Stability design for frame type structures

Stability design for frame type structures Delft University of Technology Faculty of Civil Engineering and Geoscience Section of Structural echanics aster thesis report Appendices Ing. R. P. Veerman I

Stability design for frame type structures Delft University of Technology Faculty of Civil Engineering and Geoscience Section of Structural echanics Appendices aster thesis report by Ing. R. P. Veerman Examination committee: Prof. ir. A. C. W.. Vrouwenvelder Dr. Ir. P. C. J. Hoogenboom Prof. Ir. F. S. K. Bijlaard Ir.. J.. Houben II

Preface This document is the appendices report of the master thesis: Stability design for frame type structures. This report is carried out at the section structural mechanics of Delft University of Technology. This report contains the background, the analyses and the calculations. This report is the appendix of the main report. In the main report contains general information and conclusions of this thesis. Delft, ay 009 Ing. R. P. Veerman III

Index Preface... III General information appendices... VI Appendix A Euler buckling load and deflection... VII A. Euler buckling load... VII A. Geometrical non-linear deflection... VII Appendix B Relative Slenderness... X Appendix C Differential equation buckling... XIII Appendix D Calculation example (hand-made... XIX D. Residual stress... XIX D. Section properties... XX D. Calculation for a length of 0 meter... XXII D.4 Calculation for a length of 5 meter... XXIV D.5 Calculation for a length of 0 meter... XXVI D.6 Calculation for a length of 5 meter... XXVIII D.7 Calculation for a length of 0 meter... XXXI D.8 Calculation for a length of 5 meter... XXXIV D.9 Conclusions... XXXVIII Appendix E Calculation example (atab... XXXIX E. Computer calculations in general... XXXIX E. Calculation file... X Appendix F Calculation example (atrix Frame... XVI F. Calculations in general... XVI F. Different calculations... XVI F. aximum load calculation... XVII F.4 Residual stress in atrix Frame... XVIII F.5 atrix Frame file... XVIII Appendix G Calculation according to the Dutch code... III Appendix H inear analysis portal frame... V H. In general... V H. Analysis... VI Appendix I Residual stress in the linear analysis... X I. In general... X I. Analysis if one part yields... X I. Analysis if two parts yield... XVI Appendix J Non-inear analysis portal frame... XIX J. In general... XIX J. Deflection of column the column... XXII J. Calculation of the deflections... XXIV Appendix K Residual stress in a non-inear analysis... XXVIII K. Analysis if one part yields... XXIX K. Analysis if two parts yield... XCVIII Appendix Calculation example of an unbraced portal frame... CXXII. Calculation example.... CXXII. Computer calculation file... CXXVII IV

. Calculation according to atrix Frame... CXXXII Appendix inear analysis braced portal frame...cxxxviii Appendix N Non-linear analysis braced portal frame... CXI Appendix O Residual stress in non-linear analysis... CXIX O. Analysis if one part yields... CXIX O. Analysis if two parts yield... CXII O. Conclusion... CXXVII Appendix P Calculations example of a braced portal frame... CXXIX P. Calculation example... CXXIX P. Computer calculation file... CXXXVI P. Calculation according to the Dutch code... CXCV P.4 Calculation according to atrix Frame... CXCVI Appendix Q oads of an extended Frame... CCII Q. Different loads... CCII Q. Safety rules... CCIII Q. oading... CCIII Appendix R inear analysis extended frame... CCIV Appendix S Non-linear analyses extended frame... CCXX Appendix T Residual stress in the non-linear analysis... CCXXIX T. Analysis if one part yields... CCXXIX T. Analysis if two parts yield... CCXIV T. Conclusion... CCVI Appendix U Calculations... CCIX U. anual calculations.... CCIX U. Calculation file.... CCXIX U. Calculation according to the Dutch code... CCXXX U.4 atrix Frame calculations.... CCXXXI V

General information appendices The appendices in this report are the background at the stability design for frame type structures report. The report is divided in four parts. The following appendices are related to the different parts: Appendices A till F are about the stability of a single column and is related to Chapter two. Appendices G till are about to the bearing capacity of an unbraced portal frame and are related to Chapter three. Appendices till P are about the bearing capacity of a braced portal frame and are related to Chapter four. Appendices Q till U are about the bearing capacity of a braced extended frame and are related to Chapter five. VI

Appendix A Euler buckling load and deflection In this appendix the Euler buckling load and the geometrical non-linear deflection formula have been derived. A. Euler buckling load Consider a column pinned at both ends and loaded by a compressive normal force N. At a certain load (the Euler buckling load there are two equilibrium situations possible. In the first equilibrium situation the column remains straight. In the second equilibrium situation the column deflects in a half sine shape. The amount of deflection is unknown. The Euler buckling load starts with the equilibrium between the internal and the external moment. d y internal external EI F E y dx F E is the Euler buckling load. This is an unknown value, but this will be derived. The expression of y and the second derivative of y are as follow: y y sin π x 0 d y π y 0 sin π x dx These expressions can filled in the formula what results in the following formula: π π x π x y0 sin EI FE y0 sin The sine function can be neglected. Also the signs on the left side can be neglected. π EI F E π EI FE A. Geometrical non-linear deflection In Appendix A. the Euler buckling load has been derived. Euler has used a differential equation to find a theoretical buckling load. Suppose the column is loaded horizontally and deflects in a half sine shape. If the column is loaded by a normal force, the deflection increases. The same equilibrium as Euler has used, can be used to find the total deflection. The total deflection is a summation of the original deflection and an additional deflection. The original Figure A.: Structure VII

deflection (e is the deflection due to the horizontally load. An initial deflection (imperfection can also be seen as an original deflection. The deflection increases if the column is loaded by a normal force. The additional deflection is called y (Fig. A.. d y i u Nc; s; d ( y e EI dx The expressions of y and e are the following expressions: y y sin π x 0 d y π y 0 sin π x dx π * x e e0 sin Together with the original formula this results in: π x π x π π x Nc; s; d y0 sin e0 sin y0 sin EI If the sine functions are neglected. π Nc; s; d ( y0 e0 y0 EI ( N y e y F with c; s; d 0 0 0 Writing out the formula N y N e y F c; s; d 0 c; s; d 0 0 E E F E π EI On both sides an expression e0f E will be summed up to get the total deflection as a function of e 0, F E and N c,s,d. N y N e F e y F e F c; s; d 0 c; s; d 0 E 0 0 E 0 E Separate the total deflection ( e 0 y 0 from the rest of the equation. ( ( e F y e F N 0 E 0 0 E c; s; d This results in an expression for e0 y0. FE ( e0 y0 e0 F N ( E c; s; d The denominator and the numerator can be divided by the yield load. VIII

( e y e 0 0 0 F E ( FE Nc; s; d N N c; s; d c; s; d ake one numerator and one denominator FE Nc; s; d ( y0 e0 e0 Introduce: F N E c; s; d Nc; s; d N c; s; d Use the introduced n-formula and the final result is: n y0 e0 e0 n n F N E c; s; d The total deflection is the original deflection multiplied by a factor. IX

Appendix B Relative Slenderness In Appendix A the Euler buckling load and the geometrical non-linear deflection formula have been derived. In this appendix a buckling reduction factor will be analyzed. To find the ultimate load, the yield force must be multiplied by this reduction factor. This reduction factor depends on the length and the section properties of the structure. For relative small lengths, the influence of the buckling factor is very small. The buckling reduction factor is equal to one. Due to straight hardening it is allowed to make this assumption. The analysis of the reduction factor is related to the Euler buckling stress. The Euler buckling stress is the Euler buckling load divided by the cross-section. If the reduction factor has been found, the buckling load is the multiplication of the reduction factor, the yield stress and the cross-section. FE σ E A Using the Euler buckling load this formula is: π EI buc σ E A This is the same as: ( π E σ E buc I A Using the gyration radius π E σ E buc i i I A the formula results in: Figure B.: Stress / slenderness diagram λis the slenderness of a section π E σ E λ λ i. λis used in the formula. This results in: Consider a section with a Euler buckling stress that is equal to the yield stress. The slenderness of this section is called λ e (Fig. B.. The following formula can be found. f y π E λ e X

Separate λ e from the rest of the formula. π E λ e this is equal to λe π f y E f y There are many steel grades. Every steel grade has its own yield stress. It is not practical to have a different slenderness graphic for every steel grade. This problem can be solved by using dimensionless values (Fig. B.. This results in one slenderness graphic. σ E On the y-axis: f y λ On the x-axis: λe The relative slenderness is formulated as the quotient of the slenderness and the Euler slenderness. λ λrel λ e Figure B.: Stress / slenderness diagram The known formulas can be used. λ i rel E π f y Using the formula of the gyration radius. λrel E I π f A This is the same as. λrel EI π f A y y To calculate the ultimate stress, a reduction factor must be applied. The reduction factor is called ω buc. ωbucis the Euler stress divided by the yield stress. σ E ω buc f y The value of the Euler stress is known and can be used. XI

ω buc π E λ f y The formula of the slenderness can be used. This results in the following formula. π EI ω buc f A This is equal to ω buc y EI π f y A Using the formula of relative slenderness. ω λrel Figure B.: Real failure load Only the Euler slenderness graphic has been derived. The starting position of Euler was a perfectly made section. In practice, the section is not perfectly made. The real ultimate stress is lower than the Euler buckling stress (Fig. B.. The Euler buckling stress is an upper limit of the ultimate stress. Another limit of the ultimate stress in the yield stress. To calculate a more realistic ultimate load, the reduction factor must decrease. The buckling stress depends on the dimensions of the section, the shape of the section and the way of construction. To make code calculations, all sections are divided in four groups. Inside a group, the influence of the residual stress is (more or less equal. Based on the buckling analysis of Euler, the practical experience of different types of sections and different steel grades there are made four slenderness graphics (Fig. B.4. The ultimate load is the multiplication of the reduction factor, the yield stress and the cross section. Figure B.4: Slenderness graphics XII

Appendix C Differential equation buckling n If all stresses in a section are smaller than the yield stress, the (derived in App. B n formula can be used to calculate the geometrical non-linear deflection. If the stress n increases and a part of the section starts to yield, the deflection increases and the n formula is not valid anymore. The load on the structure what result in partial yielding is called F. The total deflection of the mid-span at F is: e e FE π EI 0 F F F E E buc With e is the total deflection at force F e 0 is the initial deflection (at F 0 N F E is the Euler buckling load Due to the chosen sine shape of the deflection, the most critical cross-section is the midspan. The stresses in the midspan have been applied over the whole length of the column. In reality the ends of the column start to yield at a larger load. This assumption is a necessary approximation for performing the mathematical evolutions. After the first yield point, not the whole cross-section can be resists more loads. A part of the right is yielded. The stress in this part will not increase. The rest of the section is the effective (reduced cross-section. The effective cross-section is asymmetric. The centre of gravity is shifted. Due to extending the stress at midsection over the whole column, the centre of gravity in the whole column has shifted. The load is located in the original centre of gravity. Due to partial yielding, the original centre of gravity is not the same as the effective centre of gravity. The load is not located in the centre of gravity anymore of the effective section. For calculations it is much easier if the force is located in the centre of gravity. To change the location of the load, an eccentric moment has been introduced. The largest eccentric moment is in the middle of the column. In practice there is no eccentric moment at the end Figure C.: Eccentric moment XIII

of the column. In this analysis, the eccentric moment is located at both ends of the column (Fig. C.. An eccentric moment at the end results in an extra deflection of the column. An extra deflection results in an extra bending moment in the midsection. Due to this extra bending moment, the deflection increases again. It is important to take the eccentric moment into account at the start of the analysis. The deflection due to this eccentric moment is called e m,i. Figure C. shows the total deflection for a column. In this figure is: e total,i- the total deflection at load F i- e m,i deflection due to the eccentric moment e i the extra deflection due to F F i load extra load on the column (F total F i- z i the shift between the effective point of gravity and the original point of gravity The analyzed load case has subscript i'. The total deflection depends on the loads and on the deflection at the previous load case. The previous load case has subscript i-. The original deflection in the first load case (i is e 0. This is the initial deflection. The analysis is based on equilibrium between the internal and the external bending moments. The external moments are the load multiplied by the deflection. The internal moment is the curvature multiplies by the stiffness. The curvature is the second derivative of the deflection. The calculation is a second order differential equation. Figure C.: oads on the column extern int ern ------------------------------------------------------------------------------------------------------------ F ( e e F ( e e e F z κ EI i m, i i i total, i m, i i i i i ------------------------------------------------------------------------------------------------------------ d y Fi ( em, i ei Fi ( etotal, i em, i ei Fi zi EI i dx The expression of e i has been separate from the rest of the formula d y Fi em, i Fi y Fi ( e0 ei em, i Fi y Fi zi EI i dx The second derivative of e i and the functions of e must be calculated. d ei π π x Fi zi e sin( i dx EIi XIV

dei π π x Fi zix ei cos( C dx EI π x F z x e e C x C e i i i i sin( * EIi i sin( π x e total, i etotal, i π x em, i em, i sin( i These functions are used in the original formula. π x π x Fi zix π * x Fi em, i sin( Fi ei sin( Cx C Fi ( etotal, i em, isin( EIi F e C x C F z e EI π x Fi zix π π x Fi zi i i sin( i i i sin( i * EI i EI i C and C are integral constants. These constants can be calculated with the following boundary conditions: x 0 y 0. x y 0. The first boundary condition results in: Fi ( C Fi ( C 0 C 0 The second boundary condition results in: Fi zi Fi zi Fi C Fi C EIi EIi 0 C can be separated from the rest of the formula. Fi zi C ( Fi Fi ( Fi Fi EIi Finally this results in: Fi zi C EI i The values C and C are filled in the formula. π x π x Fi zix Fi zi π x Fi em, sin( Fi ei sin( x Fi ( etotal, i em, isin( EIi EI i F e x F z e EI π x Fi zix Fi zi π x Fi zi i i sin( i i i sin( i EIi EI i π EI i XV

This results in the following formula: π x π x Fi zix( x π x Fi em, i sin( Fi ei sin( Fi ( etotal, i em, isin( EIi π x F z x( x π x F e sin( F z e F sin( Fz i i i i i i i E, i i EIi Finally the total deflection must be calculated. The total deflection is e total, i e m, i e. The i factor etotal, i em, i ei must be separate from the rest of the formula. To start the expression e i is separate. π x π x Fi zix( x π x Fi em, i sin( Fi ei sin( Fi Fi ( etotal, i em, isin( EI π x F z x( x π x Fe sin( F F z e F sin( F z i i i i i i i i E, i i i EIi i The expressions of e i are placed on one side of the equation. All other expressions are placed on the other side of the equation. π x Fi zix( x π x Fi zix( x Fi em, i sin( Fi Fi ( etotal, i em, isin( Fi EI EI π x π x π x ei FE, i sin( Fi ei sin( Fe i i sin( i The expressions of e i are combined together. π x Fi zix( x π x Fi zix( x Fi em, i sin( Fi Fi ( etotal, i em, isin( Fi EI EI i π x π x π x ei FE, i sin( Fi sin( Fi sin( e depends on e total,i- and on e m,i. In other words: the total expression e total, i e m, i ei depends on e total,i and on e m,i. To realise this it is necessary to sum up π x π x π x ( etotal, i em, i FE, i sin( Fi sin( Fi sin( on both sides of the equation. π x Fi zix( x π x Fi zix( x Fi em, i sin( Fi Fi ( etotal, i em, isin( Fi EI EI i π x π x π x ( etotal, i em, i FE, i sin( Fi sin( Fi sin( π x π x π x ( etotal, i em, i ei FE, i sin( Fi sin( Fi sin( All expressions on the left side on the equation are written out. i i i XVI

π x F z x( x π x F z x( x F e sin( F F ( e e sin( F EI EI i i i i i m, i i i total, i m, i i i π x π x π x etotal, i FE, i sin( etotal, ifi *sin( etotal, ifi sin( π x π x π x em, ife, i sin( em, ifi sin( em, ifi sin( π x π x π x ( etotal, i em, i y0 FE, i sin( Fi sin( Fi sin( i The same expressions with different signs can be neglected. Fi zix( x Fi zix( x π x π x π x Fi Fi etotal, i FE, i sin( etotal, ifi sin( em, ife, i sin( EI EI i i π x π x π x ( etotal, i em, i ei FE, i sin( Fi sin( Fi sin( The equal parts can be combined together. Fi zix( x π x π x ( Fi Fi ( etotal, i em, i FE, i sin( etotal, i Fi sin( EI i π x π x π x ( etotal, i em, i ei FE, i sin( Fi sin( Fi sin( e e e y can be expressed in a function of e total,i- and on e m,i. 0 i m, i 0 ( etotal, i em, i ei F z x( x π x π x Fi Fi etotal, i em, i FE, i sin( etotal, i Fi sin( EIi π x π x π * x FE, i sin( Fi sin( Fi sin( i i ( ( e m,i is the deflection in the column due to the eccentric moment ( Fi z i on both sides on the column. The deflection in the midsection:. This result in the following 6EI 6EI expression: Fi zi em, i 8EI i This expression can use in the formula: Fi zix( x Fi zi π x π x ( Fi Fi etotal, i FE, i sin( etotal, i Fi sin( EIi 8EIi ( etotal, i em, i e i π x π x π x FE, i sin( Fi sin( Fi sin( The most critical cross-section (and the largest deflection is located in the midsection (x ½. The calculations are based on this location. This value of x can be applied in the formula. XVII

( etotal, i em, i ei F F F z e F z F e F F F F i i i i i i total, i E, i total, i i 8EIi 8EIi ( E, i i i By a simple check, big mistakes can be afforded. If there is no additional force there is no additional deflection. The total deflection should be equal to the total deflection in the previous load case. If F 0 than e m,i 0 and e total, i e m, i e i etotal, i e F e F total, i E, i total, i i ( etotal, i em, i ei F F E, i i ------------------------------------------------------------------------------------------------------------ etotal, i ( FE, i Fi ( etotal, i em, i ei F F E, i i ------------------------------------------------------------------------------------------------------------ e e y e Correct ( total, i m, i 0 total, i A second check is a check on the signs. A positive extra load must result in a larger deflection. Fi zi ( ( FE, i Fi Fi e total, i FE, i Fi 8EIi ( etotal, i em, i ei F F F F F The Euler buckling load ( E, i E, i i E, i i i F must be larger than the maximum load ( F F i. The Euler buckling load is an upper limit, so this is correct. A positive additional load results in a positive additional deflection. The second check is correct. As a third, the dimensions can be checked. If the dimensions are not correct, the formula could not be correct. Nmm Nmm ( N N m N mn 4 4 Nm m Nm m ( m m m N N N ------------------------------------------------------------------------------------------------------------ Nm Nm N m N Nm Nm Nm m Correct N The formula is found correctly on all checks. There are no large mistakes in the formula. i XVIII

Appendix D Calculation example (hand-made In Appendix C the differential equation for the total deflection has been made. In this Appendix the formulas ( found in App. C will be used to make some calculations. The calculations in this Appendix are made manually. ore calculations will be made in Appendix E by the computer program atab. The computer program atab has been used to make many calculations. In Appendix F the same problem is solved by a computer program (atrix Frame based on a finite element method (FE. All calculations in this Appendix are based on a HE 450A section. The difference between the calculations is the length of the column. The different lengths are 0, 5, 0, 5, 0 and 5 meter. If the ultimate loads are calculated, a reduction factor curve can be made. Due to the small amount of calculations, the reduction factor curve is not very clear.. In Appendix E more lengths are calculated (by computer calculation. This results in a flowing curve. D. Residual stress The residual stress distribution has been simplified to a rectangular residual stress distribution (Fig. D.. The moment of inertia depends mainly on the flanges. The web has just a small contribution to the moment of inertia. If (due to partial yielding the effective surface of the flanges decreases, the effective moment of inertia decreases. If the web partial yields, the moment of inertia deceases just a little bit. This reduction is neglected. For the reduction of the crosssection the effective of both the web and the flanges must taken into account. The amount of residual stress (in this report called S is a function of the yield stress. According to the NEN 677 the amount of residual stress in a HE 450A section is 0% of the yield stress. In other words: S 0.*f y 0.*55 S 06.5 N/mm Figure D.: Residual stress distribution The stress in the section cannot be larger than the yield stress. There is residual tension stress and residual compression stress. If the structure is loaded by a compression loads, the part of the section with residual compression stress yields first. This happens if σ f S. The part of the section with residual tension stress yields if σ f S. The critical stresses are: σ f y S σ f y 0. f y σ 0.7 f y σ 48.5N/mm σ f y S σ f y 0. f y σ. f y σ 46.5N/mm y y XIX

D. Section properties For the calculation of the ultimate load, it is important to know what the section properties are. It is also important to know the reduced section properties. These section properties will be calculated. I 6.7*0 8 mm 4 A 7800 mm Z 8 I ½h 6.7*0 0 Z.896*0 6 mm I I (¼bt f (½h I 6.7*0 8 *75**0 I 4.848*0 8 mm 4 A A *(¼bt f A 7800 *75* A 4650 mm Z 8 I ½h 4.848*0 0 Z.04*0 6 mm I I (¼bt f (½h I 4.848*0 8 *75**0 I.4*0 8 mm 4 A ½A A ½*7800 A 8900 mm Z 8 I ½h.4*0 0 Z.5*0 6 mm Figure D.: Profile without reduction Figure D.: Profile with reduction Figure D.4: Profile with reduction In Appendix C the following formula has been derived: ( e e e e total, i total, i m, i i F F F z e F z F e F F F F i i i i i i total, i E, i 8EIi 8EIi total, i i ( For the calculation it is important to know that: ei etotal, i etotal, i F e Fe i i i total, i i Fi σright, i σright, i Zi Ai i Fi σ left, i σ left, i Z A i i E, i i i XX

The symbols in these formulas are: e the total deflection at load case i (mm total, i e the total deflection at load F i- (mm total, i e m, i the deflection due to the eccentric moment in load case i (mm the extra deflection due to the difference in normal force (mm e i F the load in the previous load case (N F z i i i the extra load in case i (N the difference between the original point of gravity and the effective point of gravity in case i (mm I the moment of inertia in case i (mm 4 i F the Euler buckling load in case i (N E, i The starting points of this analysis where: There is an initial deflection. There is a force on the section that results in partial yielding. There is a deflection due to the original load. There is an additional force. There is a difference between the original centre of gravity and the effective point of gravity. An eccentric moment has been introduced to change the location of the load. Due to the eccentric moment the load is located in the centre of gravity of the effective section. See Appendix C for the analysis. The ultimate load calculations are made in Appendices B. till B.8. Every Appendix exists in the ultimate load calculation of one column. The difference between the Appendices is the column length. For every column three calculation methods has been used:. Calculation according to the Dutch code.. Calculation of the Euler buckling load.. Calculation according to the formulas of the analysis in Appendix C. XXI

D. Calculation for a length of 0 meter Section properties: I 6.7*0 8 mm 4 A 7800 mm Z.896*0 6 mm I 4.848*0 8 mm 4 A 4650 mm Z.04*0 6 mm I.4*0 8 mm 4 A 8900 mm Z.5*0 6 mm Ultimate load calculation according to the Dutch code. For the ultimate load calculation according to the Dutch code, the relative slenderness must be calculated. See Appendix B for more information about the relative slenderness. λ rel π λ rel.08 EI f A y π 0000 0000*6.7*0 55*7800 For the ultimate load calculation of a HE 450A section stability curve a must be used (NEN 6770 art.... λ rel.08 ω buc 0.0 F max ω buc f y A 0.0*55*7800 F max 64*0 N Calculation of the Euler buckling load. π EI π 0000*6.7*0 F max 0000 (I is the unreduced moment of inertia F max 467*0 N 8 Ultimate load calculation according to the formulas of the analysis of Appendix C. The following equation is the general equation of the residual stress method. This formula has been derived in Appendix C. Fi zi Fi z i ( Fi Fi etoatl, i FE, i etotal, ifi 8EIi 8EIi etotal, i ( etotal, i em, i e i F F F 8 E, i i i According to the NEN 677 (art...5, the maximum deflection is one over thousand times the column length. e - 000 0000 000 e - 0 mm The stress in the unloaded structure is the residual stress. The deflection of the unloaded structure is the initial deflection (imperfections. Starting the calculation the following values are used: e m,i 0 mm F - 0 N XXII

z 0 mm If these values are used, the formula of the total deflection can be simplified to the following formula: e0 FE, n etotal, e (this is also known as the F F n formula. E, π EI F E, F E, 467*0 N π 0000*6.7*0 0000 8 e i e F 0 E, 0 FE, F e F ( e e 0 0 *467 *0 467 *0 F 0 0 *467 *0 F 467 *0 F 0 *467 *0 F F σ right, Z A 467 *0 F F -48.5 N/mm 6.896*0 7800 F e 48*0 N 40 mm 48*0 F σ left, Z A left, σ 96.5 N/mm Tension 0*467 *0 467 *0 48*0 48*0 6.896 *0 7800 F 0 48*0 F 48*0 N e total, 0 40 e total, 70 mm The section starts to yield at a load of 48*0 N. The total deflection is 70 mm. This is.% of the length of the column. After partial yielding the moment of inertia decreases. The Euler buckling load can be calculated for the section with the reduced moment of inertia. The calculation continues with the new (reduced Euler buckling load. 8 π EI π 0000* 4.848*0 F E, 0000 F E, 6*0 N The new Euler buckling load is smaller than the load on the column. After first yield, the structure becomes unstable. The column fails. The maximum load is the load that results in first yielding. F E, F So the structure fails at F F F max 48*0 N XXIII

D.4 Calculation for a length of 5 meter Section properties: I 6.7*0 8 mm 4 A 7800 mm Z.896*0 6 mm I 4.848*0 8 mm 4 A 4650 mm Z.04*0 6 mm I.4*0 8 mm 4 A 8900 mm Z.5*0 6 mm Ultimate load calculation according to the Dutch code. 5000 λ rel 8 EI π 0000*6.7*0 π f A 55*7800 y λ rel.7 ω buc 0.8 F max ω buc f y A 0.8*55*7800 F max 769*0 N Calculation of the Euler buckling load. π EI π 0000*6.7*0 F max 5000 F max *0 N 8 Ultimate load calculation according to the formula of the analysis of Appendix C. Fi zi Fi z i ( Fi Fi etoatl, i FE, i etotal, ifi 8EIi 8EIi etotal, i ( etotal, i em, i e i F F F e - 000 5000 000 e - 5 mm e m,i 0 mm F - 0 N z 0 mm If these values are used, the following formula exists: e0 FE, etotal, ( e0 e F F F E, *0 N E, E, i i i e e F 0 E, 0 FE, F e F ( e e 0 5* *0 *0 F 5 5* *0 F *0 F XXIV

F σ right, Z A 5* *0 F F.896*0 7800 *0 F -48.5 N/mm 6 F e 874*0 N 96 mm 874 *0 F σ left, Z A left, σ 7.8 N/mm Tension 5* *0 *0 874*0 874 *0 6.896*0 7800 F 0 874*0 F 874*0 N e total, 5 96 e total, mm The section starts to yield at a load of 874*0 N. The total deflection is mm. This is 0.88% of the length of the column. The moment of inertia decreases. The reduced Euler buckling load becomes: π EI F E, F E, 608*0 N π 0000* 4.848*0 5000 8 The new Euler buckling load is smaller than the load on the column. After first yield, the structure becomes unstable. The column fails. The maximum load is found. F E, F So the structure fails at F F F max 874*0 N XXV

D.5 Calculation for a length of 0 meter Section properties: I 6.7*0 8 mm 4 A 7800 mm Z.896*0 6 mm I 4.848*0 8 mm 4 A 4650 mm Z.04*0 6 mm I.4*0 8 mm 4 A 8900 mm Z.5*0 6 mm Ultimate load calculation according to the Dutch code. 0000 λ rel 8 EI π 0000*6.7*0 π f A 55*7800 y λ rel.8 ω buc 0.4 F max ω buc f y A 0.4*55*7800 F max 77*0 N Calculation of the Euler buckling load. π EI π 0000*6.7*0 F max 0000 F max 0*0 N 8 Ultimate load calculation according to the formulas of the analysis of Appendix C. ( e e e e total, i total, i m, i i F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( e - 000 0000 000 e - 0 mm e m,i 0 mm F - 0 N z 0 mm If these values are used, the following formula exists. e0 FE, etotal, ( e0 e F F F E, 0*0 N E, E, i i i e e F 0 E, 0 FE, F e F ( e e 0 0 *0 *0 0 *0 F 0 0 *0 *0 F 0 *0 F XXVI

F σ right, Z A 0*0*0 F F.896*0 7800 0 *0 F -48.5 N/mm 6 F e 679*0 N 86 mm 679*0 F σ left, Z A left, σ -5.5 N/mm Compression 0*0 *0 0 *0 679 *0 679*0 6.896*0 7800 F 0 679*0 F 679*0 N e total, 0 86 e total, 06 mm The section starts to yield at a load of 679*0 N. The total deflection is 06 mm. This is 0.5% of the length of the column. The moment of inertia decreases. The reduced Euler buckling load becomes: π EI F E, F E, 5*0 N π * 00004.848*0 0000 8 The new Euler buckling load is smaller than the load on the column. After first yield, the structure becomes unstable. The column fails. The maximum load is found. F E, F So the structure fails at F F F max 679*0 N XXVII

D.6 Calculation for a length of 5 meter Section properties: I 6.7*0 8 mm 4 A 7800 mm Z.896*0 6 mm I 4.848*0 8 mm 4 A 4650 mm Z.04*0 6 mm I.4*0 8 mm 4 A 8900 mm Z.5*0 6 mm Ultimate load calculation according to the Dutch code. 5000 λ rel 8 EI π 0000*6.7*0 π f A 55*7800 y λ rel.04 ω buc 0.64 F max ω buc f y A 0.64*55*7800 F max 4044*0 N Calculation of the Euler buckling load. π EI π 0000*6.7*0 F max 5000 F max 5870*0 N 8 Ultimate load calculation according to the formulas of the analysis of Appendix C. ( e e e e total, i total, i m, i i F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( e - 000 5000 000 e - 5 mm e m,i 0 mm F - 0 N z 0 mm If the values are used, the following formula exists: e0 FE, etotal, ( e0 e F F F E, 5870*0 N E, E, i i i e e F 0 E, 0 FE, F e F ( e e 0 5*5870 *0 5870 *0 F 5 5*5870 *0 F 5870 *0 F XXVIII

F σ right, Z A 5*5870*0 F F.896*0 7800 5870 *0 F -48.5 N/mm 6 F e 578*0 N.4 mm 578*0 F σ left, Z A left, σ -5.6 N/mm Compression 5*5870 *0 5870*0 578*0 578*0 6.896*0 7800 F 0 578*0 F 578*0 N e total, 5.4 e total, 8.4 mm The section starts to yield at a load of 578*0 N. The moment of inertia decrease. The reduced Euler buckling load becomes: π EI F E, F E, 4466*0 N π 0000* 4.848*0 5000 8 The Euler buckling load is larger the load on the column. The section can resist more loads. It is possible to increase the load till another part of the section yields too. A point of attention is the shift of the centre of gravity. See Appendix C for more details. ( e e e e total, i total, i m, i i F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( E, i i i Ak zk z ½h A k z ( b *¼ b t ½ ( ½ ( ½ f t f h t f tw h bt f h t f ½h ( b *¼ b t ( h t t bt f f w f z 50**0.5 98*.5*0 00**49.5 0 50* 98*.5 00* 47 mm z Figure D.5: Shift of centre of gravity XXIX

( e e e e total, total, m, 578*0 F 47 *5000 F 47 8.4 4466*0 8* 0000* 4.848*0 8* 0000 * 4.848*0 8.4 * 578*0 4466*0 578*0 F ( F 8 8 σ F σ Z A right, right, F F e F ( e e F σ right, 48.5 Z A total, 48.5 46.5 Z A σ 578* e F *(8.4 e F 48.5-46.5 N/mm.04*0 4650 right, 6 F e 6*0 N 0.5 mm F σ left, σ left, σ left, Z A F F e F ( e e F 5.6 Z A σ left, 5.6 Z A total, 578*0 *0.5 6*0 (8.4 0.5 6*0 σ left, 5.6 6.04*0 4650 σ left, 9.4 N/mm Tension F 578*0 6*0 F 940*0 N e total, 8.4 0.5 e total, 9.9 mm In this calculation the right flange fully yields. Only the left flange can be taken as an effective cross-section. The moment of inertia has decreased too much. If the right flange fully yields, the structure fails. F max 940*0 N XXX

D.7 Calculation for a length of 0 meter Section properties: I 6.7*0 8 mm 4 A 7800 mm Z.896*0 6 mm I 4.848*0 8 mm 4 A 4650 mm Z.04*0 6 mm I.4*0 8 mm 4 A 8900 mm Z.5*0 6 mm Ultimate load calculation according to the Dutch code. 0000 λ rel 8 EI π 0000*6.7*0 π f A 55*7800 y λ rel 0.69 ω buc 0.85 F max ω buc f y A 0.85*55*7800 F max 57*0 N Calculation of the Euler buckling load: 8 π EI π 0000*6.7*0 F max 0000 F max 07*0 N (Euler buckling load is larger than the plastic yield load Ultimate load calculation according to the formulas of the analysis of Appendix C. ( e e e e total, i total, i m, i i F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( e - 000 0000 000 e - 0 mm e m,i 0 mm F - 0 N z 0 mm If these values are used, the following formula exists: e0 FE, etotal, ( e0 e F F F E, 07*0 N E, E, i i i e e F 0 E, 0 FE, F e F ( e e 0 0 *07 *0 07 *0 F 0 0 *07 *0 F 07 *0 F XXXI

F σ right, Z A 0 *07 *0 F F.896*0 7800 07 *0 F -48.5 N/mm 6 F e 406*0 N 4.4 mm F σ left, Z A σ -08. N/mm Compression left, 0*07 *0 406*0 07 *0 406*0 406*0 6.896 *0 7800 F 0 406*0 F 406*0 N e total, 0 4.4 e totqo, 4.4 mm The section starts to yield at a load of 406*0 N. The moment of inertia decreases. The reduced Euler buckling load becomes: π EI π 0000* 4.848*0 F E, 0000 F E, 0048*0 N 8 The Euler buckling load is larger than the load on the column. The column can resist more loads. It is possible to increase the load till another part of the section yields too. ( e e e e total, i total, i m, i i Ak zk z ½h A k F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( E, i i i z ( b *¼ b t ½ ( ½ ( ½ f t f h t f tw h bt f h t f ½h ( b *¼ b t ( h t t bt f f w f z 50**0.5 98*.5*0 00**49.5 0 50* 98*.5 00* 47 mm z ( e e e e total, total, m, 406*0 F 47 *0000 F 47 4.4 0048*0 8* 0000 * 4.848*0 8* 0000* 4.848*0 4.4 * 406*0 0048*0 406*0 F ( F 8 8 XXXII

σ F σ Z A right, right, F F e F ( e e F σ right, 48.5 Z A total, 48.5-46.5 Z A 4060 e F (4.4 e F σ right, 48.5-46.5 N/mm 6.04*0 4650 F e 57*0 N 7. mm σ F σ Z A left, left, σ left, 08. Z A σ left, 6 σ left, F F e F ( e e F 08. Z A total, 406*0 * 7. 57*0 (0 4.4 7. 57*0 08..04*0 4650-09.7 N/mm Compression F 406*0 57*0 F 564*0 N e total, 4.4 7. e total, 5.7 mm In this case the right flange fully yields. The column fails. F max 564*0 N XXXIII

D.8 Calculation for a length of 5 meter Section properties: I 6.7*0 8 mm 4 A 7800 mm Z.896*0 6 mm I 4.848*0 8 mm 4 A 4650 mm Z.04*0 6 mm I.4*0 8 mm 4 A 500 mm Z.5*0 6 mm Ultimate load calculation according to the Dutch code. 5000 λ rel 8 EI π 0000*6.7*0 π f A 55*7800 y λ rel 0.5 ω buc 0.97 F max ω buc f y A 0.97*55*7800 F max 69*0 N Calculation of the Euler buckling load: 8 π EI π 0000*6.7*0 F max 5000 F max 587*0 N (Euler buckling load is larger than the plastic yield load Ultimate load calculation according to the formulas of the analysis of Appendix C. ( e e e e total, i total, i m, i i F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( e - 000 5000 000 e - 5 mm e m,i 0 mm F - 0 N z 0 mm E, i i i If these values will be used, the following formula exists: e0 FE, etotal, ( e0 e F F F E, 587*0 N E, e e F 0 E, 0 FE, F e F ( e e 0 5*587 *0 587 *0 F 5 5*587 *0 F 587 *0 F XXXIV

F σ right, Z A F 5*587 *0 F.896 *0 7800 587 *0 F -48.5 N/mm 6 F e 480*0 N 0.44 mm F σ left, Z A σ -.4 N/mm Compression left, 5*587 *0 480 *0 587 *0 480 *0 480 *0 6.896 *0 7800 F 0 480*0 F 408*0 N e total, 5 0.44 e total, 5.44 mm The section starts to yield at a load of 408*0 N. The moment of inertia decrease. The reduced Euler buckling load becomes: π EI π 0000* 4.848*0 F E, 5000 F E, 409*0 N 8 The Euler load is larger than the load on the column. The column can resist more loads. It is possible to increase the load till another part of the section yields too. ( e e e e total, i total, i m, i i Ak zk z ½h A k F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( E, i i i z ( b *¼ b t ½ ( ½ ( ½ f t f h t f tw h bt f h t f ½h ( b *¼ b t ( h t t bt f f w f z 50**0.5 98*.5*0 00**49.5 0 50* 98*.5 00* 47 mm z ( e e e e total, total, m, F 47 *5000 F 47 8* 0000 * 4.848*0 8* 0000* 4.848*0 409 *0 480 *0 F ( 480*0 F 5 0.44 409*0 480 *0 8 8 ( 5 0.44 XXXV

F σ left, σ left, σ left, Z A F F e F ( e e F σ left,.4 Z A toatl,.4-48.5 Z A σ 480 e F (5.44 e F.4-48.5 N/mm.04 *0 4650 left, 6 F e 6*0 N 0.5 mm σ F σ Z A right, right, σ right, 48.5 Z A σ right, 6 F F e F ( e e F 48.5 Z A total, 480*0 * 0.5 6*0 *(5.44 0.5 6*0 48.5.04 *0 4650 σ right, -68.0 N/mm F 480*0 6*0 F 454*0 N e total, 5.44 0.5 e total, 5.95 mm The left flange starts to yield if the load is 454*0 N. The moment of inertia decreases again. The new reduced Euler buckling load becomes: π EI π 0000*.4 *0 F E, 5000 F E, 7558*0 N 8 Again the Euler load is larger than the load that is on the column. The column can resist more loads. It is possible to increase the load till also a third part of the section yields. ( e e e e total, i total, i m, i i F F F z e F z F e F F F F i i i i i i toatl, i E, i 8EIi 8EIi total, i i ( E, i i i Both flanges have a yield part. Due to the used residual stress distribution, the yielding areas in the midsection are the same in both flanges. The stress in the midsection has been generalized over the whole section. It is assumed that the yielding areas in the whole section are the same. The effective section become double symmetric again. The location of the centre of gravity in Figure D.6: New centre of gravity XXXVI

the effective section is the same as the location of the centre of gravity in the original section. The shift of the centre of gravity is zero. z 0 mm ( e e e e total, total, m, 5.95* 7558*0 5.95* 454*0 7558*0 454*0 F σ F σ Z A right, right, F F e F ( e e F σ right, 68.0 Z A total, 68.0-46.5 Z A 454*0 e F (5.95 e F σ right, 68.0-46.5 N/mm 6.5*0 8900 F e 648*0 N 0.46 mm σ F σ Z A left, left, σ left, 48.5 Z A σ left, 6 σ left, F F e F ( e e F 48.5 Z A total, 454*0 * 0.46 648*0 (5.95 0.46 648*0 48.5.5*0 8900-45. N/mm Compression F 454*0 648*0 F 689*0 N e total, 5.95 0.46 e total, 6.4 mm F max 689*0 N The right flange fully yields and even the left flange has a large compression stress. The column will fail at a load of 689*0 N. XXXVII

D.9 Conclusions For several lengths is the ultimate load is calculated. There are only small differences between the ultimate load according to Dutch code and the ultimate load according to the formulas of the analysis of Appendix C. For a better result, the residual stress distribution and the yielding zone must be changed. The residual stress distribution is taken as a rectangular distribution. The real residual stress distribution is unknown, but is a more curved residual stress distribution is close to reality. The stress in the midsection has been generalized over the whole section. For a better analysis, the section must be split in several parts. Every part has his own section properties. A finite element analysis is the result. The calculated ultimate loads are presented in Figure D.7. The purple line is the Euler buckling load The black line is the plastic yield load The red line is the ultimate load according to the Dutch code The blue line is the ultimate load according to the formulas of the analysis in Appendix C Figure D.7: Force / length diagram XXXVIII

Appendix E Calculation example (atab In Appendix D some hand-made calculations are made for the calculation of the ultimate load for a single column. The formulas which are used are derived in Appendix C. Appendix E exists in two parts. The first part is the general part. In this part the way of calculation and the results of the calculations are discussed. In the second part, the calculation file is inserted. E. Computer calculations in general To make calculations, the section properties are needed. The section properties of all HEA are inserted in the m-file. The correct section can be chosen by a number. For a good comparison a HE 450A section has been chosen. This is the 6 th section of the list. atab can calculate many iterations. any lengths can be calculated in a small time period. There has been chosen for all lengths (with steps of one meter between one and fifty meters. This corresponds to a relative slenderness between zero and three. In Appendix D, in Figure D.7 a force length diagram is given. This diagram is copied to Figure E.. As result of the computer calculations, the same force-length diagram can be made (Fig. E.. The graphic in Figure E. is a more flowing line compare with the graphic in Figure E.. Cause of this is the larger amount of ultimate load calculations. For both figures is the Euler buckling load the black line, the yield load is the purple line, the red line is according to the Dutch code and the red line is according to the formulas from the analysis of Appendix C. c r i t i c a l l o a d ( N 0 x 06 9 8 7 6 5 4 force (N 0 x 06 9 8 7 6 5 4 0 0.5.5.5 length (mm x 0 4 0 0 0.5.5.5.5 4 4.5 5 length (mm x 0 4 Figure E.: Force / length diagram Hand made calculations XXXIX

Both Figure E. and E. shows that the ultimate load calculated by the analysis (derived in appendix C is close to the ultimate load according to the Dutch code. One point of attention is the ultimate load for a column length between ten and eleven meters. At this length the flowing line has an interruption. The cause of this interruption is the change in load case. A more flowing residual stress distribution and a better yielding zone result in a more flowing line. If these simplifications will be made, the analysis becomes much more complex. E. Calculation file This is the calculation file for the computer program atab. This file is used to calculate the ultimate load for a single column. clear; clf; clc; close; E0000; fy55; FFe;%belastingsstap %input profiles HEA[.4E0.54E0.4E0.877E0 4.55E0 5.8E0 6.44E0 7.684E0 8.68E0 9.76E0.5E04.44E04.5E04.48E04.590E04.780E04.975E04.8E04.65E04.46E04.605E04.858E04.05E04.468E04;%A (cross-section.49e06 6.06E06.0E07.67E07.50E07.69E07 5.40E07 7.76E07.046E08.67E08.86E08.9E08.769E08.09E08 4.507E08 6.7E08 8.698E08.9E09.4E09.75E09.5E09.04E09 4.E09 5.58E09;%I (moment of inertia 8.0E04.95E05.75E05.45E05.49E05 4.95E05 5.685E05 7.446E05 9.98E05.E06.8E06.68E06.850E06.088E06.56E06.6E06.949E06 4.6E06 5.50E06 6.6E06 7.0E06 8.699E06.08E07.8E07;%Z plastic (Section modulus 7.76E04.06E05.554E05.0E05.96E05.886E05 5.5E05 6.75E05 8.64E05.0E06.60E06.479E06.678E06.89E06.E06.896E06.550E06 4.46E06 4.787E06 5.474E06 6.4E06 7.68E06 9.485E06.9E07;%Z elastic (Section modulus 4.055E0 4.89E0 5.74E0 6.569E0 7.448E0 8.8E0 9.70E0.005E0.097E0.86E0.74E0.58E0.440E0.5E0.684E0.89E0.099E0.99E0.497E0.69E0.875E0.58E0.69E0.996E0;%i (Gyration radius 96 4 5 7 90 0 0 50 70 90 0 0 50 90 440 490 540 590 640 690 790 890 990;%h height 00 0 40 60 80 00 0 40 60 80 00 00 00 00 00 00 00 00 00 00 00 00 00 00;%b width X

8 8 8.5 9 9.5 0.5 4 5.5 6.5 7.5 9 4 5 6 7 8 0 ;%tf thickness flange 5 5 5.5 6 6 6.5 7 7.5 7.5 8 8.5 9 9.5 0.5.5.5 4.5 5 6 6.5];%tw thickness web m6; A(,HEA(,m; I(,HEA(,m; Z(,HEA(4,m; hhea(6,m; ihea(5,m; bhea(7,m; tfhea(8,m; twhea(9,m; NpA*fy; pz*fy; if m<4 ; S0.5; ak0.4; else if m<4; S0.; ak0.; else if m<8; S0.5; ak0.49; else S0.; ak0.4; end% if end% if end% if yield-(-s*fy; yield-(s*fy; A(,A(,-*(0.5*b*tf; A(,A(,-0.5*tw*(h-*tf-*(-0.5*pi*7^; A(,4A(,-*(0.5*b*tf; I(,I(,-*(0.5*b*tf*(0.5*h^; I(,I(,; I(,4I(,-*(0.5*b*tf*(0.5*h^; Z(,*I(,/h; Z(,*I(,/h; Z(,4*I(,4/h; for k:50; (k,000*k; e0(k,(k,/000; fy(kfy*a(,; % According to NEN 677 labda(k,(k,/(pi*sqrt(e*i(,/(a(,*fy; omega(k,((ak*(labda(k,-0.labda(k,^- sqrt((ak*(labda(k,-0.labda(k,^^- 4*labda(k,^/(*labda(k,^; if omega(k,>; XI

omega(k,; else omega(k,omega(k,; end% if omega Fnen(k,A(,*fy*omega(k,; % According to Euler FE(k,pi^*E*I(,/((k,^; FE(k,pi^*E*I(,/((k,^; FE(k,pi^*E*I(,/((k,^; FE(k,4pi^*E*I(,4/((k,^; % According to residual stress method sigmatop(k,0; f0; while sigmatop(k,>yield; ff; deltaf(k,fff*f; F(k,max(deltaF(k,f; Ftotal(k,fF(k,; f(k,f(k,/ff; deltae(k,fe0(k,*fe(k,/(fe(k,-deltaf(k,f-e0(k,; etotal(k,fdeltae(k,fe0(k,; emax(k,max(etotal(k,f; e(k,max(deltae(k,:; delta(k,fdeltaf(k,f*(e0(k,deltae(k,f; sigmatop(k,-delta(k,f/z(,-deltaf(k,f/a(,; sigmatop(k,sigmatop(k,; sigmabottom(k,delta(k,f/z(,-deltaf(k,f/a(,; sigmabottom(k,sigmabottom(k,; sigmacentre(k,-deltaf(k,f/a(,; sigmacentre(k,sigmacentre(k,; Frs(k,max(deltaF(k,f; Frs(k,Frs(k,; end% while Frs(k,0; if FE(k,>Frs(k,*FF; f0; z(0.5*b*tf*0.5*tf(h-*tf*tw*0.5*hb*tf*(h-0.5*tf/(0.5*b*tf(h- *tf*twb*tf-0.5*h; while sigmatop(k,>yield & sigmacentre(k,>yield; ff; deltaf(k,ff*ff; F(k,F(k,max(deltaF(k,f; Ftotal(k,ff(k,F(k,max(deltaF(k,f; f(k,f(k,/ff; hulpz/(8*e*i(,; etotal(k,ff(k,(fe(k,*(e0(k,e(k,deltaf(k,f*hulp*(k,^- F(k,*(e0(k,e(k,/(FE(k,-F(k,-deltaF(k,f; emax(k,max(etotal(k,ff(k,; deltae(k,fetotal(k,ff(k,-e0(k,-e(k,; e(k,max(deltae(k,:e(k,; delta(k,ff(k,*deltae(k,fdeltaf(k,f*etotal(k,ff(k,; sigmatop(k,sigmatop(k,-delta(k,f/z(,- deltaf(k,f/a(,; sigmatop(k,sigmatop(k,; sigmabottom(k,sigmabottom(k,delta(k,f/z(,- deltaf(k,f/a(,; XII

sigmabottom(k,sigmabottom(k,; sigmacentre(k,sigmacentre(k,-deltaf(k,f/a(,; sigmacentre(k,sigmacentre(k,; Frs(k,max(deltaF(k,f; end% while Frs(k,Frs(k,Frs(k,; else Frs(k,Frs(k,; end% if FE(k, Frs(k,0; if FE(k,>Frs(k,*FF; f0; z(0.5*b*tf*0.5*tf(h-*tf*tw*0.5*hb*tf*(h-0.5*tf/(0.5*b*tf(h- *tf*twb*tf-0.5*h; while sigmatop(k,>yield & sigmabottom(k,>yield; ff; deltaf(k,ff*ff; F(k,F(k,max(deltaF(k,f; Ftotal(k,ff(k,F(k,max(deltaF(k,f; f(k,f(k,/ff; hulpz/(8*e*i(,; etotal(k,ff(k,(fe(k,*(e0(k,e(k,deltaf(k,f*hulp*(k,^- F(k,*(e0(k,e(k,/(FE(k,-F(k,-deltaF(k,f; emax(k,max(etotal(k,ff(k,; deltae(k,fetotal(k,ff(k,-e0(k,-e(k,; e(k,max(deltae(k,:e(k,; delta(k,ff(k,*deltae(k,fdeltaf(k,f*etotal(k,ff(k,; sigmatop(k,sigmatop(k,-delta(k,f/z(,- deltaf(k,f/a(,; sigmatop(k,sigmatop(k,; sigmabottom(k,sigmabottom(k,delta(k,f/z(,- deltaf(k,f/a(,; sigmabottom(k,sigmabottom(k,; sigmacentre(k,sigmacentre(k,-deltaf(k,f/a(,; sigmacentre(k,sigmacentre(k,; Frs(k,max(deltaF(k,f; end% while Frs(k,Frs(k,Frs(k,Frs(k,; else Frs(k,Frs(k,; end% if FE(k, Frs4(k,0; %z0; if FE(k,4>Frs(k,*FF; f0; z0; while sigmatop(k,>yield; ff; deltaf4(k,ff*ff; F4(k,F(k,max(deltaF4(k,f; Ftotal(k,ff(k,F(k,max(deltaF4(k,f; f4(k,f4(k,/ff; hulpz/(8*e*i(,; XIII

etotal(k,ff(k,(fe(k,4*(e0(k,e(k,deltaf4(k,f*hulp*(k,^- F(k,*(e0(k,e(k,/(FE(k,4-F(k,-deltaF4(k,f; emax(k,max(etotal(k,ff(k,; deltae4(k,fetotal(k,ff(k,-e0(k,-e(k,; e4(k,max(deltae4(k,:e(k,; delta4(k,ff(k,*deltae4(k,fdeltaf4(k,f*etotal(k,ff(k,; sigmatop(k,sigmatop(k,-delta4(k,f/z(,4- deltaf4(k,f/a(,4; sigmatop4(k,sigmatop(k,; sigmabottom(k,sigmabottom(k,delta4(k,f/z(,4- deltaf4(k,f/a(,4; sigmabottom4(k,sigmatop(k,; sigmacentre(k,sigmacentre(k,-deltaf4(k,f/a(,4; sigmacentre4(k,sigmacentre(k,; Frs4(k,max(deltaF4(k,f; end% while Frs(k,Frs(k,Frs(k,Frs(k,Frs4(k,; else Frs(k,Frs(k,; end% if FE(k, for kk:(frs(,frs(,frs(,frs4(,/ff; if etotal(k,kk0; etotal(k,kkemax(k,; else etotal(k,kketotal(k,kk; end% if etotal if Ftotal(k,kk0; Ftotal(k,kkFrs(k,; else Ftotal(k,kkFtotal(k,kk; end% if Ftotal end% for kk F[6070000; 604000; 605000; 5978000; 590000; 5879000; 585000; 5749000; 5657000; 5545000; 599000; 50000; 4990000; 470000; 495000; 4065000; 740000; 49000; 45000; 886000; 650000; 46000; 5000; 080000; 98000; 790000; 668000; 555000; 455000; 6000; 78000; 0000; 000; 069000; 0000; 956000; 906000; 860000; 88000; 778000; 74000; 707000; 675000; 645000; 67000; 59000; 566000; 54000; 5000; 50000]; omegars(k,frs(k,/(a(,*fy; omegaeuler(k,/(labda(k,^; omegaf(k,f(k,/(a(,*fy; end% for k %handcalculations hand[5000 0000 5000 0000 5000 0000]; fyhand[69e 69e 69e 69e 69e 69e]; FEhand[587e 07e 5870e 0e e 476e]; Frshand[689e 564e 940e 679e 874e 48e]; FNenhand[69e 57e 4044e 77e 769e 64e]; labdahand[5000/(pi*sqrt(e*i(,/(fy*a(, 0000/(pi*sqrt(E*I(,/(fy*A(, 5000/(pi*sqrt(E*I(,/(fy*A(, 0000/(pi*sqrt(E*I(,/(fy*A(, 5000/(pi*sqrt(E*I(,/(fy*A(, 0000/(pi*sqrt(E*I(,/(fy*A(,]; XIV

omegahandfrshand/(fy*a(,; %plot(,fnen,'red';hold on;plot(,fe(:,,'black';plot(,frs,'blue';axis([0 50000 0 e7];xlabel('length (mm';ylabel('force (N';grid; %plot(labda,omegaf,'green';hold on;plot(labda,omegaeuler,'black';plot(labda,omega,'red';plot(labda,omegar s,'blue';axis([0 0.];xlabel('relative slenderness';ylabel('reduction factor';grid; %plot(hand,fyhand,'black';hold on; plot(hand,fehand,'m';plot(hand,frshand,'r';plot(hand,fnenhand,'blue'; axis([5000 0000 0 e7];xlabel('length (mm';ylabel('critical load (N';grid clear k; %clear kk; %for x:695;if Ftotal(5,x<Ftotal(5,(x-;Ftotal(5,xFtotal(5,(x- ;end;end %for x:695;if etotal(5,x<etotal(5,(x-;etotal(5,xetotal(5,(x- ;end;end %plot(etotal(5,:,ftotal(5,: %plot(etotal(5,:,ftotal(5,:;xlabel('deflection in the midsection (mm';ylabel('critical load (N';axis([0 50 0 e6];title('ength is 5 meter';grid XV

Appendix F Calculation example (atrix Frame In Appendix D and in Appendix E some ultimate load calculations are made. anually calculations in Appendix D and computer calculations (with the program atab in Appendix E. All these ultimate load calculations are based on the formulas derived in Appendix C. In this Appendix the ultimate load of the same structures are made by the computer program atrix Frame. atrix Frame is a computer program based on the finite element method (FE. The calculations in this appendix are made as a reference. F. Calculations in general atrix Frame can only calculate straight sections. To calculate with an initial deflection the column is split in four parts. The middle of the column (K has a horizontal displacement of. The horizontal displacement of point K and point 000 K4 is 0.8 (Fig. F., the horizontal displacement is elongate to get a clear view. 000 These horizontal deflections correspond to a curved column. atrix Frame has a four calculations method. These calculations methods are: Geometrical inear elastic Geometrical Non-inear elastic Physical Non-inear Physical and Geometrical Non-inear Figure F. Input column F. Different calculations Four different calculations have been made for a column with a length of ten meters and a HE450 A section. The different calculations are: geometrical linear elastic, geometrical nonlinear elastic, physical non-linear and physical and geometrical non-linear. The geometrical linear elastic calculations results in a linear relation between the load on the column and the deflection of the column. The linear elastic calculation does not take yielding into account. There is no load limit in this calculation method. n The geometric non-linear elastic calculation is based on the formula. This formula is n derived in Appendix A.. There is a non-linear relation between the load and the deflection. The limit of the load is the Euler buckling load (F E 07*0 N. The physical non-linear calculation (method does take yielding into account. Due to partial the effective section properties decreases. The deflection increases. The calculation stops if the whole mid-section yields. The physical and geometrical non-linear calculation methods take both the geometrical and yielding influence into account. The normal force multiplied by the deflection results in a XVI