes, see figure D6. The distance between the two centres of gravity is called the plastic lever arm and denoted by sz.

[ ]W A s mmy pl z,. .= ⋅ = ⋅ = ⋅

223718

2118 9 141 103 3 Eq. D11

The plastic section capacity due to bending can be determined by: My;pl = fy ⋅ Wy,pl = 295.7 ⋅ 141⋅103 = 41.7 [kNm] Eq. D12 However there is also a shear force present and the plastic section capacity must be reduced. This is done by rewriting equation D12 to: My;pl = σpl ⋅ Wy,pl Eq. D13 In which 22

ypl 3f τ−=σ

Expanding equation D13 and factoring for Fpl gives:

( )( )

( )

2

0

0

y

002pl;y

22

2pl;y

2y

pl

22

0

0

y

00pl2pl;y

2pl;y

2y

2pl

pl;y

2

0

0

y

00pl2ypl;y

2

y

yz2ypl

pl;y22

yplpl;y

b4h

1I2hb

w3L

wfF

b4h

1I2hbF

w3wfLF

wb4

h1

I2hbF

3fwstIsSV

3fLF

w3fLFM

⋅

+⋅

⋅+⋅α

⋅=⇒

⋅

+⋅

⋅−⋅=⋅α⋅⇒

⋅

⋅

+⋅

−=⋅

−=⋅α⋅

⋅τ−=⋅α⋅=

Eq. D14

sz

��

��

��

-fy

fy

figure D5: plastic stress distribution

Torsion due to shear force Torsion is originated if a shear force is applied eccentric to the shear centre. Figure D6 shows that if a channel section is loaded on the web torsion is induced in the member. This figure also shows the assumed positive directions of the axes.

s.c.F1

F2 c.g.F1

F2

experiment 2

e

e+φy

xz

figure D6: origination of torsion

Mx1= -F⋅e Mx2= F⋅e��

α ⋅ L

L

warping free to occur,axial rotation prevented��

��

��

β ⋅ L

figure D7: position of torsional loads along the axis In [6] a solution of the differential equation of torsion is given for a single concentrated torsional load:

0 ≤ x < α⋅L : ( ) ( )( ) ( ) ( ) ( )

t

x

GILM

Lx1

LxsinhLcosh

LtanhLsinhx

α−+

λλ

αλ−

λαλ=φ Eq. D15

β⋅L ≤ x < L: ( )( )( ) ( ) ( ) ( )

( )t

x

GILM

LxL

L

xcoshLsinhxsinhLtanhLsinh

x

α−+

λ

λαλ−λλ

αλ

=ϕ Eq. D16

In which λ is the section parameter for torsion which is defined by: w

t

EIGI

=λ

From figure D6 it follows that the angular rotation is negative with regard to the chosen co-ordinate system, therefore the torsional moment is Mx = - F ⋅ e. The eccentricity ‘e’ is 25.23 [mm], see Appendix A, and the force for which these figures are made is 50 [kN], thus the torsional moment is Mx = 1.26 [kNm].

Appendix Page 23

0 700 1400 2100 28000.15

0.07

0

superposed angular rotationangular rotation due to Mx;t1

angular rotation due to Mx;t2

Angular rotation

Span L [mm]

φ[rad]

figure D8: angular rotation due to Mx;t1, Mx;t2 and superposed rotation By superposing equations D15 and D16 for a torsional load at α⋅L on the same equations for a load at β⋅L the angular rotation for experiment 2 is found. The superposed equation can be given in three parts, equations D17 through D19. 0 ≤ x < α⋅L :

φ ( )x ..sinh( )..α λ Ltanh( ).λ L

cosh( )..α λ L sinh( ).x λ.λ L

.sinh( )..β λ Ltanh( ).λ L

cosh( )..β λ L sinh( ).x λ.λ L

.xL

( )2 α β.Mx L

.G It Eq. D17

α⋅L≤ x < β⋅L :

φ ( )x ..sin ( )..βλLtan( ).λL

cos( )..β λL sin( ).xλ.λL

.sin ( )..αλLtan( ).λL

sin ( ).xλ .sin ( )..λLα cos( ).xλ

.λLα

.Mx L.GIt Eq. D18

β⋅L ≤ x < L:

Eq. D19

φ( )x ..sinh( )..α λ L

tanh( ).λ Lsinh( ).x λ .sinh( )..λ L α cosh( ).x λ

.λ L

.sinh( )..β λ Ltanh( ).λ L

sinh( ).x λ .sinh( )..λ L β cosh( ).x λ

.λ Lα β .x

L( )α β

.Mx L.GIt

By differentiating the expression for the angular rotation expressions for the Saint-Venant torsion, bi-moment and warping restraint torsion can be obtained. The relation with the angular rotation is given by equation D20.

3

3

wwx;

2

2

wx

ttx;

dxdEIMmoment restraint Warping

dxdEIB moment -Bi

dxdGIM Torsion Venant -Saint

φ−=

φ−=

φ=

Eq. D20

In figure D9 through D11 the distribution of Saint-Venant torsion, bi-moment and warping restraint torsion is given.

0 700 1400 2100 28001.1 106

0

1.1 106 Saint-Venant torsion

Span [mm]

Mx;

t [N

mm

]

figure D9: distribution of Saint-Venant torsion

0 700 1400 2100 2800

2 108

0Bi-moment

Span [mm]

Bx

[Nm

m^2

]

figure D10: distribution of bi-moment

0 700 1400 2100 28001 106

0

1 106 Warping restraint torsion

Span [mm]

Mx;

w

[Nm

m]

figure D11: distribution of warping-restraint torsion

axial stress The axial stresses are comprised of stress due to bending and due to the bi-moment. The sum of these two components may not exceed the yield stress, see equation D21

yBMx fxy

≤σ+σ=σ Eq. D21

with: σMy stands for stress due to bending moment My

σBx stands for stress due to the bi-moment Bx. Both My and Bx are largest at x = α⋅L and x = β⋅L, see figures D1 and D10.

Appendix Page 25

y-axis

z-axis

x-axis

+ =

σx = 105

σx = 122

σx = - 82

σx = -174

σx = - 256

σx = -69

σx = -30

σBx σMy

B*

C*

figure D12 : axial stress due to warping restraint torsion and due to bending. The rightmost section gives the superposed

axial stress. The depicted sections are shown in isometric projection, the proportion in x-, y- and z-direction are 1:1:1. The relation between drawing units and stress is one to ten.

In the cross section the combination of these stresses is largest at the corners B* and C*, see figure D12. In table A2 values for the z-ordinate and the warping function ‘ω’ are listed. The axial stress at B* due to bending is given by equation D22 and due to the bi-moment by equation D23.

2h

ILF

IzM

yy

yMy

−⋅α⋅==σ Eq. D22

( )

( )( ) ( ) ( ) ( )

( ) ( ) ( )

W

*B

B

w

xB

I

LsinhLcoshLtanhLsinhLsinhLcosh

LtanhLsinheF

IsB

x

x

ω⋅

αλ

βλ−

λβλ+αλ

αλ−

λαλ

λ⋅

=σ

=ω

=σ

Eq. D23

The expression for Bx in equation D23 is found by differentiating equation D17 twice, multiplying it by –EIw and substituting F⋅e for Mx. The expressions for the axial stress are now such that factoring for F is quite easy which will come in handy later. The stresses in figure D12 are obtained from equations D22 and D23 in which ‘z’ and ‘ω’ were altered. The depicted cross section is at x = α⋅L = 0.7 m. The force for which the stresses have been determined is 30 [kN].

9.1.1 Shear stress The shear stresses are comprised of stress due to bending, due to Saint Venant torsion and due to the restrained warping. The sum of these components may not exceed the yield shear stress, see equation D24.

3

f yMMVxs w;xt;xz

≤τ+τ+τ=τ Eq. D24

with: τvz stands for shear stress due to the shear force Vz

τMx;w for shear stress due the Saint Venant torsional moment Mx;t τMx;w stands for shear stress due to the warping restraint torsional moment Mx;w.

The shear force is constant from 0 < x < α⋅L and β⋅L < x < L, see figure D1. The Saint Venant-torsion is largest at the supports, see figure D9, and the warping restraint torsion is largest at x = α⋅L and x = β⋅L, see figures D11. The largest combination is found at x = α⋅L and x = β⋅L. In § Elastic section capacity the point in the cross section at which the largest combination of axial- and shear stresses occurs depends on the shear stress distribution. However, due to the presence of torsion, the point at which the largest combination of stresses will occur depends now on the axial stress since it is a factor larger than the shear stress is. The axial stress is largest at point B* and C*. In figure D13 the distribution of shear stress is given. ��

��

��

��

τmax

shear stress due to bending

Mx

τ

shear stress due toSaint Venant torsion

Mx;t

��

��

��

��

τmax

shear stress due to warpingrestraint torsion

��

��

Mx;w

figure D13: shear stress distribution at centre lines. The arrows give the direction of the shear stress at x= α⋅L. The shear

stress due to bending causes the torsional moment in the member. The shear stresses due to Saint Venant torsion and warping restraint torsion balance it.

The components of equation D24 are expanded in equation D25 through D27

( )( ) y

z

ry

f00z

y

yzV I

V3.2962

tI2thbV

stIsSV

z⋅=

⋅==τ Eq. D25

( ) =±=τ stI

M

t

t;xM t;x

Eq. D26

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )r

tM t

I

2LcoshLcoshLtanhLsinhLcoshLcosh

LtanhLsinheF

t;x

β−α−+αλ

βλ−

λβλ+αλ

αλ−

λαλ⋅⋅−

±=τ

The expression for Mx;t in equation D26 is found by differentiating equation D17, multiplying it by GIt, substituting F⋅e for Mx and entering α⋅L for ‘x.’

( )( ) ==τ ω

stIsSM

w

w;xM w;x

Eq. D27

( )( ) ( ) ( ) ( )

( ) ( ) ( )

rwM tI

SLcoshLcoshLtanhLsinhLcoshLcosh

LtanhLsinheF

w;x

ω⋅

αλ

βλ−

λβλ−αλ

αλ+

λαλ−⋅⋅−

=τ

The expression for Mx;w in equation D27 is found by differentiating equation D17 three times, multiplying it by –EIw, substituting F⋅e for Mx and entering α⋅L for ‘x.’

Appendix Page 27

0 141.75 283.55000

0

5000

ω( )s

b0 b0 h0

s

0 141.75 283.58 104

4 104

0

4 104

b0 b0 h0

s

0 141.75 283.58 105

4 105

0

4 105

b0 b0 h0

s [mm]

∫ω ds

( ) ( )∫ω=ω dsstsS

warping function along s

elastic section modulusfor warping restraint torsion

A B C D

[mm2]

[mm3]

[mm4]

figure D14: warping function along s, integrated warping function and elastic section modulus for warping restraint torsion From figure D13 it can be seen what sign the respective shear stresses must have and from equations D25 through D27 the respective stresses can be determined. The elastic section modulus for warping restraint torsion Sω is used in equation D27. This dimension is found by integrating the warping function of figure A3. Thus elastic section modulus for warping restraint torsion is the volume under the warping function. The warping function at the centre lines, which is called contour warping, represents the average warping over the thickness. One can integrate the contour warping as a line function instead of the area of the full warping function. In figure D14 the warping function is given along the centre line (along the ordinate s, see figure D4b), below that the integrated warping function is given. To incorporate the thickness of the section this function is then multiplied by the thickness of respectively the flanges and web. Because of the difference between the thickness of web and flanges a discontinuity occurs in the function of Sω.at s = b0. The actual function will be fluent because the difference in thickness is bridged by the root. Thus the average of Sω at the flange and the web is taken in equation D27, which is shown by the dashed line in figure D14.

Von Mises criteria The combination of axial stress and shear stress at point B is largest and should not exceed the Von Mises yield criteria, see equation D7. To write the Von Mises criterion with all the above shown equations will not fit on this pages, thus just the numbering of the equations that is referred to is given, see equation D28.

( ) ( ) ( ) ( ) y2222 fD27 D26 D25 .3D23 D21 stressshear 3stress axial ≤++⋅++=⋅+ Eq. D28

Equations D21, D23, D25, D26 and D27 are written such that the load F can be factored easily and an expression for the elastic section capacity for torsion and bending, Fel,torsion, can be obtained.

Ultimate load approximated To determine the ultimate load for a combination of bending and torsion is even more complicated then determining the elastic section capacity for torsion and bending. For members subjected to bending only, the ultimate load can be obtained by multiplying the elastic section capacity by the sectional form factor α. The form factor for a UPE 160 section is defined by equation D29.

α = = =WW

y;plastic

y;elastic

141004120670

117. Eq. D29

To approximate the ultimate load for sections subjected to bending and torsion it is suggested that the same procedure is used. This procedure is just to get an idea of the magnitude of the ultimate load. F pl;torsie = α ⋅ F el;torsie = 1.17 ⋅ F el;torsie Eq. D30

Appendix Page 29

E. Introducing load on section Controle van de krachtsinleiding bij de oplegging en puntlast NEN 6770: § 14.2 Krachtsinleiding zonder verstijvingen De krachtsinleiding wordt gecontroleerd ter plaatse van de puntlasten en ter plaatse van de reactiekrachten. De grootste voorspelde kracht is F max;s;d en is gelijk aan 41.7 [kN]. De reactie kracht is in grootte gelijk en richting tegengesteld aan deze belasting. In de toets moet de grootste kracht ten gevolgen van de belasting ingevuld worden en is aangeduid met Fs;d . Fu;d is de kleinste waarde van Fu;1;d , Fu;2;d of Fu;3;d Deze waarden worden bepaald met art. 14.2.1, 14.2.2 en 14.2.3 en zijn bepaald voor de puntlast en de reactiekracht. Toets: Eindoplegging: de kleinste waarde is Fu;2;d = 65.26 [kN]

164.026.657.41

FF

d;u

d;s ≤== voldoet

Puntlast: de kleinste waarde is Fu;1;d = 123.3 [kN]

169.07.415.1

2.293.1235.1

7.41M5.1

MF5.1

F

d;u;y

d;s;y

d;u

d;s ≤=⋅

+⋅

=⋅

+⋅

voldoet

met: My;u;d = My;pl = 41.7 [kNm] (zie Eq. D12) My;s;d is de rekenwaarde van het buigend moment om de y-as ten gevolge van de belasting.

My;s;d = F max;s;d · a · L = 41.7 · 0.25 · 2.8 = 29.2 [kNm]

NEN 6770: § 14.2.1 Vloeien van het lijf Fu;1;d = (c+d1)• tw• fy;d

( ) ( ) [ ]c t t r mmw f= + + ⋅ ⋅ − = + + ⋅ − =2 1 2 6 5 10 2 12 1 2 23512

12. .

Zie figuur E1 voor de bepaling van ‘c’. Eindoplegging:

d t bt

ff ff

f

w

y;f d

y;w d

f d

y;f d1

2

1= −

;

;

;

;

σ

Het buigend moment is nul bij de oplegging, dus σf;d = 0 ; het profiel is uit één materiaal, dus fy;f;d = fy;w;d en d1 wordt:

[ ]d t bt

mmff

w1 10 70

6 532 8= = =

..

Fu;1;d = (c+d1)• tw• fy;d = (23.5+32.8) • 6.5 • 295.7 = 108.2 [kN] Puntlast:

d t bt

ff ff

f

w

y;f d

y;w d

f d

y;f d1

2

2 1= ⋅ −

;

;

;

;

σ

De gekozen overspanning is 2800 millimeter. De langsspanning σf;d in de hoek van het lijf met de flens wordt gegeven door:

σα

xy

y

h

y

M zI

F LI

Nmm

= =⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅⋅

=

23

6 240 10 0 25 2800 80

9 65 102321.

..

d1

2

2 10 706 5

29572957

1 2321295 7

40 659= ⋅ −

=.

.

...

.

Fu;1;d = (c+d1)• tw• fy;d = (23.5 + 40.659) • 6.5• 235 = 123.3 [kN]

tw

tf

r

c

figuur E1: krachtsinleiding in de hoek lijf-flens

Appendix Page 31

NEN 6770: § 14.2.2 Lokaal plooien van het lijf Eindoplegging:

[ ]F t E f tt

tt

ch t

kNu d w y;df

w

w

f f; . .2;

20125 32

65 26= ⋅ ⋅ + ⋅

−

=

Puntlast:

[ ]F t E f tt

tt

ch t

kNu d w y;df

w

w

f f; .2;

20 5 32

261= ⋅ ⋅ + ⋅

−

=

En er moet gelden: ch tf−

= <2

017 0 2. . voldoet.

NEN 6770: § 14.2.2 Globaal plooien van het lijf Eindoplegging en puntlast: Het lijf wordt beschouwd als een gedrukte staaf met een breedte bef en een lengte lbuc = h, zie figuur E3.

Voor de effectieve breedte geldt: b h x hef = + ≤2

Invullen geeft bef = 80 + 100 =180 > 160 dus bef =h Aan de hand van NEN 6770 art. 12.1.1. kan de staaf getoetst worden:

[ ]I b t mmy;lijf ef w= = ⋅112

36 103 3 4.

[ ]A b t mmlijf ef w= = ⋅104 103 2.

[ ]iIA

mmy;lijfy;lijf

lijf= = 188.

[ ]λ ybuc

y;lijf

li

= = −85 27. [ ]λ πcy;d

Ef

= = −83721.

λλλ

ωrely

cbucknik curve c= = ⇒ ⇒ =1019 0 529. .

Fu;3;d = ωbuc• Alijf • fy;d = 0.529 • 1040 • 295.7 = 163 [kN]

F

h

x =100

½h

bef = h

figuur E2: krachts-inleiding in het lijf

F

lbuc= h

bef = h

figuur E3: globaal plooiien van het lijf

F. Test Rig

Plans

3 Left view

1 Top view

A A'

C

C'

2 Front view

B

B'

0 600300

scale900

figure F1: test rig (dimensions are in millimetres on given scale)

Appendix Page 33

1a Section B-B'

1 Top View

x - a

xis

y - axis

support point of load application

0 600300

scale900

figure F2: view from the top and section B-B’

2 Front view

2a Section C-C' jack

hinge

x - a

xis

z - axis

0 600300

scale900

figure F3: Frontal view and cross section at C-C’

Appendix Page 35

3a Section A-A'3 Left view

hinge

jack

hinge

anker

prestressing cable

anker

load cellz - axis

y - axis

0 600300scale

900

figure F4: view from the left and section A-A’

Photographs

figure F5: test rig in laboratory (photograph: Ben Elfrink)

Appendix Page 37

figure F6: bird eye view of test rig (photograph: Ben Elfrink)

Load bearing yokes

Load

at s

hear

cen

tre (1

B)

Load

at t

op fl

ange

(2A

)Lo

ad a

t lin

e of

sym

met

ry (2

B)

fron

tal v

iew

yok

e

AA

'

sect

ion

A-A

'

line

of w

ork

figure F7: the three load bearing yokes in frontal view and cross section

Appendix Page 39

figure F8: photographs of the three load bearing yokes

Details

Experiment 1B

figure F9: plans of fitting for applying load at the shear centre

load applied at shear centre

displacments measuredat shear centre

wire for measuringlateral displacement

wire for measuringvertical displacement

figure F10: load applied at the shear centre

Appendix Page 41

Experiment 2B

figure F11: plan of load pin for applying load at the line of symmetry

figure F12: plan of pin holder

figure F13: applying load at the line of symmetry

9.1.2 Experiment 2A

(a) (b) figure F14: applying load at the ‘top’ flange (in experiments under)

Appendix Page 43

Fittings to sections

Experiment 1B

figure F15: fittings to section for experiment 1B

Experiment 2A

figure F16: fittings to section for experiment 2A

Appendix Page 45

Experiment 2B

figure F17: fittings to section for experiment 2B

Experiment 2C

figure F18: fittings to section for experiment 2C

Appendix Page 47

G. Supports

centre point of rotation

centre of gravity of section teflon foil

figure G1: plan of support

joint

rod

figure G2: support (photograph: Ben Elfrink)

figure G3: parameters for determining friction in ball and socket joint Figure G3 is obtained from: Elges, Afmetingenkatalogus, K 234 NL Gewrichtslagers, Stangkoppen INA Naaldalger Mij. B.V., Barneveld, 1995

Appendix Page 49

H. Calibration

Used equipment table H1: used recorders and their specification measured at channel:

manufacturer type of recorder id. range

deviation of linearity [%]

0 WS10-250-R10K-L10-SD4

982844732 0-100 [mm]

1 WS10-250-R10K-L10-SD4

982844730 0-100 [mm]

2 WS10-250-R10K-L10-SD4

982844733 0-100 [mm]

3 -20 – 20 [º] 4 -20 – 20 [º] 5 -20 – 20 [º] 6 Solarton AXR/1/s M922497 A848-03 0-2 [mm] 0.4 7 Solarton AXR/5/s M921980 A826-05 0-10 [mm] 0.4 8 Solarton AXR/5/s M921980 A826-03 0-10 [mm] 0.4 9 Solarton AXR/1/s M922497 A848-01 0-2 [mm] 0.4 60 HBM W 20 K 4437 0-40 [mm] 0.2 61 HBM W 20 K 3099 0-40 [mm] 0.2 62 HBM W 20 K 3100 0-40 [mm] 0.2 63 HBM W 10 K 14354 0-20 [mm] 0.2 64 HBM W 10 K 13588 0-20 [mm] 0.2 70 HBM W 10 TK 14062 0-10 [mm] 0.2 71 HBM W 10 TK 14063 0-10 [mm] 0.2 72 HBM W 10 TK (unreadable) 0-10 [mm] 0.2 73 HBM W 10 TK 0-10 [mm] 0.2 90 TUE Load cell 0-80 [kN] 1 91 TUE Load cell 0-80 [kN] 1 In table H1 all used recorders are listed. The range in which they were used is given and the deviation of linearity is given. The precision of the load cells is much higher than the listed one, however, they have been calibrated in a testing bench on which the reading has a deviation of 1 % linearity. Thus, the precision of the load cell can not exceed that of the testing bench. On the next pages the calibration of all recorders is given. In case of displacement recorders, the displacement is imposed on the recorder and is therefore the independent variable (the x-value) in the calibration. The current is the output (the y-value). Using the method of least squares a linear relation between displacement and current is found. This displacement is displayed in the graphs, in which the number preceding ‘x’ is the calibration factor. The value of R2 indicates the precision of the linear line fitted through the observed data. The closer R2 is to the value of ‘1’ the better the fit. The value of R2 is displayed in the graphs as well. In case of the rotation recorders the independent variable is the imposed angle and for the load cell it is the applied load.

Calibration of recorders on channels 0, 1 and 2:

serie nr. serie nr. serie nr. channel 0 982844732 Channel 1 982844730 Channel 2 982844733 Displacement Current Displacement Current Displacement Current [mm] [mV] [mm] [mV] [mm] [mV]

0 -3615.3 0 -3594.2 0 3597.6 5.0035 -3248.6 5.0565 -3233.5 5.0135 3240.4

*) *) 10.0155 -2879.9 10.029 2880.7 15.0625 -2526.9 15.2235 -2509.3 15.0335 2516.7 20.013 -2170.9 20.0035 -2164.3 20.0125 2157.3

25.2645 -1791.3 25.1235 -1792.2 25.0115 1796 30.028 -1449.4 30.0115 -1437.2 30.014 1435.7

35.0225 -1088.1 35.1495 -1071.8 35.0075 1077.2 40.051 -729.1 40.0185 -726.8 40.0125 722.7 45.068 -370.4 45.01 -367.4 45.008 359.6

50.0095 -6.4 50.0195 -6.6 50.011 -3.7 55.0285 357.1 55.048 354.2 55.03 -364.9 60.018 716 60.0155 712 60.0325 -722.9

65.0245 1079.5 65.1385 1076.8 65.011 -1084.9 70.0245 1437.9 70.1565 1432.9 70.0185 -1447.9 75.0237 1799.4 75.043 1785.9 75.021 -1806.4

80.15 2165.9 80.055 2151.1 80.0145 -2171.3 85.144 2532.9 85.0425 2513.8 85.0125 -2531.3

*) *) 90.053 2871.9 90.0275 -2884.6 95.053 3236 95.068 3230.3 95.047 -3245.1

100.006 3594.3 100.046 3509.7 100.033 -3604 *) data omitted in calibration

Ca libra tion of Cha nne l 0

y = 72.112x - 3613.4R2 = 1

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

0 50 100

Dis p lace m e nt [m m ]

Cur

ren

t [m

V]

measuredLinear


y = 71.608x - 3591.7R2 = 0.9999

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

0 50 100


Cur

ren

t [m

V]

measuredLinear


y = -72.057x + 3600.3R2 = 1

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

0 50 100


Cur

ren

t [m

V]

measuredLinear

Appendix Page 51


Channel 3 Channel 4 Channel 5 Angle Current Current Current [º] [mV] [mV] [mV]

-20 7714.8 7207.8 6743.2 -15 7221.6 6627.8 6201.2 -10 6691.8 6034.8 5664.2 -5 6092.6 5393.8 5078.2 0 5425 4708 4439.6 5 4883.8 4161.6 3939.4

10 4303.8 3611 3424.8 15 3755.8 3098.4 2941.4 20 3183.8 2596 2470.2


y = -115.02x + 5474.8R2 = 0.9993

2000

3000

4000

5000

6000

7000

8000

-20 -10 0 10 20

Angle [º]

Cur

rent

[mV

]

measured

Linear


y = -117.05x + 4826.6R2 = 0.998

2000

3000

4000

5000

6000

7000

8000

-20 -10 0 10 20

Angle [º]

Cur

rent

[mV

]

measured

Linear


y = -108.3x + 4544.7R2 = 0.9984

2000

3000

4000

5000

6000

7000

-20 -10 0 10 20

Angle [º]

Cur

rent

[mV

]

measured

Linear

Calibration of recorders on channels 6, 7, 8 and 9:

Channel 7 Channel 8 Channel 6 Channel 9 Displacement Current Current Displacement Current Current [mm] [mV] [mV] [mm] [mV] [mV]

0 8799 8596 0 9667.1 9330 1.001 6790.5 6583 0.201 7656.3 7300.9

2 4794.2 4579.6 0.4 5658.2 5303.9 3 2784 2563.7 0.6 3664.5 3293.6 4 783.5 556.5 0.8 1666.7 1291.1 5 -1211 -1146.2 1 -325.7 -706

6.003 -3216.6 -3452.8 1.204 -2355.4 -2745.2 7.001 -5211.2 -5446.4 1.4 -4330.9 -4722.1 8.002 -7209.5 -7444.2 1.6 -6331 -6717.2 9.001 -9202.9 -9436.8 1.8 -8355.8 -8706

2 -10277.8 -10666


y = -1999.9x + 8791.1R2 = 1

-12000

-10000

-8000

-6000

-4000

-2000

0

2000

4000

6000

8000

10000

0 5 10

Dis p lace m e n t [m m ]

Cur

rent

[mV

]

measured

Linear


y = -2001.9x + 8605.2R2 = 0.9998

-12000

-10000

-8000

-6000

-4000

-2000

0

2000

4000

6000

8000

10000

0 5 10


Cur

rent

[mV

]

measured

Linear


y = -9988.8x + 9660.2R2 = 1

-15000

-10000

-5000

0

5000

10000

15000

0 0.5 1 1.5 2


Cur

rent

[mV

]

measured

Linear


y = -10006x + 9306.9R2 = 1

-15000

-10000

-5000

0

5000

10000

15000

0 0.5 1 1.5 2


Cur

rent

[mV

]

measured

Linear

Appendix Page 53


Channel 60 Channel 61 Channel 62 Displacement Current Current Current [mm] [mV] [mV] [mV]

0 51.12 -72.62 -82.66 2.5005 43.72 -64 -73.78 5.0035 36.16 -56.78 -64.66 7.5025 28.58 -48.88 -55.58 10.006 20.06 -40.6 -46.36 12.507 13.02 -32.76 -36.98

15.0055 5.02 -24.56 -27.42 20.011 -11.02 -8.08 -8.06

25.0105 -27.12 8.32 11.14 27.503 -34.98 16.42 20.6 30.004 -42.72 24.46 29.96 32.551 -50.44 32.54 39.3

35.0205 -57.92 40.46 48.42 37.5105 -65.48 48.56 57.68 40.0085 -72.02 56.2 66.38


y = -3.1202x + 51.5R2 = 0.9998

-80

-60

-40

-20

0

20

40

60

0 10 20 30 40

Dis place m e nt [m m ]

Cu

rre

nt [m

V]

measured

Linear


y = 3.2325x - 72.77R2 = 1

-80

-60

-40

-20

0

20

40

60

80

0 10 20 30 40


Cu

rre

nt [m

V]

measured

Linear


y = 3.7623x - 83.418R2 = 0.9999

-100

-80

-60

-40

-20

0

20

40

60

80

0 10 20 30 40


Cu

rre

nt [m

V]

measured

Linear

Calibration of recorders on channels 63 and 64: Displacement Channel 63 Displacement Channel 64 [mm] [mV] [mm] [mV]

0 -74.6 0 -102.78 2.5675 -69.88 2.5045 -101.94 5.2845 -59.48 5.0015 -96.4 7.514 -46.16 7.5025 -84.66

10.0025 -29.44 10.0015 -64.36 12.547 -12.12 12.5025 -49.56

15.0015 4.56 15.0045 -31.94 17.512 21.08 17.513 -14.34

20.0135 39 20.006 3.28 22.516 56.04 22.502 20.84 25.006 73.16 25.01 38.46

Ca libra tion of cha nne l 63

y = 6.7665x - 96.613R2 = 0.9998

-80

-60

-40

-20

0

20

40

60

80

5 10 15 20 25


Cur

ren

t [m

V]

measured

Linear


y = 6.8406x - 133.61R2 = 0.9989

-120

-100

-80

-60

-40

-20

0

20

40

60

5 10 15 20 25


Cur

ren

t [m

V]

measured

Linear

Appendix Page 55

Calibration of recorders on channels 70, 71, 72 and 73:

Channel 70 71 72 73 Displacement Current Current Current Current [mm] [mV] [mV] [mV] [mV]

0 0.76 9.2 -0.68 -5.12 0.5015 3.8 12.38 2.66 -2.5 1.008 6.94 15.66 6.22 1.32

1.5405 10.2 19.06 9.92 5.4 2.2015 14.24 23.28 14.5 10.48 2.9075 *) 27.8 19.4 15.06 3.601 22.82 32.3 24.24 21.14 4.318 27.2 36.96 29.1 26.54

5.0075 31.5 41.46 33.96 31.7 5.703 35.82 46 38.78 36.82 6.411 40.3 50.64 43.78 42 7.112 44.68 55.18 48.7 47.02

7.8035 49.06 59.68 53.6 51.96 8.5415 53.7 64.4 58.8 57.16 9.0065 56.62 67.22 62.1 60.44 9.5385 59.9 67.22 **) 65.9 64

10.0635 63.1 67.22 **) 69.66 67.56 *) data omitted in calibration **) out or range, data omitted


y = 6.2058x + 0.5964R2 = 1

0

10

20

30

40

50

60

70

0 5 10


Cu

rre

nt [

mV

]

meas ured

Linear


y = 6.4679x + 9.1011R2 = 1

0

10

20

30

40

50

60

70

80

0 5 10


Cu

rre

nt [

mV

]

meas ured

Linear


y = 6.9872x - 0.8897R2 = 1

0

10

20

30

40

50

60

70

80

0 5 10


Cu

rre

nt [

mV

]

meas ured

Linear


y = 7.3481x - 5.6211R2 = 0.9996

-10

0

10

20

30

40

50

60

70

80

0 5 10


Cu

rre

nt [

mV

]

meas ured

Linear

Calibration of load cell prior to experiments Load cell 90 Load cell 91 Load Current Linear fit Residues Load Current Linear fit Residues [kN] [mV] [mV] [mV] [kN] [mV] [mV] [mV]

0.03 -0.048 -0.05374 0.005743 0.001 0.102 0.112497 -0.0105 5 0.459 0.4527 0.0063 5 -0.407 -0.4009 -0.0061

10 0.965 0.9622 0.0028 10 -0.918 -0.9144 -0.0036 15 1.474 1.4717 0.0023 15 -1.426 -1.4279 0.0019 20 1.98 1.9812 -0.0012 20 -1.938 -1.9414 0.0034 25 2.491 2.4907 0.0003 25 -2.452 -2.4549 0.0029 30 2.993 3.0002 -0.0072 30 -2.964 -2.9684 0.0044

37.5 3.762 3.76445 -0.00245 35 -3.483 -3.4819 -0.0011 40 4.017 4.0192 -0.0022 40 -3.993 -3.9954 0.0024

47.5 4.775 4.78345 -0.00845 45 -4.509 -4.5089 -1E-04 50 5.03 5.0382 -0.0082 50 -5.02 -5.0224 0.0024 55 5.542 5.5477 -0.0057 55 -5.535 -5.5359 0.0009 60 6.055 6.0572 -0.0022 60 -6.053 -6.0494 -0.0036 65 6.564 6.5667 -0.0027 65 -6.571 -6.5629 -0.0081 70 7.08 7.0762 0.0038 70 -7.06 -7.0764 0.0164 75 7.591 7.5857 0.0053 75 -7.606 -7.5899 -0.0161 80 8.108 8.0952 0.0128 80 -8.121 -8.1034 -0.0176

Ca libta tion of cha nne l 90

y = 0.1019x - 0.0568R2 = 1

012345678

0 20 40 60 80

Load [k N]

Cur

rent

[V

]

measured

Linear

Ca libta tion of cha nne l 91

y = -0.1027x + 0.1126R2 = 1

-8

-7

-6

-5

-4

-3

-2

-1

00 20 40 60 80

Load [k N]

Cur

rent

[V

]

measured

Linear

Re sidue s of cha nne l 90

-0.01

-0.005

0

0.005

0.01

0.015

0 20 40 60 80 100

Load [k N]

Cur

rent

[V

]


-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

0 20 40 60 80 100

Load [k N]

Cur

rent

[V]

Appendix Page 57

Calibration of load cell after experiments Load cell 90 Load cell 91 Load Current Linear fit Residues Load Current Linear fit Residues [kN] [mV] [mV] [mV] [kN] [mV] [mV] [mV]

1 0.046 0.0489 -0.0029 0.061 0.032 0.01669 0.01531 2 0.157 0.1511 0.0059 2 -0.188 -0.1807 -0.0073 5 0.468 0.4577 0.0103 5 -0.496 -0.4861 -0.0099

10 0.973 0.9687 0.0043 10 -1.019 -0.9951 -0.0239 15 1.478 1.4797 -0.0017 15 -1.504 -1.5041 0.0001 20 1.975 1.9907 -0.0157 20 -2 -2.0131 0.0131 25 2.502 2.5017 0.0003 25 -2.5 -2.5221 0.0221 30 3 3.0127 -0.0127 30 -3.039 -3.0311 -0.0079 35 3.539 3.5237 0.0153 35 -3.523 -3.5401 0.0171 40 4.034 4.0347 -0.0007 40 -4.041 -4.0491 0.0081 45 4.53 4.5457 -0.0157 45 -4.569 -4.5581 -0.0109 50 5.059 5.0567 0.0023 50 -5.057 -5.0671 0.0101 55 5.57 5.5677 0.0023 55 -5.589 -5.5761 -0.0129 60 6.09 6.0787 0.0113 60 -6.083 -6.0851 0.0021


y = 0.1022x - 0.0533R2 = 1

0

1

2

3

4

5

6

7

0 20 40 60

Load [k N]

Cur

rent

[V

]

measured

Linear


y = -0.1018x + 0.0229R2 = 1

-7

-6

-5

-4

-3

-2

-1

00 20 40 60

Load [k N]

Cur

rent

[V]

measured

Linear


-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

0 20 40 60 80

Load [k N]

Cur

rent

[V]


-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0 20 40 60 80

Load [k N]

Cur

rent

[V]

I. Dimensions

Restrictions on dimensions In draft code NEN EN 10279 [5] restrictions on tolerance and shape deviation are given for hot rolled channel sections, see figure I5. In figure I1 it is shown on which dimensions restrictions are given and in table I1 the tolerance levels for an UPE 160 are given. In table I2 the results of measurements of the width and the thickness of the flanges are given. These results of width and thickness are well within the tolerance levels. The height of the section has not been measured systematically. However, from the observations made it can be said that it is within ± 0.5 millimetres, well within the prescribed tolerance level of ± 2 millimetres. table I1: restrictions on dimensions of an UPE 160 according to [5] property

dimension tolerance [mm]

height h = 160 + 2.0 -2.0 width b = 70 + 2.0 -2.0 thickness of web tw = 6.5 + 0.5 -0.5 thickness of flange tf = 10 Tolerance limited

by weight -0.5 heel radius r3 0.3 t out of square (k + k1) 2.0 straightness qxx 0.2% L = 5.6 qyy 0.3% L = 8.4 mass per unit length Kg/m18.6 +/- 4% standard Length L = 12000 +100 - 0 table I2: results of measurements on width and thickness of flanges width of flange thickness of flange bottom- top- bottom- top- b [mm] b [mm] t [mm] t [mm] mean 69.37 69.6 9.93 9.9 member 1 standard deviation 0.075 0.11 0.075 0.069 number of observ. 12 12 12 12 mean 69.34 69.48 9.92 9.89 member 2 standard deviation 0.137 0.114 0.071 0.085 number of observ. 16 16 16 16 mean 69.54 69.68 9.9 9.88 member 3 standard deviation 0.104 0.177 0.064 0.108 number of observ. 16 16 16 16 From visual inspection it can be concluded that the out of squareness of the sections was well within the set limits of table I1. The straightness of the cut sections has been measured. It was clear that the sections were completely straight as far as the straightness qxx was concerned. After a few random checks these measurements were abandoned. The straightness about the other main axis, qyy, has been measured from a flat table, as shown in figure I2. At four positions along the section the distance from the table to the top surface of the section, b* has been measured. The values of b* at the supports, position ‘0’ and ‘3’, are entered in the formula depicted in figure I2. By subtracting the value of the width of the flanges, ‘b’, from the value of the expression for y(x) the straightness can be determined, see table I3 and I4.

h

b/2

b

r3

t

s

figure I1: dimensions on which tolerance restrictions are given by [5]

Appendix Page 59

��

L

L/2L/4 L/4

position: 0 1 2 3

qyy

flat table

b0* b3

*

x

( ) *0

*0

*3 bx

Lbbxy +

−=

b0

figure I2: measuring straightness qyy In table I3 the results for sections of experiment 2 are given and they are plotted in figure I3. From the table and the graphs it can be determined that the sections are almost completely straight.

Straigh tne s s q -yyExpe r im e nt 2A1

-1

-0.5

0

0.5

1

0 700 1400 2100 2800

Le ngth [m m ]

offs

et [m

m]

top flange

bottom flange

Straigh tne s s q -yyExpe r im e nt 2B2

-1

-0.5

0

0.5

1

0 700 1400 2100 2800

Le ngth [m m ]

offs

et [m

m]

top flange

bottom flange

Straigh tne s s q -yyExpe r im e nt 2C1

-1

-0.5

0

0.5

1

0 700 1400 2100 2800

Le ngth [m m ]

offs

et [m

m]

top flange

bottom flange

figure I3: straightness qyy for specimen 2A1, 2B2 and 2C1 table I3:determining of straightness of specimen 2A1, 2B2 and 2C1

top flange bottom flange experiment position b [mm] b* [mm] b*-b x = y(x) = qyy b [mm] b* [mm] b*-b x = y(x) = qyy

2A1 0 69.6 71 1.4 0 1.4 0 69.8 70.55 0.75 0 0.75 0 (from 1 69.5 70 0.5 700 1.238 0.738 69.5 70.85 1.35 700 0.6875 -0.7

member 3 ) 2 69.7 70.1 0.4 2100 0.913 0.513 69.7 70.8 1.1 2100 0.5625 -0.5 3 69.5 70.25 0.75 2800 0.75 0 69.6 70.1 0.5 2800 0.5 0

2B2 0 69.4 69.4 0 0 0 0 69.5 69.6 0.1 0 0.1 0 (from 1 69.4 69.7 0.3 700 0.087 -0.212 69.6 70.3 0.7 700 0.0875 -0.6

member 1) 2 69.4 69.8 0.4 2100 0.262 -0.137 69.6 70.25 0.65 2100 0.0625 -0.6 3 69.4 69.75 0.35 2800 0.35 0 69.7 69.75 0.05 2800 0.05 0

2C1 0 69.7 69.7 0 0 0 0 70 70 0 0 0 0 (from 1 69.45 69.6 0.15 700 0.225 0.075 69.8 70.2 0.4 700 0.075 -0.3

member 3) 2 69.4 69.4 0 2100 0.675 0.675 69.8 69.8 0 2100 0.225 0.22 3 69.5 70.4 0.9 2800 0.9 0 69.4 69.7 0.3 2800 0.3 0

In table I4 the results for sections of experiment 1 are given and they are plotted in figure I4. All four sections are clearly bent one way, due to the heat brought into the section while welding on the perpendicular plates. The curvature could also be seen quite well. The plates were welded on at the points of load application, which corresponds with position 1 and 2 of figure I2. Most of the curvature was concentrated around the weld. One could argue that the sections is bent about these points and that in between the section is straight.

Straigh tne s s q-yyExpe r im e nt 1B1

0

1

2

3

4

0 700 1400 2100 2800Le ngth [m m ]

offs

et [m

m]

top flange

bottom flangeStraigh tne s s q-yy

Expe r im e nt 1B2

0

1

2

3

4

0 700 1400 2100 2800Le ngth [m m ]

offs

et [m

m]

top flange

bottom flange


0

1

2

3

4

0 700 1400 2100 2800Le ngth [m m ]

offs

et [m

m]

top flange

bottom flange


0

1

2

3

4

0 700 1400 2100 2800Le ngth [m m ]

offs

et [m

m]

top flange

bottom flange

figure I4: straightness qyy for specimen 1B1, 1B2, 1B3 and 1B4 table I4:determining of straightness of specimen

top flange Bottom flange experiment position b [mm] b* [mm] b*-b x = y(x) = qyy b [mm] b* [mm] b*-b x = y(x) = qyy

1B1 0 69.4 73.6 4.2 0 4.2 0 69.6 74 4.4 0 4.4 0 1 69.4 70 0.6 700 4.05 3.45 69.7 70.7 1 700 3.925 2.93 2 69 69.6 0.6 2100 3.75 3.15 69.4 69.15 -0.25 2100 2.975 3.23 3 69.4 73 3.6 2800 3.6 0 69.5 72 2.5 2800 2.5 0

1B2 0 69.4 73.8 4.4 0 4.4 0 69.5 73.5 4 0 4 0 1 69.35 69.45 0.1 700 3.95 3.85 69.5 69.8 0.3 700 3.375 3.08 2 69.4 69.5 0.1 2100 3.05 2.95 69.5 69.5 0 2100 2.125 2.13 3 69.4 72 2.6 2800 2.6 0 69.5 71 1.5 2800 1.5 0

1B3 0 69.25 72.9 3.65 0 3.65 0 69.3 73.25 3.95 0 3.95 0 1 69.2 69.25 0.05 700 3.563 3.513 69.25 69.4 0.15 700 3.6125 3.46 2 69.3 69.2 -0.1 2100 3.388 3.488 69.35 69.5 0.15 2100 2.9375 2.79 3 69.6 72.9 3.3 2800 3.3 0 69.4 72 2.6 2800 2.6 0

1B4 0 69.3 71.95 2.65 0 2.65 0 69.5 71.3 1.8 0 1.8 0 1 69.2 69.65 0.45 700 2.563 2.113 69.5 69.65 0.15 700 2 1.85 2 69.5 69.6 0.1 2100 2.388 2.288 69.5 70 0.5 2100 2.4 1.9 3 69.3 71.6 2.3 2800 2.3 0 69.6 72.2 2.6 2800 2.6 0

Appendix Page 61

table I5: measurements of dimensions of member 1 member 1 width of flange thickness of flange

bottom- top- bottom- top- experiment position b [mm] b [mm] t [mm] t [mm]

2B1 0 69.3 69.5 9.95 9.9 1 69.5 69.6 10 9.9 2 69.4 69.4 9.8 9.8 3 69.4 69.5 9.9 10

2B2 0 69.4 69.5 10 9.9 1 69.4 69.6 10 9.9 2 69.4 69.6 10 10 3 69.4 69.7 10 9.8

2B3 0 69.2 69.65 9.8 9.9 1 69.4 69.7 9.9 9.85

2 69.3 69.8 9.9 9.9 3 69.35 69.6 9.95 10

number of observations 12 12 12 12 mean 69.37 69.60 9.93 9.90 standard deviation 0.075 0.110 0.075 0.069

table I6: measurements of dimensions fo member 2 and 3 member 2 width of flange thickness of flange member 3 width of flange thickness of flange

bottom- top- bottom- top- Bottom- top- bottom- top- experiment position b [mm] b [mm] t [mm] t [mm] experiment position b [mm] b [mm] t [mm] t [mm]

1B1 0 69.4 69.6 9.9 9.9 2A1 0 69.6 69.8 10 9.9 1 69.4 69.7 10 10 1 69.5 69.5 9.9 10 2 69 69.4 10 9.8 2 69.7 69.7 9.9 9.9 3 69.4 69.5 9.8 9.9 3 69.5 69.6 10 10.1

1B2 0 69.4 69.5 9.8 9.8 2A2 0 69.6 69.6 9.9 9.9 1 69.35 69.5 9.9 9.8 1 69.5 69.9 9.8 9.8 2 69.4 69.5 9.9 9.8 2 69.4 69.6 9.9 10 3 69.4 69.5 10 9.9 3 69.7 69.75 9.8 9.85

1B3 0 69.25 69.3 10 10 2C1 0 69.7 70 9.9 9.7 1 69.2 69.25 10 10 1 69.45 69.8 9.9 9.75 2 69.3 69.35 9.9 9.9 2 69.4 69.8 9.9 9.8 3 69.6 69.4 10 9.8 3 69.5 69.4 9.9 9.75

1B4 0 69.3 69.5 9.9 9.9 2C2 0 69.55 69.6 10 9.9 1 69.2 69.5 9.85 10 1 69.6 69.9 9.95 9.9 2 69.5 69.5 9.9 10 2 69.4 69.6 9.8 10 3 69.3 69.6 9.9 9.8 3 69.5 69.4 9.9 9.9

number of observations 16 16 16 16 number of observations 16 16 16 16 mean 69.34 69.48 9.92 9.89 mean 69.54 69.68 9.90 9.88 standard deviation 0.137 0.114 0.071 0.085 standard deviation 0.104 0.177 0.064 0.108

figure I5: tolerances for parallel flange channels, obtained from [5]

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