CE 579: STRUCTRAL STABILITY AND DESIGN

Transcription

1 8/5/4 CE 579: STRUCTRA STABIITY AND DESIGN Amit H. Varma rofessor School of Civil Engineering urue Universit h. No. (765) Chapter. Introuction to Structural Stabilit OUTINE Definition of stabilit Tpes of instabilit Methos of stabilit analses Eamples small eflection analses Eamples large eflection analses Eamples imperfect sstems Design of steel structures

2 8/5/4 STABIITY DEFINITION Change in geometr of a structure or structural component uner compression resulting in loss of abilit to resist loaing is efine as instabilit in the boo. Instabilit can lea to catastrophic failure must be accounte in esign. Instabilit is a strength-relate limit state. Wh i we efine instabilit instea of stabilit? Seem strange! Stabilit is not eas to efine. Ever structure is in euilibrium static or namic. If it is not in euilibrium, the bo will be in motion or a mechanism. A mechanism cannot resist loas an is of no use to the civil engineer. Stabilit ualifies the state of euilibrium of a structure. Whether it is in stable or unstable euilibrium. STABIITY DEFINITION Structure is in stable euilibrium when small perturbations o not cause large movements lie a mechanism. Structure vibrates about it euilibrium position. Structure is in unstable euilibrium when small perturbations prouce large movements an the structure never returns to its original euilibrium position. Structure is in neutral euilibrium when we cant ecie whether it is in stable or unstable euilibrium. Small perturbation cause large movements but the structure can be brought bac to its original euilibrium position with no wor. Thus, stabilit tals about the euilibrium state of the structure. The efinition of stabilit ha nothing to o with a change in the geometr of the structure uner compression seems strange!

3 8/5/4 STABIITY DEFINITION BUCKING Vs. STABIITY Change in geometr of structure uner compression that results in its abilit to resist loas calle instabilit. Not true this is calle bucling. Bucling is a phenomenon that can occur for structures uner compressive loas. The structure eforms an is in stable euilibrium in state-. As the loa increases, the structure suenl changes to eformation state- at some critical loa cr. The structure bucles from state- to state-, where state- is orthogonal (has nothing to o, or inepenent) with state-. What has bucling to o with stabilit? The uestion is - Is the euilibrium in state- stable or unstable? Usuall, state- after bucling is either neutral or unstable euilibrium 3

4 8/5/4 BUCKING < cr = cr > cr BUCKING Vs. STABIITY Thus, there are two topics we will be intereste in this course Bucling Suen change in eformation from state- to state- Stabilit of euilibrium As the loas acting on the structure are increase, when oes the euilibrium state become unstable? The euilibrium state becomes unstable ue to: arge eformations of the structure Inelasticit of the structural materials We will loo at both of these topics for Columns Beams Beam-Columns Structural Frames 4

5 8/5/4 TYES OF INSTABIITY Structure subjecte to compressive forces can unergo:. Bucling bifurcation of euilibrium from eformation state- to state-. Bifurcation bucling occurs for columns, beams, an smmetric frames uner gravit loas onl. Failure ue to instabilit of euilibrium state- ue to large eformations or material inelasticit Elastic instabilit occurs for beam-columns, an frames subjecte to gravit an lateral loas. Inelastic instabilit can occur for all members an the frame. We will stu all of this in this course because we on t want our esigne structure to bucle or fail b instabilit both of which are strength limit states. TYES OF INSTABIITY BIFURCATION BUCKING Member or structure subjecte to loas. As the loa is increase, it reaches a critical value where: The eformation changes suenl from state- to state-. An, the euilibrium loa-eformation path bifurcates. Critical bucling loa when the loa-eformation path bifurcates rimar loa-eformation path before bucling Seconar loa-eformation path post bucling Is the post-bucling path stable or unstable? 5

6 8/5/4 SYMMETRIC BIFURCATION ost-bucling loa-eform. paths are smmetric about loa ais. If the loa capacit increases after bucling then stable smmetric bifurcation. If the loa capacit ecreases after bucling then unstable smmetric bifurcation. ASYMMETRIC BIFURCATION ost-bucling behavior that is asmmetric about loa ais. 6

7 8/5/4 INSTABIITY FAIURE There is no bifurcation of the loa-eformation path. The eformation stas in state- throughout The structure stiffness ecreases as the loas are increase. The change is stiffness is ue to large eformations an / or material inelasticit. The structure stiffness ecreases to ero an becomes negative. The loa capacit is reache when the stiffness becomes ero. Neutral euilibrium when stiffness becomes ero an unstable euilibrium when stiffness is negative. Structural stabilit failure when stiffness becomes negative. INSTABIITY FAIURE FAIURE OF BEAM-COUMNS M K= M K< K M No bifurcation. Instabilit ue to material an geometric nonlinearit 7

8 8/5/4 INSTABIITY FAIURE Snap-through bucling Snap-through INSTABIITY FAIURE Shell Bucling failure ver sensitive to imperfections 8

9 8/5/4 Chapter. Introuction to Structural Stabilit OUTINE Definition of stabilit Tpes of instabilit Methos of stabilit analses Eamples small eflection analses Eamples large eflection analses Eamples imperfect sstems Design of steel structures METHODS OF STABIITY ANAYSES Bifurcation approach consists of writing the euation of euilibrium an solving it to etermine the onset of bucling. Energ approach consists of writing the euation epresg the complete potential energ of the sstem. Analing this total potential energ to establish euilibrium an eamine stabilit of the euilibrium state. Dnamic approach consists of writing the euation of namic euilibrium of the sstem. Solving the euation to etermine the natural freuenc (w) of the sstem. Instabilit correspons to the reuction of w to ero. 9

10 8/5/4 STABIITY ANAYSES Each metho has its avantages an isavantages. In fact, ou can use ifferent methos to answer ifferent uestions The bifurcation approach is appropriate for etermining the critical bucling loa for a (perfect) sstem subjecte to loas. The eformations are usuall assume to be small. The sstem must not have an imperfections. It cannot provie an information regaring the post-bucling loaeformation path. The energ approach is the best when establishing the euilibrium euation an eamining its stabilit The eformations can be small or large. The sstem can have imperfections. It provies information regaring the post-bucling path if large eformations are assume The major limitation is that it reuires the assumption of the eformation state, an it shoul inclue all possible egrees of freeom. STABIITY ANAYSIS The namic metho is ver powerful, but we will not use it in this class at all. Remember, it though when ou tae the course in namics or earthuae engineering In this class, ou will learn that the loas acting on a structure change its stiffness. This is significant ou have not seen it before. M a a M a 4E I a E I Mb b M b What happens when an aial loa is acting on the beam. The stiffness will no longer remain 4EI/ an EI/. Instea, it will ecrease. The reuce stiffness will reuce the natural freuenc an perio elongation. You will see these in our namics an earthuae engineering class.

11 8/5/4 STABIITY ANAYSIS FOR ANY KIND OF BUCKING OR STABIITY ANAYSIS NEED TO DRAW THE FREE BODY DIAGRAM OF THE DEFORMED STRUCTURE. WRITE THE EQUATION OF STATIC EQUIIBRIUM IN THE DEFORMED STATE WRITE THE ENERGY EQUATION IN THE DEFORMED STATE TOO. THIS IS CENTRA TO THE TOIC OF STABIITY ANAYSIS NO STABIITY ANAYSIS CAN BE ERFORMED IF THE FREE BODY DIAGRAM IS IN THE UNDEFORMED STATE BIFURCATION ANAYSIS Alwas a small eflection analsis To etermine cr bucling loa Nee to assume bucle shape (state ) to calculate Eample Rigi bar supporte b rotational spring Rigi bar subjecte to aial force Rotationall restraine at en Step - Assume a eforme shape that activates all possible.o.f. cos (-cos)

12 8/5/4 BIFURCATION ANAYSIS cos (-cos) Write the euation of static euilibrium in the eforme state M o For small eformations cr Thus, the structure will be in static euilibrium in the eforme state when = cr = / When < cr, the structure will not be in the eforme state. The structure will bucle into the eforme state when = cr BIFURCATION ANAYSIS Eample - Rigi bar supporte b translational spring at en Assume eforme state that activates all possible.o.f. Draw FBD in the eforme state O cos (-cos)

13 8/5/4 BIFURCATION ANAYSIS Write euations of static euilibrium in eforme state O M o cos For small eformations cr (-cos) ( ) Thus, the structure will be in static euilibrium in the eforme state when = cr =. When <cr, the structure will not be in the eforme state. The structure will bucle into the eforme state when = cr BIFURCATION ANAYSIS Eample 3 Three rigi bar sstem with two rotational springs A B C D Assume eforme state that activates all possible.o.f. Draw FBD in the eforme state A ( ) C B ( ) Assume small eformations. Therefore, = D 3

14 8/5/4 4 BIFURCATION ANAYSIS Write euations of static euilibrium in eforme state A D B C ( ) ( ) A B +( - ) ) ( ) ( M B ( - ) D C ( ) ( - ) ) ( ) ( M C BIFURCATION ANAYSIS Euations of Static Euilibrium Therefore either an are eual to ero or the eterminant of the coefficient matri is eual to ero. When an are not eual to ero that is when bucling occurs the coefficient matri eterminant has to be eual to ero for euil. Tae a loo at the matri euation. It is of the form [A] {}={}. It can also be rewritten as ([K]-l[I]){}={} ) ( ) (

15 8/5/4 BIFURCATION ANAYSIS This is the classical eigenvalue problem. ([K]-l[I]){}={}. We are searching for the eigenvalues (l) of the stiffness matri [K]. These eigenvalues cause the stiffness matri to become gular Singular stiffness matri means that it has a ero value, which means that the eterminant of the matri is eual to ero. ( ) ( ) ( ) (3 ) ( ) cr 3 or Smallest value of cr will govern. Therefore, cr =/ BIFURCATION ANAYSIS Each eigenvalue or critical bucling loa ( cr ) correspons to a bucling shape that can be etermine as follows cr =/. Therefore substitute in the euations to etermine an ( ) et cr ( ) All we coul fin is the relationship between an. Not their specific values. Remember that this is a small eflection analsis. So, the values are negligible. What we have foun is the bucling shape not its magnitue. The bucling moe is such that = Smmetric bucling moe A ( ) = et cr ( ) D B C 5

16 8/5/4 BIFURCATION ANAYSIS Secon eigenvalue was cr =3/. Therefore substitute in the euations to etermine an ( ) et 3 cr ( ) 3 ( ) All we coul fin is the relationship between an. Not their specific values. Remember that this is a small eflection analsis. So, the values are negligible. What we have foun is the bucling shape not its magnitue. The bucling moe is such that =- Antismmetric bucling moe C et 3 cr ( ) 3 A =- B D Chapter. Introuction to Structural Stabilit OUTINE Definition of stabilit Tpes of instabilit Methos of stabilit analses Bifurcation analsis eamples small eflection analses Energ metho Eamples small eflection analses Eamples large eflection analses Eamples imperfect sstems Design of steel structures 6

17 8/5/4 ENERGY METHOD We will currentl loo at the use of the energ metho for an elastic sstem subjecte to conservative forces. Total potential energ of the sstem epens on the wor one b the eternal forces (W e ) an the strain energ store in the sstem (U). = U - W e. For the sstem to be in euilibrium, its total potential energ must be stationar. That is, the first erivative of must be eual to ero. Investigate higher orer erivatives of the total potential energ to eamine the stabilit of the euilibrium state, i.e., whether the euilibrium is stable or unstable ENERGY METHD The energ metho is the best for establishing the euilibrium euation an eamining its stabilit The eformations can be small or large. The sstem can have imperfections. It provies information regaring the post-bucling path if large eformations are assume The major limitation is that it reuires the assumption of the eformation state, an it shoul inclue all possible egrees of freeom. 7

18 8/5/4 ENERGY METHOD Eample Rigi bar supporte b rotational spring Assume small eflection theor Rigi bar subjecte to aial force Rotationall restraine at en Step - Assume a eforme shape that activates all possible.o.f. cos (-cos) ENERGY METHOD SMA DEFECTIONS cos (-cos) Write the euation representing the total potential energ of sstem U We U We ( cos ) ( cos ) Foreuilibrium ; Therefore, For small eflections ; Therefore, cr 8

19 8/5/4 ENERGY METHOD SMA DEFECTIONS The energ metho preicts that bucling will occur at the same loa cr as the bifurcation analsis metho. At cr, the sstem will be in euilibrium in the eforme. Eamine the stabilit b consiering further erivatives of the total potential energ This is a small eflection analsis. Hence will be ero. In this tpe of analsis, the further erivatives of eamine the stabilit of the initial state- (when =) ( cos ) When cr When cr When cr Stableeuilibrium Unstableeuilibrium Not sure ENERGY METHOD SMA DEFECTIONS In state-, stable when < cr, unstable when > cr No iea about state uring bucling. No iea about post-bucling euilibrium path or its stabilit. Ineterminate Unstable cr Stable 9

20 8/5/4 ENERGY METHOD ARGE DEFECTIONS Eample arge eflection analsis (rigi bar with rotational spring) U We U We ( cos ) ( cos ) Foreuilibrium ; Therefore, Therefore, The post bucling for euilibrium relationship is givenabove cos (-cos) ENERGY METHOD ARGE DEFECTIONS arge eflection analsis See the post-bucling loa-isplacement path shown below The loa carring capacit increases after bucling at cr cr is where. Rigi bar with rotational spring oa /cr for euilibrium cr En rotation

21 8/5/4 ENERGY METHOD ARGE DEFECTIONS arge eflection analsis Eamine the stabilit of euilibrium ug higher orer erivatives of, ).,. ( ) tan ( cos, cos ) cos ( for But Alwas STABE valuesof all e i Alwas But ENERGY METHOD ARGE DEFECTIONS At =, the secon erivative of =. Therefore, inconclusive. Consier the Talor series epansion of at = Determine the first non-ero term of, Since the first non-ero term is >, the state is stable at = cr an = n n n n !... 4! 3!! cos cos ) cos ( cos !

22 8/5/4 ENERGY METHOD ARGE DEFECTIONS. Rigi bar with rotational spring STABE STABE.8 oa /cr.6.4 STABE En rotation ENERGY METHOD IMERFECT SYSTEMS Consier eample but as a sstem with imperfections The initial imperfection given b the angle as shown below cos( ) The free bo iagram of the eforme sstem is shown below ( ) cos (cos -cos)

23 8/5/4 ENERGY METHOD IMERFECT SYSTEMS ( ) U W U ( ) W (cos cos ) e ( ) (cos cos ) ( ) Foreuilibrium ; Therefore, ( ) Therefore, e ( ) for euilibrium The euilibrium relationship is givenabove cos (cos -cos) ENERGY METHOD IMERFECT SYSTEMS. ( ) cr relationsh Rigi bar ipswith for rotational ifferent valuesof spring shownbelow: oa /cr En rotation

24 8/5/4 ENERGY METHODS IMERFECT SYSTEMS As shown in the figure, eflection starts as soon as loas are applie. There is no bifurcation of loa-eformation path for imperfect sstems. The loa-eformation path remains in the same state through-out. The smaller the imperfection magnitue, the close the loaeformation paths to the perfect sstem loa eformation path The magnitue of loa, is influence significantl b the imperfection magnitue. All real sstems have imperfections. The ma be ver small but will be there The magnitue of imperfection is not eas to now or guess. Hence if a perfect sstem analsis is one, the results will be close for an imperfect sstem with small imperfections ENERGY METHODS IMERFECT SYSTEMS Eamine the stabilit of the imperfect sstem ug higher orer erivatives of ( ) (cos cos ) ( ) cos Euilibrium path will be stable if i. e., if cos i. e., if cos ( ) i. e., if cos i. e., tan Which is alwas true, hence alwas in STABE EQUIIBRIUM 4

25 8/5/4 ENERGY METHOD SMA DEFECTIONS Eample - Rigi bar supporte b translational spring at en Assume eforme state that activates all possible.o.f. Draw FBD in the eforme state O cos (-cos) ENERGY METHOD SMA DEFECTIONS Write the euation representing the total potential energ of sstem U W U ( ) W ( cos ) e e ( cos ) Foreuilibrium ; Therefore, O cos (-cos) For small Therefore, eflections ; cr 5

26 8/5/4 ENERGY METHOD SMA DEFECTIONS The energ metho preicts that bucling will occur at the same loa cr as the bifurcation analsis metho. At cr, the sstem will be in euilibrium in the eforme. Eamine the stabilit b consiering further erivatives of the total potential energ This is a small eflection analsis. Hence will be ero. In this tpe of analsis, the further erivatives of eamine the stabilit of the initial state- (when =) ( cos ) cos For small eflections an When, When, When STABE UNSTABE INDETERMINATE ENERGY METHOD ARGE DEFECTIONS Write the euation representing the total potential energ of sstem U W U ( ) W ( cos ) e ( cos ) cos Foreuilibrium ; Therefore, cos Therefore, e cos for euilibrium The post bucling relationship is givenabove O cos (-cos) 6

27 8/5/4 ENERGY METHOD ARGE DEFECTIONS arge eflection analsis See the post-bucling loa-isplacement path shown below The loa carring capacit ecreases after bucling at cr cr is where. Rigi bar with translational spring cos cos cr for euilibrium oa /cr En rotation ENERGY METHOD ARGE DEFECTIONS arge eflection analsis Eamine the stabilit of euilibrium ug higher orer erivatives of ( cos ) cos cos cos For euilibrium cos cos cos (cos ) cos AWAYS. HENCE UNSTABE 7

28 8/5/4 8 ENERGY METHOD ARGE DEFECTIONS At =, the secon erivative of =. Therefore, inconclusive. Consier the Talor series epansion of at = Determine the first non-ero term of, Since the first non-ero term is <, the state is unstable at = cr an = n n n n !... 4! 3!! cos cos ) cos ( 3 3 whenbuclingoccurs UNSTABE at 3 4 cos cos ENERGY METHOD ARGE DEFECTIONS Rigi bar with translational spring En rotation oa /cr UNSTABE UNSTABE UNSTABE

29 8/5/4 ENERGY METHOD - IMERFECTIONS Consier eample but as a sstem with imperfections The initial imperfection given b the angle as shown below cos( ) The free bo iagram of the eforme sstem is shown below O cos (cos -cos) ENERGY METHOD - IMERFECTIONS U W U ( ) W (cos cos ) e e ( ) (cos cos ) ( ) cos Foreuilibrium ; Therefore, ( ) cos Therefore, cos ( ) for euilibrium The euilibrium relationship is givenabove O cos (cos -cos) 9

30 8/5/4 ENERGY METHOD - IMERFECTIONS. cos ( ) ma ( ) 3 cos ma Rigi bar with translational spring cr cos ( ) 3 Envelope of pea loas ma oa /cr En rotation ENERGY METHOD - IMERFECTIONS As shown in the figure, eflection starts as soon as loas are applie. There is no bifurcation of loa-eformation path for imperfect sstems. The loa-eformation path remains in the same state through-out. The smaller the imperfection magnitue, the close the loaeformation paths to the perfect sstem loa eformation path. The magnitue of loa, is influence significantl b the imperfection magnitue. All real sstems have imperfections. The ma be ver small but will be there The magnitue of imperfection is not eas to now or guess. Hence if a perfect sstem analsis is one, the results will be close for an imperfect sstem with small imperfections. However, for an unstable sstem the effects of imperfections ma be too large. 3

31 8/5/4 3 ENERGY METHODS IMERFECT SYSTEMS Eamine the stabilit of the imperfect sstem ug higher orer erivatives of cos ) (cos cos ) ( ) cos (cos ) ( m euilibriu For ) cos ( cos cos cos cos cos ) (cos 3 3 cos cos ) ( cos cos ) ( cos ma 3 ma an When an ENERGY METHOD IMERFECT SYSTEMS Unstable when Stable when ma ma 3 cos cos ) ( cos ma an When

32 8/5/4 Chapter. Secon-Orer Differential Euations This chapter focuses on eriving secon-orer ifferential euations governing the behavior of elastic members. First orer ifferential euations. Secon-orer ifferential euations. First-Orer Differential Euations Governing the behavior of structural members Elastic, Homogenous, an Isotropic Strains an eformations are reall small small eflection theor Euations of euilibrium in uneforme state Consier the behavior of a beam subjecte to bening an aial forces 3

33 8/5/4. First-Orer Differential Euations Assume tensile forces are positive an moments are positive accoring to the right-han rule ongituinal stress ue to bening A M I M I This is true when the - ais sstem is a centroial an principal ais sstem. A A I A A A; an I A A A A are principalmoment of inertia A A I ; A Centroialais A I M A I M I The corresponing strain is. First-Orer Differential Euations A E M E I M E I M If =M =, then E I lane-sections remain plane an perpenicular to centroial ais before an after bening The measure of bening is curvature f which enotes the change in the slope of the centroial ais between two point apart tanf For small eformations f M f E I M E I f tanf f an similarlm E I f 33

34 8/5/4. First-Orer Differential Euations Shear Stresses ue to bening V t I V t I s O s O t s t s. First-Orer Differential Euations Differential euations of bening Assume principle of superposition Treat forces an eformations in - an - plane seperatel Both the en shears an act in a plane parallel to the - plane through the shear center S V M V M ( E I f ) E I f 34

35 8/5/4. First-Orer Differential Euations Differential euations of bening E I f v f ( v) E I For small eflections f v v iv SimilarlE I u eflectionin 3 / u iv positive irection v eflectionin positive irection Fourth-orer ifferential euations ug firstorer force-eformation theor Torsion behavior ure an Warping Torsion Torsion behavior uncouple from bening behavior Thin walle open cross-section subjecte to torsional moment This moment will cause twisting an warping of the cross-section. The cross-section will unergo pure an warping torsion behavior. ure torsion will prouce onl shear stresses in the section Warping torsion will prouce both longituinal an shear stresses The internal moment prouce b the pure torsion response will be eual to M sv an the internal moment prouce b the warping torsion response will be eual to M w. The eternal moment will be euilibriate b the prouce internal moments M Z =M SV + M W 35

36 8/5/4 ure an Warping Torsion M Z =M SV + M W Where, M SV = G K T f an M W = - E I w f" M SV = ure or Saint Venant s torsion moment K T = J = Torsional constant = f is the angle of twist of the cross-section. It is a function of. I W is the warping moment of inertia of the cross-section. This is a new cross-sectional propert ou ma not have seen before. M Z = G K T f - E I w f" (3), ifferential euation of torsion ure Torsion Differential Euation ets loo closel at pure or Saint Venant s torsion. This occurs when the warping of the cross-section is unrestraine or absent r f f r r f G r f M M SV SV where, K r A G f r A G K T T f J A r A A A For a circular cross-section warping is absent. For thin-walle open cross-sections, warping will occur. The out of plane warping eformation w can be calculate ug an euation I will not show. 36

37 8/5/4 ure Torsion Stresses The torsional shear stresses var linearl about the center of the thin plate SV G r f ( ) G t f SV ma sv Warping eformations The warping prouce b pure torsion can be restraine b the: (a) en conitions, or (b) variation in the applie torsional moment (non-uniform moment) The restraint to out-of-plane warping eformations will prouce longituinal stresses ( w ), an their variation along the length will prouce warping shear stresses ( w ). 37

38 8/5/4 Warping Torsion Differential Euation ets tae a loo at an approimate erivation of the warping torsion ifferential euation. This is vali onl for I an C shape sections. h u f f whereu flangelateral isplacement M V f E I E I M f M M M Shear force in the flange W V where I moment in the flange f f u M u V W W W f f f f h E I E I E I W f f f W f u h f h f f borrowing. e. of bening is warpingmoment of inertia newsection propert Torsion Differential Euation Solution Torsion ifferential euation M Z =M SV +M W = G K T f - E I W f This ifferential euation is for the case of concentrate torue G K f E I f M T w Z G KT M Z f f E IW E IW M M Z f l f f C C coshl C3 hl E I l E I W Torsion ifferential euation for the case of istribute torue MZ mz G K f E I T iv f G K E I T W w iv f m mz f E I W Z f C4 C5 C6 coshl C7 hl iv mz f l f E IW The coefficients C... C 6 can be obtaine ug en conitions W m G K T 38

39 8/5/4 Torsion Differential Euation Solution Torsionall fie en conitions are given b f f These impl that twisting an warping at the fie en are full restraine. Therefore, eual to ero. Torsionall pinne or simpl-supporte en conitions given b: f f These impl that at the pinne en twisting is full restraine (f=) an warping is unrestraine or free. Therefore, W = f = Torsionall free en conitions given b f =f = f = These impl that at the free en, the section is free to warp an there are no warping normal or shear stresses. Results for various torsional loaing conitions given in the AISC Design Guie 9 can be obtaine from m private site Warping Torsion Stresses Restraint to warping prouces longituinal an shear stresses E W where, W NormalieUnit Warping Sectionr opert S W W n W n f t E S W f Warping StaticalMoment Sectionr opert The variation of these stresses over the section is efine b the section propert W n an S w The variation of these stresses along the length of the beam is efine b the erivatives of f. Note that a major ifference between bening an torsional behavior is The stress variation along length for torsion is efine b erivatives of f, which cannot be obtaine ug force euilibrium. The stress variation along length for bening is efine b erivatives of v, which can be obtaine ug force euilibrium (M, V iagrams). 39

40 8/5/4 Torsional Stresses Torsional Stresses 4

41 8/5/4 Torsional Section roperties for I an C Shapes f an erivatives for concentrate torue at mispan 4

42 8/5/4 Summar of first orer ifferential euations E I E I T v M u M G K f E I f M W () () (3) NOTES: () Three uncouple ifferential euations () Elastic material first orer force-eformation theor (3) Small eflections onl (4) Assumes no influence of one force on other eformations (5) Euations of euilibrium in the uneforme state. Chapter. Secon-Orer Differential Euations This chapter focuses on eriving secon-orer ifferential euations governing the behavior of elastic members. First orer ifferential euations. Secon-orer ifferential euations 4

43 8/5/4. Secon-Orer Differential Euations Governing the behavior of structural members Elastic, Homogenous, an Isotropic Strains an eformations are reall small small eflection theor Euations of euilibrium in eforme state The eformations an internal forces are no longer inepenent. The must be combine to consier effects. Consier the behavior of a member subjecte to combine aial forces an bening moments at the ens. No torsional forces are applie eplicitl because that is ver rare for CE structures. Member moel an loaing conitions Member is initiall straight an prismatic. It has a thin-walle open cross-section Member ens are pinne an prevente from translation. The forces are applie onl at the member ens These consist onl of aial an bening moment forces, M TX, M TY, M BX, M BY Assume elastic behavior with small eflections Right-han rule for positive moments an reactions an assume positive. 43

44 8/5/4 Member isplacements (cross-sectional) Consier the mile line of thinwalle cross-section an are principal coorinates through centroi C Q is an point on the mile line. It has coorinates (, ). Shear center S coorinates are ( o, ) Shear center S isplacements are u, v, an f Member isplacements (cross-sectional) Displacements of Q are: u Q = u + a f a v Q = v a f cos a where a is the istance from Q to S But, a = ( -) / a cos a = ( -) / a Therefore, isplacements of Q are: u Q = u + f ( -) v Q = v f ( ) Displacements of centroi C are: u c = u + f ( ) v c = v - f ( ) 44

45 8/5/4 Internal forces secon-orer effects Consier the free bo iagrams of the member in the eforme state. oo at the eforme state in the - an - planes in this Figure. The internal resisting moment at a istance from the lower en are: M = - M BX + R + v c M = - M BY + R - u c The en reactions R an R are: R = (M TY + M BY ) / R = (M TX + M BX ) / Therefore, Internal forces secon-orer effects M M M M BX BY ( M M ) ( v f ) TX ( M M ) ( u f ) TY BX BY 45

46 8/5/4 Internal forces in the eforme state In the eforme state, the cross-section is such that the principal coorinate sstems are change from -- to the h sstem f u c v c h M B R R M BY f u c v c h 46

47 8/5/4 M ς M ξ a σ+ σ M η σ M R R M Internal forces in the eforme state The internal forces M an M must be transforme to these new h aes Since the angle f is small M M + f M M h = M f M M M M M M M h BX BY M M BX BY ( M M ) ( v f ) TX ( M M ) ( u f ) TY ( M M ) v f M ( M M ) TX ( M M ) u f M ( M M ) TY BX BY BX BY BY BX TY TX BY BX 47

48 8/5/4 Twisting component of internal forces Twisting moments M are prouce b the internal an eternal forces There are four components contributing to the total M () Contribution from M an M M () Contribution from aial force M (3) Contribution from normal stress M 3 (4) Contribution from en reactions R an R M 4 The total twisting moment M = M + M + M 3 + M 4 Twisting component of 4 u v Twisting moment ue to M & M M = M (u/) + M (v/) Therefore, ue to small angles, M = M u/ + M v/ M = M u + M v 48

49 8/5/4 Twisting component of 4 u v The aial loa acts along the original vertical irection In the eforme state of the member, the longituinal ais is not vertical. Hence will have components proucing shears. These components will act at the centroi where acts an will have values as shown above assuming small angles Twisting component of 4 These shears will act at the centroi C, which is eccentric with respect to the shear center S. Therefore, the will prouce seconar twisting. M = ( u/ v/) Therefore, M = ( u v ) 49

50 8/5/4 Twisting component 3 of 4 The en reactions (shears) R an R act at the shear center S at the ens. But, along the member ens, the shear center will move b u, v, an f. Hence, these reactions will also have a twisting effect prouce b their eccentricit with respect to the shear center S. M 4 + R u + R v = Therefore, M 4 = (M TY + M BY ) v/ (M TX + M BX ) u/ Wagner s effect or contribution complicate. Two cross-sections that are apart will warp with respect to each other. The stress element A will become incline b angle (a f/) with respect to ais. Twist prouce b each stress element about S is eual to M 3 M 3 a ( A) f Twisting component 4 of 4 f a a A A 5

51 8/5/4 Twisting component 4 of 4 et, a A K M M A 3 3 f K f K for small angles Twisting component 4 of 4 et, a A K M M A 3 3 f K f K for small angles 5

52 8/5/4 Total Twisting Component M = M + M + M 3 + M 4 M = M u + M v M = ( u v ) M 4 = (M TY + M BY ) v/ (M TX + M BX ) u/ M 3 = -K f Therefore, M M u + M v + ( u v ) (M TY + M BY ) v/ (M TX + M BX ) u/-k f While, M M M h M BX BY ( M M ) v f M ( M M ) TX ( M M ) u f M ( M M ) TY BX BY BY BX TY TX BY BX Total Twisting Component M = M + M + M 3 + M 4 M = M u + M v M = ( u v ) M 3 = -K f M 4 = (M TY + M BY ) v/ (M TX + M BX ) u/ Therefore, v u M M u M v ( u v) ( MTY M BY ) ( MTX M BX ) K f v u M ( M ) u ( M ) v ( MTY M BY ) ( MTX M BX ) K f But, M M BX ( M BX MTX ) ( v f ) an, M M BY ( M BY MTY ) ( u f ) M ( M BX ( M BX MTX ) ) u ( M BY ( M BY M TY ) ) v v u ( MTY M BY ) ( MTX M BX ) K f 5

53 8/5/4 Internal moments about the h aes Thus, now we have the internal moments about the h aes for the eforme member cross-section. M M BX ( MTX M BX ) v f M BY ( MTY M BY ) Mh M BY ( M TX M BX ) u M BX ( TY M BY ) TY +M BY f M TX +M BX M ( M BX ( M BX MTX ) ) u ( M BY ( M BY MTY ) ) v v u ( MTY M BY ) ( MTX M BX ) K f h Internal Moment Deformation Relations The internal moments M, M h, an M will still prouce fleural bening about the centroial principal ais an twisting about the shear center. The fleural bening about the principal aes will prouce linearl varing longituinal stresses. The torsional moment will prouce longituinal an shear stresses ue to warping an pure torsion. The ifferential euations relating moments to eformations are still vali. Therefore, M = - E I v..(i = I ) M h = E I h u..(i h = I ) M = G K T f E I w f 53

54 8/5/4 Internal Moment Deformation Relations Therefore, M E I v M BX ( M TX M BX ) v f M BY ( M TY M BY ) Mh E I u M BY ( MTX M BX ) u M BX ( TY M BY ) TY +M BY f M TX +M BX M G KT f E Iw f ( M BX ( M BX M TX ) ) u v u ( M BY ( M BY M TY ) ) v ( M TY M BY ) ( M TX MBX ) K f 3 Therefore, Secon-Orer Differential Euations You en up with three couple ifferential euations that relate the applie forces an moments to the eformations u, v, an f. E I v v f M BY ( M TY M BY ) M BX ( M TX M BX ) E I u u f M BX ( M TY M BY ) M BY ( TX M BX ) TX +M BX M TY +M BY E Iw f ( G KT K) f u ( M BX ( M BX M TX ) ) v u v ( M BY ( M BY M TY ) ) ( M TY M BY ) ( M TX M BX ) These ifferential euations can be use to investigate the elastic behavior an bucling of beams, columns, beam-columns an also complete frames that will form a major part of this course. 54

55 8/5/4 Chapter 3. Structural Columns 3. Elastic Bucling of Columns 3. Elastic Bucling of Column Sstems Frames 3.3 Inelastic Bucling of Columns 3.4 Column Design rovisions (U.S. an Abroa) 3. Elastic Bucling of Columns Start out with the secon-orer ifferential euations erive in Chapter. Substitute = an M TY = M BY = M TX = M BX = Therefore, the secon-orer ifferential euations simplif to: 3 ( ) ( ) E I v v f E I u u f E I f ( G K K) f u ( ) v ( ) w T This is all great, but before we procee an further we nee to eal with Wagner s effect which is a little complicate. 55

56 8/5/4 Wagner s effect for columns K f a f A where, A M Mh EWn f A I I M ( v f ) M ( u f ) h ( v f ) ( u f ) K f E Wn f f a A A I A I ( v f ) ( u f ) Neglecting higher orer terms; K f f a A A K f E Wn f f a A A I I A A Wagner s effect for columns But, a ( ) ( ) a A ( ) ( ) A A A a A A A A a A A A A A A A A A A A A a A ( ) A I I A Finall, K f ( ) A I I A f I I K f ( ) f A I I et r ( ) A K f r f 56

57 8/5/4 Secon-orer ifferential euations for columns Simplif to: 3 ( ) ( ) E I v v f E I u u f E I f ( r G K ) f u ( ) v ( ) w T Where I r I A Column bucling oubl smmetric section For a oubl smmetric section, the shear center is locate at the centroi o = =. Therefore, the three euations become uncouple 3 E I v v E I u u E I f ( r G K ) f w Tae two erivatives of the first two euations an one more erivative of the thir euation. 3 E I v E I u iv iv v u iv E I f ( r G K ) f w T, v u f E I E I E Iw et F F F T T r G K 57

58 8/5/4 Column bucling oubl smmetric section 3 v u iv iv iv f F v v F u u F f f All three euations are similar an of the fourth orer. The solution will be of the form C l + C cos l + C 3 + C 4 Nee four bounar conitions to evaluate the constant C..C 4 For the simpl supporte case, the bounar conitions are: u= u =; v= v =; f= f = ets solve one ifferential euation the solution will be vali for all three. Column bucling oubl smmetric section v iv v 4 F v Solution is v C F C cos F C C v v 3 4 v C F F C F cos F v v v v Bounar conitions : v() v() v( ) v( ) C C v() C v() C F C cos F C C v( ) v v 3 4 C F F C F cos F v( ) v v v C C Fv cos Fv C 3 F cos C4 v Fv Fv Fv v The coefficient matri F F v F v v v F n n Fv E I n E I Smallest value of n : EI 58

59 8/5/4 Column bucling oubl smmetric section Similarl, F u u F n n Fu E I n E I EI Smallest value of n : Summar EI EI EI f w G K T r 3 Similarl, F f F n f r G K n T Ff E Iw n f E I w G K Smallest value of n : n f E I w G K T T r r Column bucling oubl smmetric section Thus, for a oubl smmetric cross-section, there are three istinct bucling loas,, an. The corresponing bucling moes are: v = C (/), u =C (/), an f = C 3 (/). These are, fleural bucling about the an aes an torsional bucling about the ais. As ou can see, the three bucling moes are uncouple. You must compute all three bucling loa values. The smallest of three bucling loas will govern the bucling of the column. 59

60 8/5/4 Column bucling bounar conitions Consier the case of fi-fi bounar conitions: v iv F v v Solution is v C F C cos F C C v v 3 4 v C F cos F C F F C 4 v 3 v v v v 3 Bounar conitions : v() v() v( ) v( ) C C v() C F C v() C F C cos F C C v( ) v v 3 4 v cos v v v 3 C F F C F F C v( ) C F C v Fv cos Fv C 3 F cos C4 v Fv Fv Fv The coefficient matri F F cos F v v v Fv Fv Fv v cos F F v n n Fv 4 n E I Smallest value of n : E I E I (.5 ) ( K ) Column Bounar Conitions The critical bucling loas for columns with ifferent bounar conitions can be epresse as: EI ( K ) EI ( K ) EI f ( K ) G K r w T 3 Where, K, K, an K are functions of the bounar conitions: K= for simpl supporte bounar conitions K=.5 for fi-fi bounar conitions K=.7 for fi-simple bounar conitions 6

61 8/5/4 Column bucling eample. Consier a wie flange column W7 84. The bounar conitions are: v=v =u=u =f=f = at =, an v=v =u=u =f=f = at = For fleural bucling about the -ais simpl supporte K =. For fleural bucling about the -ais fie at both ens K =.5 For torsional bucling about the -ais pin-fi at two ens - K =.7 E I E A r EA ( K ) ( K ) K r E I E A r EA r ( K ) ( K ) r K r E I E I A G K G K r w w f K r T T ( K ) r r ( I I ) Column bucling eample. E A E Y AY K Y K r r r r r E r r EA ( / ) ( / ) 79. Y AY K Y K r r r A f EIw A G K T r Y r I I K r ( ) f EIw G K T r Y r ( I I ) Y K r f Y r Y 6

62 8/5/4 Column bucling eample. Critical bucling loa / iel loa ( cr / Y ) Yiel loa Y Cannot be eceee Torsional bucling about -ais governs Fleural bucling about -ais Fleural bucling about -ais governs r (Slenerness Ratio) Fleural bucling about -ais Torsional bucling about -ais - fleural bucling - fleural bucling - torsional bucling Column bucling eample. When is such that /r < 3; torsional bucling will govern r =.69 in. Therefore, /r = 3 =338 in.=8 ft. Tpical column length = 5 ft. Therefore, tpical /r =. 6.8 Therefore elastic torsional bucling will govern. But, the preicte loa is much greater than Y. Therefore, inelastic bucling will govern. Summar Tpicall must calculate all three bucling loa values to etermine which one governs. However, for common steel builings mae ug wie flange sections the minor (-ais) fleural bucling usuall governs. In this problem, the torsional bucling governe because the en conitions for minor ais fleural bucling were fie. This is ver rarel achieve in common builing construction. 6

63 8/5/4 Column Bucling Singl Smmetric Columns Well, what if the column has onl one ais of smmetr. ie the - ais or the -ais or so. C S As shown in this figure, the ais is the ais of smmetr. The shear center S will be locate on this ais. Therefore =. The ifferential euations will simplif to: 3 E I v v ( ) E I u u f E I f ( r G K ) f u ( ) w T Column Bucling Singl Smmetric Columns The first euation for fleural bucling about the -ais (ais of non-smmetr) becomes uncouple. E I v v () iv E I v v iv v F v v where, F v v C F C cos F C C F v v v 3 4 EI ( K ) EI Bounar conitions Bucling mo v C F v Euations () an (3) are still couple in terms of u an f. 3 ( ) E I u u f E I f ( r G K ) f u ( ) w These euations will be satisfie b the solutions of the form u=c (/) an f=c 3 (/) T 63

64 8/5/4 Column Bucling Singl Smmetric Columns f ( ) () E I u u E I f ( r G K ) f u ( ) (3) w T ( ) iv E I u u f iv E I f ( r G K ) f u ( ) w T et, u C ; f C3 Therefore, substituting these in euations an 3 4 E I C C C3 4 3 T 3 E Iw C ( r G K ) C C Column Bucling Singl Smmetric Columns E I C C3 an E Iw ( r G KT) C3 C EI EI w et, an G K f T r C C3 f r C3 C C ( ) r C f 3 ( ) r f 64

65 8/5/4 Column Bucling Singl Smmetric Columns ( )( ) r r f ( f) r ( ) ( ) f f f ( ) ( ) 4 ( ) f f f r ( ) r 4 f ( ) r ( f) ( ) f ( f ) ( ) r 4 f ( ) ( f ) r ( ) f ( ) r Thus, there are two roots for Smaller value will govern 4 f ( ) ( f ) r ( f ) ( ) r Column Bucling Singl Smmetric Columns The critical bucling loa will the lowest of an the two roots shown on the previous slie. If the fleural torsional bucling loa govern, then the bucling moe will be C (/) C 3 (/) This bucling moe will inclue both fleural an torsional eformations hence fleural-torsional bucling moe. 65

66 8/5/4 Column Bucling Asmmetric Section No aes of smmetr: Therefore, shear center S ( o, o ) is such that neither o not o are ero. ( ) ( ) E I v v f () E I u u f () E I f ( r G K ) f u ( ) v ( ) (3) w T For simpl supporte bounar conitions: (u, u, v, v, f, f =), the solutions to the ifferential euations can be assume to be: u = C (/) v = C (/) f = C 3 (/) These solutions will satisf the bounar conitions note above Column Bucling Asmmetric Section Substitute the solutions into the.e. an assume that it satisfie too: 3 E I C C C 3 E I C C C 3 E Iw C3 cos ( r G KT ) C3 cos C cos C cos E I C E I C ( ) 3 cos E Iw r G KT C 66

67 8/5/4 Column Bucling Asmmetric Section C C f r C3 cos where, EIw f T EI EI G K r Either C, C, C 3 = (no bucling), or the eterminant of the coefficient matri = at bucling. Therefore, eterminant of the coefficient matri is: o ro ro o ( )( )( f ) ( ) ( ) Column Bucling Asmmetric Section o ro ro o ( )( )( f ) ( ) ( ) This is the euation for preicting bucling of a column with an asmmetric section. The euation is cubic in. Hence, it can be solve to obtain three roots cr, cr, cr3. The smallest of the three roots will govern the bucling of the column. The critical bucling loa will alwas be smaller than,, an f The bucling moe will alwas inclue all three eformations u, v, an f. Hence, it will be a fleural-torsional bucling moe. For bounar conitions other than simpl-supporte, the corresponing,, an f can be moifie to inclue en conition effects K, K, an K f 67

68 8/5/4 Column Bucling - Inelastic A long topic Effects of geometric imperfection EI v v EI u u v o o v v o eas to bifurcation bucling of perfect oubl-smmetric columns v M M (v v o ) EI v (v v o ) v F v (v v o ) v F v v F v v o v F v v F v ( o ) Solution v c v p v c A (F v ) Bcos(F v ) v p C Dcos 68

69 8/5/4 Effects of Geometric Imperfection Solve for C an D first v p F v v p F v o C Dcos F v C Dcos F v o C F v C F v o cos D F v D C F C F an v v o D F v D F C v o F v Solutionbecomes an D F v A (F v ) Bcos(F v ) v o F v Geometric Imperfection Solve for A an B Bounarconitionsv() v() v() B v() A F v A Solutionbecomes F v v o F v F v v F v o o E E v E o E TotalDeflection v v o E o o E E o o E E A F o A F = amplification factor 69

70 8/5/4 Geometric Imperfection A F E amplificationfactor M (v v o ) M A F ( o ) i.e., M A F (moment ue to initialcrooe) 8 Increases eponentiall imit A F for esign imit / E for esign 6 4 Value use in the coe is.877 This will give A F = 8.3 Have to live with it / E Resiual Stress Effects 7

71 8/5/4 Resiual Stress Effects Histor of column inelastic bucling Euler evelope column elastic bucling euations (burie in the million other things he i). Tae a loo at: An amaing mathematician In the 75s, I coul not fin the eact ear. The elastica problem of column bucling inicates elastic bucling occurs with no increase in loa. /v= 7

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77 8/5/4 Histor of Column Inelastic Bucling Engesser etene the elastic column bucling theor in 889. He assume that inelastic bucling occurs with no increase in loa, an the relation between stress an strain is efine b tangent moulus E t Engesser s tangent moulus theor is eas to appl. It compares reasonabl with eperimental results. T =E T I / (K) Histor of Column Inelastic Bucling In 895, Jas pointe out the problem with Engesser s theor. If /v=, then the n orer moment (v) will prouce incremental strains that will var linearl an have a ero value at the centroi (neutral ais). The linear strain variation will have compressive an tensile values. The tangent moulus for the incremental compressive strain is eual to E t an that for the tensile strain is E. 7

78 8/5/4 Histor of Column Inelastic Bucling In 898, Engesser correcte his original theor b accounting for the ifferent tangent moulus of the tensile increment. This is nown as the reuce moulus or ouble moulus The assumptions are the same as before. That is, there is no increase in loa as bucling occurs. The correcte theor is shown in the following slie Histor of Column Inelastic Bucling The bucling loa R prouces critical stress R = r /A During bucling, a small curvature f is introuce The strain istribution is shown. The loae sie has an The unloae sie has U an U ( ) f U ( ) f E t ( ) f U E( ) f 73

79 8/5/4 Histor of Column Inelastic Bucling f v E t ( ) v U E( ) v But, the assumptionis U A A ( ) E( ) A ES E t S where, S E t ( ) A ( ) ( ) A an S ( ) A ( ) Histor of Column Inelastic Bucling S an S are the statical moments of the areas to the left an right of the neutral ais. Note that the neutral ais oes not coincie with the centroi an more. The location of the neutral ais is calculate ug the euation erive ES - E t S = M v M U ( ) A M v v (EI E t I ) where, I ( ) A ( ) A ( ) an I ( ) A ( ) 74

80 8/5/4 Histor of Column Inelastic Bucling M v v (EI E t I ) v (EI E t I ) v v v EI E t I v F v v where, F v EI E t I E I an E E I I E t I I R E I (K) E is the reuce or ouble moulus R is the reuce moulus bucling loa Histor of Column Inelastic Bucling For 5 ears, engineers were face with the ilemma that the reuce moulus theor is correct, but the eperimental ata was closer to the tangent moulus theor. How to resolve? Shanle eventuall resolve this ilemma in 947. He conucte ver careful eperiments on small aluminum columns. He foun that lateral eflection starte ver near the theoretical tangent moulus loa an the loa capacit increase with increag lateral eflections. The column aial loa capacit never reache the calculate reuce or ouble moulus loa. Shanle evelope a column moel to eplain the observe phenomenon 75

81 8/5/4 Histor of Column Inelastic Bucling Histor of Column Inelastic Bucling 76

82 8/5/4 Histor of Column Inelastic Bucling Histor of Column Inelastic Bucling 77

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84 8/5/4 Column Inelastic Bucling Three ifferent theories Tangent moulus Reuce moulus Shanle moel /v= Tangent moulus theor assumes erfectl straight column Ens are pinne Small eformations No strain reversal uring bucling Elastic bucling analsis T Slope is ero at bucling = with increag v v Tangent moulus theor Assumes that the column bucles at the tangent moulus loa such that there is an increase in (aial force) an M (moment). The aial strain increases everwhere an there is no strain reversal. T Strain an stress state just before bucling T T = T /A M - v = Strain an stress state just after bucling v v M T T T T =E T T Curvature = f = slope of strain iagram T f T h T f h where istance from centroi T f h E T 79

85 8/5/4 Tangent moulus theor Deriving the euation of euilibrium M A A T T T f( h / ) E T M ( T f( h / )E T ) A A M T A E T f A fh / )E T M E T f I M E T I v A A ( ) A A The euation M - T v= becomes -E T I v - T v= Solution is T = E T I / Eample - Aluminum columns Consier an aluminum column with Ramberg-Osgoo stressstrain E si curve E.. E. n. n n n. ne n n. E. ne n.. E E. ne E n T si n 8.55 E T E T.E+ ifferences euation.98e E E E E E E E E E E E E E E E E E E E E E E E E

86 8/5/4 Tangent Moulus Bucling 6 Ramberg-Osgoo Stress-Strain Stress-tangent moulus relationship Strain (in./in.) Stress (si) ET ifferences ET euation Tangent Moulus Bucling (K/r) cr Column Inelastic Bucling Curve K/r T E T I T A E T I T E T A K/ r K/ r ( ) cr E T T ( ) 8

87 8/5/4 Resiual Stress Effects b Consier a rectangular section with a simple resiual stress istribution Assume that the steel material has elastic-plastic stress-strain curve. rc rc Assume simpl supporte en conitions.5 /b rt.5 Assume triangular istribution for resiual stresses.5 E One major constrain on resiual stresses is that the must be such that r A Resiual Stress Effects.5 b b /.5 b / b.5 b.5 b b b b 8 b 8 Resiual stresses are prouce b uneven cooling but no loa is present 8

88 8/5/4 Resiual Stress Effects Response will be such that - elastic behavior when b.5 EI an EI Yielingoccurswhen.5 i.e.,.5 Y ab ab Inelasticbuclingwilloccurafter.5 Y Y b ab Y ( a) Y Y /b Y Resiual Stress Effects Total aial force corresponingto the ielesec tion Y ( b ab) Y Y ( a) ab Y ( a)b Y ( a)ab Y b ab Y Y ab a b Y Y b( a ) Y ( a ) If inelasticbuclingwere to occurat thisloa cr Y ( a ) a cr Y 83

89 8/5/4 If inelasticbuclingoccursabout ais cr T E (ab) 3 ab ab T EI a T T T Y Y cr Y T Y T Y T Y l T l T Y T Y Y cr T et, Y l E Y r K If inelasticbuclingoccursabout ais cr T E (ab) 3 T EI T ( a) 3 3 cr Y T T Y T Y T Y Y T Y l T Y T l Y 3 T cr T et, Y l E Y ab r K ab Y 84

90 8/5/4 Resiual Stress Effects / Y l l Normalie column capacit Column Inelastic Bucling amba 85

91 8/5/4 Tangent moulus bucling - Numerical A fib fib Centroial ais Discretie the cross-section into fibers Thin about the iscretiation. Do ou nee the flange To be iscretie along the length an with? For each fiber, save the area of fiber (A fib ), the istances from the centroi fib an fib, I -fib an I -fib the fiber number in the matri. 3 Discretie resiual stress istribution 4 Calculate resiual stress ( r-fib ) each fiber 5 Chec that sum( r-fib A fib )for Section = ero Tangent Moulus Bucling - Numerical 6 Calculate effective resiual strain ( r ) for each fiber r = r /E 4 Calculate the critical (K) X an (K) Y for the (K) X-cr = srt [(EI) T /] (K) -cr = srt [(EI) T /] 7 Assume centroial strain 3 Calculate the tangent (EI) TX an (EI) TY for the (EI) TX = sum(e T-fib { fib A fib +I -fib }) (EI) T = sum(e T-fib { fib A fib + I -fib }) Calculate average stress = = /A 8 Calculate total strain for each fiber tot =+ r Calculate Aial Force = Sum ( fib A fib ) 9 Assume a material stress-strain curve for each fiber Calculate stress in each fiber fib 86