Theory of Elasticity Ct 5141 Direct Methods

Transcription

1 Delf Universi of Technolog Facul of Civil Engineering and Geosciences Theor of Elasici C 54 Direc Mehods Prof.dr.ir. J. Blaauwendraad June 003 Las updae: April 004 C 54

2 Acknowledgemen In wriing and updaing hese lecure noes a number of co-workers have pleasanl suppored me. The ladies ir. W. Yani Sujadi and Ir. Madelon A. Burgmeijer have devoedl aken care of he Duch e and formulas. Their suggesions o improve he e have been welcomed. Man hanks also o m fellow eacher Dr.ir. Pierre C.J. Hoogenboom for his posiive conribuion in he conen of he lecures and wriing nice eams. Finall I am indebed o Dr.ir. Co Siers. He ranslaed he eising Duch language lecure noes ino hese English ones and adaped he illusraions where necessar. J. Blaauwendraad April 003

3 Inroducion o he course When he civil engineering suden chooses for he course Theor of Elasici, (s)he is alread eensivel familiarised wih he mahemaical descripion of srucural behaviour b means of differenial equaions. Wih his in mind reference can be made o he courses Elasosaics and Elasic Plaes and Slabs. This course coninues along he same line and eends on i. During he compilaion of hese lecure noes, he course maerial has been resrucured. The reason for his is ha hree main objecives are aimed for:. Hisoricall a number of essenial subjecs in srucural mechanics eis, which have o be deal wih. The soluion of hese problems is incorporaed in he basic knowledge of a srucural engineer enering he building pracice. This par of he course is direced a resuls. The course maerial conains a number of known soluions for plae problems and orsional problems for beams wih solid or hollow cross-secions.. The engineer also should be capable of finding soluions for enirel new problems. For his aspec he course should give direcions along which line a soluion can be obained. I appears ha wo sraegies can be followed, in order o arrive a a consisen analical modelling. These sraegies will be discussed. In his par of he course he emphasis lies on he developmen of modelling skills for new problems. I also will be demonsraed ha he classic problems (menioned under objecive ) can be fied in as applicaions or eamples. The eercises and eamples are mainl focussed on srucural ssems ha can be regarded as one-dimensional. 3. The grea meri of he Theor of Elasici is ha via an analical approach eac soluions could be obained for coninuous problems of mechanics, long before numerical mehods were developed and became available on a large scale as he are oda. However, i should be noiced ha he realised eac soluions were usuall limied o a cerain class of problems, someimes wih a relaivel academic geomer or for a limied number of no o comple boundar condiions. Afer he inroducion of he compuer, he number of possibiliies is increased enormousl, which made i possible o compue soluions for coninua wih an arbirar geomer. In his respec especiall he Finie Elemen Mehod (FEM) is imporan. Wih his mehod approimae soluions are generaed and he basis of he mehod lies in he applicaion of energ principles. Since energ principles also pla a role in he eac formulaion of coninua, he possibili eiss o make a smooh ransiion from he heor of elasici o he numerical mehods. This par of he course can be considered as an inroducion o he course abou he Finie Elemen Mehod In view of hese hree objecives he following se-up of he course is selaced. The lecure noes consiss ou of wo pars. The firs par deals wih he firs wo objecives and direc mehods will be used. In he oher par aenion is paid o he hird objecive and energ principles and variaional mehods will be discussed. Chaper of he firs par sars wih a recapiulaion of he force and displacemen mehods for discree bar srucures. The formulaion can be presened ver compac wih he mari

4 noaion. Deliberael an approach is followed ha makes i possible o pinpoin clearl which sraegies are used. In chaper a sar is made wih he analsis of coninuous srucures. The differenial equaion and boundar condiions are derived for a simple one-dimensional elemen, i.e. he beam. For a large par his is revision of previousl obained knowledge. On basis of his well-known coninuous srucural elemen, he analog will be demonsraed wih he discree approach. Wih combinaions of he differen elasic cases (eension, bending, shear, orsion, elasic suppor) several srucural ssems can be modelled, such as can be found in high-rise buildings. The derived equaions are alwas ordinar parial differenial equaions of he second order, fourh order or higher. Afer ha wo-dimensional problems will be addressed. Again wih he same sraeg he differenial equaions are derived for in is plane loaded plaes (chaper 3) and ransversel loaded plaes (chaper 4). In hese chapers for a number of classic problems he soluion of he parial differenial equaions is worked ou or provided direcl (objecive ). The following wo chapers deal wih he heor of hree-dimensional coninua. Chaper 5 focuses on he basic equaions. In chaper 6 a specific classic problem is formulaed and worked ou, namel orsion in beams wih a solid or hollow cross-secion (objecive ). This concludes he par Direc Mehods. The par Energ principles and variaional mehods will be offered as a separae se of lecure noes. In hese noes he following subjecs are addressed: work, energ principles, variaional mehods and approimae soluions. The validi of he derivaions eends o general hree-dimensional coninua. However, he derivaions iself will be worked ou for one-dimensional cases. This is also he case for eamples and applicaions. 3

5 Table of conens Acknowledgemen... Inroducion o he course... Table of conens... 4 Recapiulaion for discree bar srucures Basic equaions for saicall deerminae srucures Sraegies for he analsis of saicall deerminae srucures....3 Basic equaions for saicall indeerminae srucures Sraegies for he analsis of saicall indeerminae srucures Force mehod Displacemen mehod....5 Summar for discree bar srucures... Coninuous beam Saicall deerminae beam subjeced o eension Force mehod Displacemen mehod Saicall indeerminae beam subjeced o eension Force mehod Displacemen mehod Elaboraion of an eample Saicall deerminae beam subjeced o bending Force mehod Displacemen mehod Elaboraion of an eample The sign difference beween B and B Saicall indeerminae beam subjeced o bending Force mehod Displacemen mehod Elaboraion of an eample Beams subjeced o orsion or shear Summar for load cases of coninuous slender beams Walls coupled b springs subjeced o a emperaure load Force mehod Displacemen mehod Epilogue Walls coupled b springs subjeced o a wind load Plaes loaded in-plane Basic equaions Applicaion of he force mehod Soluions in he form of polnomials Soluion for a deep beam Aismmer for plaes subjeced o eension

6 3.5. Thick-walled ube Curved beam subjeced o a consan momen Descripion in polar coordinaes of plaes subjeced o eension Poin load on a half-plane Brazilian spliing es Hole in large plae under uniaial sress... 4 Plaes in bending Recangular plaes The hree basic equaions in an orhogonal coordinae ssem Force mehod Displacemen mehod Aismmer for plaes subjeced o bending Basic equaions for aismmer Differenial equaion Aismmeric applicaions Simpl suppored circular plae wih boundar momen Resrained circular plae wih uniforml disribued load Simpl suppored circular plae wih uniforml disribued load Circular plae wih poin load in he cenre General soluion procedure for circular plaes Theor of elasici in hree dimensions Basic equaions Soluion procedures and boundar condiions Alernaive formulaion of he consiuive equaions Separae laws of Hooke for he change of volume and shape Hooke s law for oal deformaions and sresses The displacemen mehod in he descripion of Lamé Torsion of bars Problem definiion Basic equaions and boundar condiions Displacemen mehod Force mehod Eac soluion for an ellipic cross-secion Membrane analog Numerical approach Cross-secion wih holes Thin-walled ubes wih one cell Thin-walled ubes wih muliple cells Cross-secion buil up ou of differen maerials Torsion wih prevened warping

7 Recapiulaion for discree bar srucures. Basic equaions for saicall deerminae srucures In he lecures ha preceded his course, a large amoun of aenion was paid o he calculaion of bar srucures. A number of basic ideas from ha lecure maerial will be summarised briefl. In his summar wo sraegies are highlighed, which are especiall suiable for he analsis of problems, namel he displacemen mehod and he force mehod. Moreover, he basic ideas will be addressed in such a manner ha i becomes clear ha he same sraeg can be used for he analsis of coninuous bodies. For his purpose a simple fla russ is considered. Fig.. shows a saicall deerminae russ. The srucure consiss ou of N, e F u N, e F u 4 4 N, e 5 5 N, e 3a 3 3 N, e 4a F u F3 u3 F u 6 6 N, e 4a F u 3 3 Fig..: Saicall deerminae russ wih relevan quaniies. si members and hree movable nodes, which are numbered from up o 3. Quaniies associaed wih he nodes ge a subscrip equal o he node number. The bars are numbered from o 6. Quaniies associaed wih he bars receive a superscrip wih he elemen number. The heigh and lengh of he russ are 3 a and 8a, respecivel. Each nodal poin has wo degrees of freedom, a displacemen u in -direcion and a displacemen u in -direcion. In hese respecive direcions eernal forces F and can F be applied. The degrees of freedom of all nodes combined, form he vecor u and all he forces form he vecor f. Sresses are generaed inside he srucure, ogeher wih he corresponding srains. In his case he sress resulan N of each bar is used, ogeher wih he associaed change of lengh (eension) e of he bar. The form he vecor of generalised sresses or sress resulans N and he generalised deformaions or shorl deformaions e, respecivel. The signconvenion for he eernal quaniies u, u, F and F differs from he sign-convenion for he inernal quaniies N and e. For he eernal quaniies a vecor sign-convenion is used. When he are poining in posiive - or -direcion he are defined posiive. For he 6

8 inernal quani N a sress sign-convenion is applicable. A posiive sign is chosen for ensile forces. Likewise e is assumed posiive if i concerns an elongaion. Essenial in his course is he wa in which he several quaniies are defined. The differen degrees of freedom are idenified and deermined. Then i is also known which eernal loads can be applied. Separael i is ascerained which inernal (generalised) sresses will appear, afer ha he are idenified ogeher wih he corresponding (generalised) deformaions. The eernal vecors u and f mus provide eacl he performed eernal work, and he inernal vecors e and N deermine he inernal deformaion work. In he coming chapers his approach will be applied o coninuous srucures oo. In previous courses, i alread has been discussed ha hree basic relaions deermine he behaviour of srucures. This riple is: he kinemaic equaions he consiuive equaions he equilibrium equaions The kinemaic equaions provide he relaion beween he displacemens u and he deformaions e. The consiuive equaions relae he deformaions e o he sress resulans N. And he equilibrium equaions prescribe how he sress resulans N are conneced wih eernal work inernal work u u, u, u 3 u, u, u 3 e e e 6 N f F, F, F N N F, F, F kinemaic equaions consiuive equaions equilibrium equaions Fig..: Diagram displaing he relaions beween he quaniies plaing a role in he analsis of a russ. he eernal load f. The scheme in Fig.. provides an overview of all hese ineracing relaions. Now, his riple of equaions will be worked ou in deail for he eample of he russ as shown in Fig... Kinemaic relaions Considering he sign-convenion for he displacemens and deformaions, for each of he bars he following relaions can be derived (see Fig..3): 7

9 e u e u 4 e 5 e 3a 3 e u 6 u 3 e 4a u u 3 4a Fig..3: Relaions eis beween deformaions e and displacemens u. e u 4 3 = = + u 4 = = = + 3 e u u e =+ e u u e u u u u e u u In mari noaion his becomes: u e u e e u = e u e 0 0 u e u3 This resul can be rewrien briefl b he inroducion of he kinemaic mari B : 3 e = Bu (kinemaic equaions) (.) Consiuive relaions For each bar a siffness relaion eiss beween he normal force N and he deformaion e : N = EA e l The fleibili formulaion provides he inverse form: 8

10 e= l N EA B inroducion of he abbreviaions: EA l D = ; C l = E A he formulaion of he consiuive equaions becomes: N D e N D e N D e 4 = 4 4 N D e N D e N D e or briefl: ( consiuive relaions in siffness formulaion) N = De (.)a ( siffness) and: e C N e C N 3 3 e C 3 N 4 = 4 4 e C N 5 5 e C 5 N e C N or briefl: ( consiuive relaions in fleibili formulaion) e = C N (.)b ( compliance ) I will become clear ha he siffness formulaion of he consiuive equaions is used in he displacemen mehod and ha he fleibili formulaion is used in he force mehod. Equilibrium equaions The ne hree pairs of equilibrium equaions are obained from he equilibrium of all nodes in he direcion of he respecive degrees of freedom (see Fig..4): 9

11 N F F 4 N 5 N 3a N 4 N 5 N 4a 3 N F F 3 N 6 6 F F 3 4a N Fig..4: For each node equilibrium eis beween he normal forces N and he eernal loads f. + N N N N N = F N N = F + N + N N = F + + = F + N + N = F + = F 3 3 In mari form his reads: N F N N F 3 4 = N F N F N F 3 F Comparison of his mari wih he previousl found kinemaic mari eacl he ransposed of B. Therefore i can be wrien: B shows ha i is = T B N f (equilibrium equaions) (.3) where he superscrip T is he inernaionall acceped smbol o indicae he ransposed of a mari. 0

12 . Sraegies for he analsis of saicall deerminae srucures In he previous secion he following se of basic equaions have been found: e = Bu N = De or e = C N (.4) (.5) T B N = f (.6) Hisorical The firs sep is he calculaion of he sress quaniies N. In he case of a saicall deerminae russ he vecors N and f have he same number of componens, which means T ha he mari B in he equilibrium equaion (.6) is square. Therefore, he sress quaniies T can be deermined direcl b inversion of B : T N = B f ( Cremona ) (.7) This is he mahemaical formulaion of he classical graphical mehod involving he drawing of Cremona diagrams. The second sep is he calculaion of he deformaion quaniies e. When he normal forces N are known, he changes in member lengh e direcl follow from he fleibili formulaion of he consiuive equaions e = C N. Then in he hird sep, he displacemens can be obained from he kinemaic relaions given in (.4). In a saicall deerminae russ he vecors e and u again have he same number of componens. So, he mari B is square and can be invered. The required displacemens subsequenl can be obained from: u= B e ( Willio ) (.8) This is he mahemaical descripion of he classical graphical mehod, in which he displacemens are deermined from he changes in bar lengh b consrucion of a Willio diagram. The described compuaional mehod in hese lecure noes conains he same consecuive phases, which also sudens hisoricall have o follow during he learning process of applied mechanics. Firs, he force ransmission and he equilibrium are horoughl discussed. Then he concep of deformaions is inroduced and hirdl he displacemens are calculaed. The riple of equaions as lised below is evaluaed in he order indicaed b he arrow: Hisorical equilibrium equaions consiuive equaions (fleibili formulaion) kinemaic equaions

13 Numerical Afer he inroducion of he firs compuers, algorihms have been developed ha lierall followed above procedure, and basicall onl replaced he graphical elemen b a numerical echnique. However, simulaneousl he displacemen mehod or siffness mehod came ino use, which appeared o be more suiable for compuer analsis. In his mehod he riple of equaions is solved in he reversed order: Numerical equilibrium equaions consiuive equaions (siffness formulaion) kinemaic equaions When he kinemaic equaions (.4) are subsiued in he consiuive equaions (.5), followed b a subsiuion ino he equilibrium equaions (.6), he resul becomes: T B DB u= f (equaions) (.9) Each of hese marices is square. Since D is a smmerical mari and he pre-muliplicaion T mari B is he ransposed of he pos-muliplicaion mari B, he final produc will be a square and smmerical mari. This mari is indicaed b K and is called he siffness mari, i.e.: T K = B DB (siffness mari) (.0) The ssem of equaions can now be summarised as follows: Ku= f (.) From his ssem he displacemens can be solved. In he sandard displacemen mehod he mari K is assembled from he individual siffness marices of he several members. In his course inenionall for anoher derivaion of K is chosen, as an inroducion on he ne chapers dealing wih coninuous srucures. In he displacemen mehod firs he displacemens are calculaed. Afer ha, he deformaions can be obained from he kinemaic relaions and finall from he consiuive equaions he sress quaniies can be deermined. This means ha he riple of equaions is considered again in he same order. The formulaion of he equaions given b (.9) will be considered again during he T discussion of coninuous srucures. I is quie obvious, ha he produc of B, D and B has o deliver a smmerical mari. For linear-elasic srucures his follows direcl from T Mawell s law of reciprocal deflecions. Conversel, i can be concluded ha B alwas has o be he ransposed of B, irrespecive of he srucure considered.

14 .3 Basic equaions for saicall indeerminae srucures The formulaion of he basic equaions as described in secion. will be repeaed for a saicall indeerminae srucure. The same russ is considered wih hree free nodes, however wih an era sevenh member as shown in Fig..5. This makes he srucure saicall indeerminae o he firs degree. N, e F u 7 7 N, e N, e F u 4 4 N, e 5 5 N, e 3a 3 3 N, e 4a F u F3 u3 F u 6 6 N, e 4a F u 3 3 Fig..5: Saicall indeerminae russ wih relevan quaniies. There sill are wo imes si eernal quaniies (degrees of freedom u and forces f ), bu inernall he number is larger. Seven normal forces in N and seven corresponding elongaions in e are presen. Again he riple of equaions will be formulaed. Afer he deailed analsis of secion. his can be done briefl. Kinemaic equaions u e u e 3 e u e = u e u e u e Again his can be wrien as: e = Bu (.) Now he mari B has seven rows and si columns and herefore is no square anmore. The firs si rows are idenical o he mari B of secion.. The sevenh row is an eension due o he era member. 3

15 Consiuive equaions Also in his case, he siffness and fleibili formulaions of he consiuive equaions read: N = De ; e = C N (.3) Now N and e boh conain seven componens and D and C are square marices wih seven rows and seven columns. Equilibrium equaions The si nodal equilibrium equaions are now epressed in seven sress quaniies: N F N F 5 5 N N F = 3 4 F 5 N 4 5 F 5 N F 5 N 3 7 Also in his case, he mari is he ransposed of mari wrien: B. Therefore, i briefl can be T B N = f (.4) Again he firs si columns are equal o eension. T B from secion., he sevenh columns is an.4 Sraegies for he analsis of saicall indeerminae srucures On basis of he basic equaions i formall will be described how he force mehod and he displacemen mehod will work ou..4. Force mehod s sep: Equilibrium Normall, he firs sep would have been he soluion of he equilibrium equaions. However, his is no possible because he number of unknowns eceeds he number of equaions b one. In such a case, he old and well-ried mehod of making he srucure saicall deerminae can be used, where one of he members is cu and a redundan is inroduced on he cuing face. In he eample of Fig..6, bar 7 is cu and he redundan φ is inroduced. 4

16 φ F φ F 3a F F 3 4a F F 3 4a Fig..6: Inroducion of redundan φ. Doing so a saicall deerminae main ssem is creaed. On op of he si eernal componens of f, also he redundan φ has o be considered as an eernal load (sill unknown) on he main ssem. This means ha his main ssem is subjeced o wo load vecors: 4 F 5 F F 0 ; φ F 0 F 3 0 F 3 0 The second vecor conains he componens of he redundan φ acing in he hree nodes. In secion. i has been shown, how b (.7) he normal forces can be calculaed of he saicall deerminae main ssem. For he firs load vecor his provides: F 4 N F N N F 3 4 = N F N F N F 3 3 (.5) T where he mari is he inverse of he mari B from secion.. When in (.5) he vecor wih eernal forces is replaced b he second vecor wih he componens of he redundan φ, he normal forces in he main ssem resuling from φ can be found. The mari-vecor produc hen resuls in: 5

17 4 N 5 N 3 4 N 5 φ 4 = 3 N 5 5 N 0 6 N 0 (.6) The sevenh normal force is independen from he si eernal forces and is direcl equal o he redundan, i.e.: 7 N = φ (.7) The sum of hese hree inermediae resuls delivers he normal forces for he eernal load ogeher wih he redundan. This sum can be wrien as: B T 4 4 N F F N N F 4 3 N = F N F N F 3 7 N φ f φ (.8) P f P The columns of his mari associaed wih he load vecor f form he mari P f and he column working on he redundan φ is called P. For cases wih more han one redundan, P will conain more columns and φ more han one componen. So, mari relaion (.8) can briefl be wrien as: N = Pf f + P φ (equilibrium ssem) (.9) These sress resulans form an equilibrium ssem and herefore will saisf he equilibrium equaions (.4) from secion.3. nd sep: Consiuion The force mehod uilises he consiuive relaions in he fleibili formulaion: e = C N (consiuion) (.0) 6

18 3 rd sep: Compaibili The soluion process coninues as follows. In Fig..6 i was alread shown ha he bar ends on he cuing face can move independenl wih respec o each oher. The overlap (or more generall he gap) is he resul of boh he eernal forces and he redundan. For he deerminaion of he sill unknown redundan φ he so-called compaibili condiion is required, which is given b: =0 (.) In words: he gap caused b eernal forces has o be eliminaed b he gap resuling from he redundan. For he russ of he eample, i will be checked how he gap is relaed o he deformaions of he members. This has o be a purel kinemaic relaion, which is governed b he geomer of he srucure. For he calculaion of i has o be clear how much he disance is reduced beween he nodes and 4. This reducion will be called,4. Is magniude is: 4 3 4, = 5 u + 5 u The value of can direcl be epressed in e : u u = e The magniude of u = u e he displacemen 4 u u is no direcl known. However, i is known ha: of which direcl can be epressed in he deformaions: u 5 4 = e e Wih hese resuls, he gap,4 can be wrien as: 4 4 5, = e + e e e The gap a he posiion of he redundan is equal o his resul increased b he change of lengh of bar 7: Thus: = + 7 4, e = e + e e e + e 5 In mari noaion his becomes:

19 4 4 { } 3 = e e 3 e 4 e 5 e 6 e 7 e Closer inspecion reveals ha he row-mari is jus equal o he ransposed of he mari P. Therefore he gap equals: T = Pe (.) The gap is here wrien as a vecor, because for saicall indeerminae srucures o higher T degrees more han one gap is presen. In ha case, he mari P will conain more rows. The compaibili condiion can now briefl be formulaed as he requiremen ha: T Pe = 0 (compaibili) (.3) In he eample of he russ his is: 4 e + e + e e + e = Now, formall he recipe for he compaibili condiion has been derived on bases of kinemaic consideraions, which has been done in such a manner ha a phsical inerpreaion can be given. From a mahemaical poin of view, condiion (.3) can also direcl be derived from he kinemaic relaions (.): e = Bu which conain seven equaions wih si unknown displacemens. This means ha one dependen relaion beween he seven deformaions can be formulaed. In order o find his relaion, he displacemens have o be eliminaed. This can be done b linear combinaion of he rows of B in such a wa ha a row of zeros is creaed. The weigh facors wih which he T rows have o be muliplied jus form he row-mari P. This formal recipe: he eliminaion of he displacemens from he kinemaic relaions in order o find he compaibili condiion, shall be applied again o coninuous srucures. 8

20 Ssem of equaions Wih he hree inermediae resuls for equilibrium (.9), consiuion (.0) and compaibili (.3) a ssem of equaions is creaed for he calculaion of he redundan(s) φ. Subsiuion of (.9) ino (.0) leads o: ( f ) e = C P f + P φ (.4) Combinaion of his resul wih he compaibili condiion (.3) ields: T T PCPf PCP φ (.5) f + =0 The firs erm in his relaion is he gap resuling from he eernal load f. This is a known erm and will be called f. Therefore, he redundan(s) φ can be obained from he equaions: T PCP (equaions) (.6) φ = f This formulaion will be used again for coninuous srucures. The produc of he hree T T marices P, C and P delivers a smmerical mari because C is smmerical and P is he ransposed of P. The produc mari has o be smmerical since i has o saisf Mawell s law, which proves T ha P is alwas he ransposed of P. When for he produc of he hree marices he oal fleibili mari F is inroduced: F = T PCP The ssem of equaions o be solved becomes: F φ = f Remark I alread was saed ha he sresses given b (.9) saisf he equilibrium equaions (.4). Subsiuion of hese sresses ino he equilibrium equaions provides he condiion: T T B P f B Pφ f f + = Since boh f and φ have o be differen from zero, from his relaion i can be concluded: T B Pf T = I B P = 0 (.7) 9

21 Here I is he uni mari and 0 is he zero mari. In his manner he compaibili condiion can be derived as well. When in he kinemaic T relaion (.) boh he lef-hand and righ-hand sides are muliplied b P, i follows: T P e = T P Bu T T From (.7) i can be seen ha B P is a zero mari. The mari P B is is ransposed and herefore a zero mari oo. This means ha he righ-hand side of he above equaion is equal o zero, reducing i o he alread obained compaibili condiion (.3). Remark In he case of a saicall deerminae srucure, he mari no here (in ha case i has zero columns). P f equals B and he mari P is Calculaion of sress quaniies When he redundans are obained from he equaions (.6), he sress quaniies (he normal forces) can be calculaed from (.9): N = Pf f + P φ Calculaion of displacemens T The inermediae resul B Pf = I of remark, can be used o calculae he displacemens. From he calculaed sress quaniies N, firs he deformaions are obained from (.0): e = C N Afer ha he displacemens u can be calculaed. For ha purpose boh righ-hand and lefhand sides of he kinemaic relaions (.) are muliplied b P, i.e.: e = Bu P e = P Bu T f T f T f T Because B P f is he uni mari, is ransposed equaion can be simplified o: T Pf B is he uni mari oo. Therefore above T Pe f = u (calculaion of he displacemens) (.8) Which demonsraes how he displacemens can be obained from he deformaions. Summar of he force mehod Overseeing he sraeg of he force mehod, i is eviden ha he riple of basic equaions is evaluaed wo imes in he order as shown below. Firs, he ssem of equaions is buil up, he redundans of which are solved. Afer ha, successivel he sresses, deformaions and displacemens are deermined. Force mehod equilibrium equaions consiuive equaions (fleibili formulaion) kinemaic equaions 0

22 .4. Displacemen mehod The displacemen mehod for saicall indeerminae srucures is eacl he same as he one for saicall deerminae srucures. The basic equaions (.), (.3) and (.4) - being a bi differen in his case - are evaluaed wo imes in he order given below. Displacemen mehod equilibrium equaions consiuive equaions (siffness formulaion) kinemaic equaions During he firs ccle again a ssem of equaions is derived: T B DBu= f (.9) T In his case he mari B is no square and has more rows han columns. Naurall, B conains more columns han rows. The mari muliplicaion resuls in a square smmerical T siffness mari K for he srucure wih he same number of rows as B and he same number of columns as B, as shown in he scheme below: T u = f B D B 6 In he second ccle, successivel he deformaions are calculaed from he kinemaic relaions and he sresses are obained from he found deformaions using he consiuive equaions. 6.5 Summar for discree bar srucures In his chaper for saicall deerminae and saicall indeerminae srucures he basic equaions have been derived and wo sraegies have been discussed. For saicall deerminae srucures he words hisorical and numerical indicaed hese sraegies and for he saicall indeerminae srucures he erms force mehod and displacemen mehod were used. The numerical mehod for saicall deerminae srucures is compleel idenical o he displacemen mehod for saicall indeerminae srucures. The word numerical is a bi misleading. I does no mean ha he oher mehods are no suiable o be implemened on a compuer. I onl indicaes ha he displacemen mehod is he mos appropriae one. The hisorical mehod for saicall deerminae srucures fis ino he scheme of he displacemen mehod. However, i is a special version of i, because going hrough wo ccles is no necessar. Wihou compaibili condiions direcl all required quaniies can be deermined.

23 s nd N = Pf f + Pφ equilibrium equaions T B N = f EQUILIBRIUM T B DB u = f e = CN consiuive equaions N = De COMPATIBILITY Fφ = f T P e = 0 T u= P e f kinemaic equaions nd e = Bu s FORCE METHOD DISPLACEMENT METHOD Fig..7: Soluion schemes for force and displacemen mehods. In view of all he consideraions made, i is possible o creae one compac overview for all srucures. In he cenral column of his scheme as shown in Fig..7, he hree basic equaions are lised. In he righ column i has been summarised in wha form hese basic equaions are used for he displacemen mehod. The wo ccles are numbered b s and nd. In he lef column he same has been done for he force mehod. However, differen formulaions of he kinemaic equaions are used in he wo ccles. Clarificaion. In he scheme of he force mehod, for saicall deerminae srucures onl one ccle is T T required. In ha case he mari P f is equal o B and P f ransforms ino B.. In he scheme of he force mehod wo kinemaic relaions are lised. For saicall indeerminae srucures he lef relaion is used in he firs ccle and he righ relaion in he second ccle (which is he firs ccle for saicall deerminae srucures). Two main aspecs The sraeg of he displacemen mehod does no require an clarificaions. In he sraeg of he force mehod aenion is focussed on wo main aspecs:. The firs one is he consrucion of a sress field N ha saisfies he equilibrium equaions and in which he (o be deermined) redundans are incorporaed.. The second one is he derivaion of he compaibili equaions. These are epressions in he deformaions e. The are found b eliminaion of he displacemens from he kinemaic equaions. Remark The saicall indeerminae russ of he eample was inernall saicall indeerminae. For bar srucures he calculaion procedure remains he same for eernall saicall deermined srucures.

24 Coninuous beam The sraegies considered in chaper will be applied for he soluion of coninuous problems. In his chaper hese sill are beam srucures. Secion. will focus on saicall deerminae beam srucures and in secion. saicall indeerminae srucures will be highlighed. In hese secions an aiall loaded bar is considered, in which onl a normal force is generaed. This phenomenon also can be called a bar loaded in eension. In he secions.3 and.4 he discussion will be repeaed for a beam problem wih bending. The ne secion.5 will briefl focus on problems wih shear deformaion and orsion.. Saicall deerminae beam subjeced o eension Fig.. shows he considered srucure. In his srucure, he onl degree of freedom is he displacemen u ( ) in he direcion of he bar ais. The displacemen is defined posiive if i akes place in posiive -direcion. An eernal disribued load f ( ) corresponds wih his degree of freedom. For his load, he same sign convenion applies. f ( ) u( ) F l N ε N N f d d dn N + d d Fig..: Bar subjeced o eension wih relevan quaniies. Ne o he wo eernal quaniies, wo inernal ones are presen as well. The are he (generalised) sress being he normal force N( ) and a specific srain ε ( ), which is caused b ha normal force. Wih he choice of hese wo inernal quaniies he deformaion work is uniquel deermined. In he scheme of Fig.. i is depiced, which quaniies eis and wha relaions can be esablished. The hree basic equaions now are: du ε = ( kinemaic equaion) d N = EA ε or ε = N ( consiuive equaion) EA dn + f = 0 ( equilibrium equaion) d (.) Wih inroducion of he wo operaors B and B given b: d d B = ; B = (.) d d 3

25 eernal work inernal work u ( ) ε( ) N ( ) f( ) kinemaic equaion consiuive equaion equilibrium equaion Fig..: Diagram displaing he relaions beween he quaniies plaing a role in he analsis of a bar subjeced o eension. and he siffness D and fleibili C : D = EA( ) ; C = EA ( ) (.3) he basic equaions can be reformulaed as: ε = B u ( kinemaic equaion) N = D ε or ε= C N ( consiuive equaion) B N=f ( equilibrium equaion) (.4) Comparison wih he basic equaions (.), (.) and (.3) from secion. shows a large analog. For he soluion of a concree problem, he force mehod as well as he displacemen mehod can be applied. Boh mehods will be discussed in his chaper. Remark Tha in his case a separae operaor B = B has been inroduced has a special reason. When above relaions are discreised b he Finie Elemen Mehod or Finie Difference Mehod hese operaors are replaced b marices; hen operaor B is replaced b mari B and T operaor B is replaced b mari B. The sign difference beween he operaors also can be found in heir mari counerpars as shown in secion Force mehod The saring poin is he equilibrium equaion. This single equaion conains one unknown sress quani N, which confirms ha he problem is saicall deermined. So, b inegraion he normal force N( ) can direcl be deermined from he eernal load f ( ). In chaper his boiled down o a mari-inversion problem. Also inegraion (in a generalised sense) can be regarded as an inversion of differeniaion. Then he consiuive equaion in is fleibili formulaion can be used o calculae he srains ε ( ). Afer ha, inegraion of he kinemaic equaion in combinaion wih he boundar condiion direcl delivers he displacemen field u ( ). 4

26 Remark The mehod of analsis is compleel analogous o he one for saicall deerminae russes. Remark The seleced srucure in he eample of Fig.. is boh inernall and eernall saicall deerminae. From he equilibrium equaion i onl can be esablished ha he problem is inernall saicall deerminae. For a conclusion abou he eernal deerminac of he problem he boundar condiions have o be inspeced... Displacemen mehod In he displacemen mehod he kinemaic equaion and he consiuive equaion (in siffness formulaion) are subsiued ino he equilibrium equaion. Doing so in (.), he following second order differenial equaion is obained: d d EA u f d d = (.5) In he case of a prismaic bar, he eensional siffness equaion reduces o: EA is consan and he differenial du EA = f (.6) d For he subsiuion process he operaor equaions (.) can be used oo. Then he differenial equaion appears in he form: B DB u = f (.7) Now he analog wih equaion (.9) of he comparable russ problem is quie clear. Elaboraion of an eample The srucure of Fig.. is considered wih he values of EA and f assumed consan. In he force mehod successivel he hree basic equaions are evaluaed ogeher wih he wo boundar condiions given b: = 0 u = 0 ; = l N =0 Inegraion of he equilibrium equaion delivers: For N( ) = N(0) f N(0) herefore: = l he normal force has o be zero, so ha for N(0) i holds: = l f 5

27 ( ) N( ) = l f Applicaion of he consiuive equaion ields: l ε( ) = EA f Finall, from he kinemaic equaion i hen follows: ( l ) u = u + d= f ( ) (0) 0 ε EA In he las erm, he boundar condiion u (0) = 0 alread has been incorporaed. The graphical represenaions of N( ) and u ( ) can be found in Fig..3. N u Fig..3: Normal force and displacemen as a funcion of. Wih he displacemen mehod he same resul has o be obained b solving he differenial equaion: du EA = d f ogeher wih he following boundar condiions: = 0 u = 0 ; = l N =0 The second boundar condiion can be rewrien as a condiion for he displacemen field: du N = EAε = EA = 0 d Wih hese wo boundar condiions i indeed is found (also see he course Elasosaics of slender srucures ): u = ( l ) EA f From which for N( ) i follows: 6

28 du N( ) = EA = ( l ) f d Remark The discussion in his secion appears o be ver rivial. This has been done on purpose o achieve wo goals, firs o epress he analog wih he discree approach of chaper and second o highligh he correspondence wih ne chapers in which he analsis appears o be less obvious.. Saicall indeerminae beam subjeced o eension The problem wih he bar of previous secion is eended wih a ssem of disribued springs. These springs are conneced o he bar a is ais and can deform onl in he direcion of his ais. The forces of he springs ac in ha direcion oo. Fig..4 shows he se-up of his new problem. The springs are depiced as leaf springs resrained a he boom and hingeconneced o he bar a he op. l f ( ) u( ) F N ε N e s N f d d sd dn N + d d Fig..4: Saicall indeerminae bar wih relevan quaniies. Also in his case, he displacemen field is fied wih one degree of freedom u ( ). Therefore, here is eacl one componen of eernal disribued load f ( ). Wih respec o he inernal sress quaniies and corresponding deformaions he siuaion is differen compared o he previous eample. Ne o he bar elemen here is a spring elemen. Deformaion energ can be accumulaed in boh of hem, such ha for each of he elemens separael a generalised sress and a generalised deformaion occur. Therefore i makes sense o inroduce separae smbols for hese quaniies. For he bar elemen hese are again he normal force N( ) and he specific srain ε ( ). In he spring elemen he force per uni lengh in -direcion is indicaed b s ( ) and he deformaion of he spring b e ( ). The scheme of Fig..5 displas all quaniies ogeher wih he governing relaions. 7

29 eernal work inernal work ε ( ) N( ) u ( ) e( ) = s( ) = f( ) e ( ) s ( ) kinemaic equaions consiuive equaions equilibrium equaion Fig..5: Diagram displaing he relaions beween he quaniies plaing a role in he analsis of a spring-suppored bar subjeced o eension. Governing equaions The normal force N has he dimension of a force and herefore is indicaed b a capial. The load s in he springs is a force per uni of lengh, so i has a differen dimension. For ha reason a lower case leer is used. The kinemaic equaions for his case are: du ε = d e= u (kinemaic equaions) (.8) The consiuive equaions are: N = EAε or ε = N EA s= k e or e= s k (consiuive equaions) (.9) where EA is he eensional siffness and k he spring modulus. For convenience sake, boh parameers are aken consan. The equilibrium equaion for a small secion of he bar now becomes: dn s + f = 0 (equilibrium equaion) (.0) d These hree ses of basic equaions can be reformulaed b using operaors, which are defined b: 8

30 d d d B = ; B = d (.) ε N e = ; s = e s (.) EA 0 0 EA D= ; C= 0 k 0 k (.3) Again, he hree basic equaions can be wrien in a brief manner as discussed in chaper : e = B u s = De B s = f ( kinemaic equaions) ( consiuive equaions) ( equilibrium equaion) (.4) Noice ha he operaor B is almos he ransposed of he operaor B. During ransposiion he derivaive changes of sign. Again, i will be discussed how hese basic equaions are used in he force and displacemen mehods... Force mehod In his mehod, he equilibrium equaion (.0) is he firs equaion o be evaluaed. This is one equaion wih wo unknown sress quaniies N and s, which confirms ha he problem is saicall indeerminae. Therefore, one redundan φ ( ) has o be inroduced. This means ha here is onl one compaibili condiion oo. In his case a horizonal cu is made beween he bar and he springs. The disribued load a boh faces of he cu hen becomes he redundan. In Fig..6a a posiive φ ( ) has been drawn. φ ( ) φ ( ) Fig..6a: Selecion of he redundan in a spring-suppored bar subjeced o eension. I makes sense o inroduce a separae smbol for he redundan. In he simple bar problem ha is under invesigaion here, he redundan φ ( ) is equal o he spring load s ( ), bu his is no necessaril alwas he case. Wih his choice for he redundan from he equilibrium 9

31 equaion (.0), he following relaions for he inernal sress quaniies N and s can be found: dn = φ d s = φ f (equilibrium) (.5) Afer deerminaion of he redundan φ, he force N and he load s can be obained from he relaions (.5). The second sep in he force mehod is he evaluaion of he consiuive equaions in fleibili formulaion: ε = N EA e = φ k (consiuion) (.6) The hird sep is he derivaion of he compaibili condiion. This condiion is an equaion describing he relaion beween ε and e ha has o be saisfied. The compaibili condiion can be obained from he kinemaic equaions (.8) b eliminaion of he displacemen u (noice wha has been saed in he summar of chaper ). This can be done b differeniaion of he second equaion wih respec o. Then boh righ-hand sides are equal o du d, which afer subracion of he wo equaions disappears and a relaion resuls onl conaining ε and e. This is he compaibili condiion and i is given b: de ε = 0 (compaibili) (.7) d The fourh sep is he derivaion of he differenial equaion for he redundan φ. Analogousl as done for he russ in chaper, firs he subsiuion is required of he equilibrium ssem (.5) ino he consiuive equaions (.6). For his purpose he firs equaion of (.6) is differeniaed wih respec o. The wo relaions hen become: dε = d dn EA d ; e = φ k Subsiuion of he equilibrium ssem (.5) hen provides: dε = ( φ f ) ; e = φ (.8) d EA k Ne his resul has o be subsiued ino he compaibili condiion (.7), which firs is differeniaed once: 30

32 dε = d d d e 0 Now (.8) easil can be subsiued resuling in: d φ φ kd EA EA + = f (.9) This is a second-order differenial equaion wih respec o he redundan φ. For he soluion of his equaion wo boundar condiions have o be formulaed. Similari wih bar srucures The provided derivaion of he differenial equaion for φ did no show clearl he analog wih he force mehod for bar srucures. The recognizabili is increased when he derivaion is carried ou a bi differenl. Again, in he firs sep i is sared wih he equilibrium equaion (.0), bu now an inegraion is carried ou. The following equilibrium ssem is found: N = φ d fd ; s= φ (.0) 0 0 The inegraion consan is no considered, because i is no imporan in his case. Togeher wih he consiuive equaion in he fleibili formulaion (.9), an epression for he deformaions is found: ε = φ d f d ; e= φ (.) EA EA k 0 0 Now, he compaibili condiion is deermined b eliminaion of he displacemen from he kinemaic equaions (.8). Inegraion of he firs one followed b subracion of he second one gives: ε d e = 0 (.) 0 The phsical inerpreaion of his resul is shown in Fig..6b. Subsiuion of (.) ino (.) provides he compaibili condiion from which φ can be calculaed: φ + φ d d = f d d (.3) k EA EA

33 u e = u e= ε d e 0 Fig..6b: Visual represenaion of he gap o be neuralised in a spring-suppored bar subjeced o eension. B differeniaion wice, a more suiable equaion is obained: d φ φ kd EA EA + = f (.4) Now i is clear ha he inclusion of inegraion consans in (.0) and (.) makes no sense, because hese consans would have disappeared anwa b he wo differeniaions. The differenial equaion (.4) is idenical o he previousl found equaion (.9). Noaion wih operaors Now, i will be demonsraed how above resul can also be obained wih he use of operaors. The similari wih chaper will become clear. The inernal sress quaniies in (.0) are: d d N 0 0 = f + φ s 0 Or briefl, analogousl o (.9): s = P + Pφ f f where: P f d d 0 0 = ; P = 0 For he srains i hen follows: e=c ( Pf f + Pφ ) 3

34 where C is given in (.3). The compaibili condiion (.) can be wrien as: ε d = 0 0 e Inroducion of he suiable operaor P given b: P = 0 d provides he following brief noaion: P e = 0 In analog wih (.3) from chaper, he operaor P is he ransposed of P wih as era addiion he sign difference in he inegraion erm. Subsiuion of he mari equaion for he srains e changes his equaion ino: P CP φ = (.5) f where f is he incompaibili resuling from he eernal load f, i is given b: = P CP (.6) f f f Wrien in his form he previousl found differenial equaions (.9) and (.4) can be compared o he resuls obained in chaper... Displacemen mehod In he displacemen mehod he consiuive equaions (.9) in siffness formulaion and he kinemaic equaions (.8) are direcl subsiued ino he equilibrium equaion (.0). This leads o: du EA + k u = f (.7) d Inroducion of he operaor definiions of (.) and he mari D given in (.3) provides he following mari noaion of he differenial equaion: B DB u = f (.8) In his form, he analog wih chaper becomes clear again. 33

35 ..3 Elaboraion of an eample A soluion is provided for he problem of Fig..7 (also see Fig..4), where f ( ) has a consan value f and where EA and k are consans oo. f l Fig..7: Spring-suppored bar, which is uniforml loaded in aial direcion. In he force mehod he differenial equaion is: φ + = d φ kd EA EA f and in he displacemen mehod he equaion reads: du EA + k u = f d In he differenial equaion of he displacemen mehod, he siffnesses EA and k appear as consan coefficiens. In he equaion of he force mehod hese are he compliances k and EA. A he posiion where he siffness k appears in one case, he compliance EA appears in he oher case and vice versa. Boh differenial equaions are of he second order. For boh he displacemen mehod and he force mehod, wo boundar equaions are required. The are: = 0 u = 0 ( kinemaic) = l N = 0 ( dnamic) Force mehod In he force mehod he boundar condiions are ransformed ino condiions epressed in φ. Use is made of he kinemaic relaion e= u. The kinemaic boundar condiion u = 0 becomes e = 0 and herefore φ = 0. The dnamic boundar condiion N = 0 becomes EAdu d = 0. Since e is equal o u, i also holds de d = 0. In view of he direc relaion beween s and e, he condiion becomes ds d = 0. Bu s is equal o φ, so ha for he dnamic boundar condiion i has o hold dφ d = 0. Therefore, he following problem has o be solved: 34

36 φ + = f (differenial equaion) d φ kd EA EA = 0 φ = 0 = l dφ = 0 d (boundar condiions) A paricular soluion is: φ = f A soluion for he homogeneous equaion (righ-hand side zero) equals r φ = Ae The equaion for he deerminaion of he roos r reads: r r + e = 0 k EA B inroducion of he characerisic lengh λ : λ = EA k he characerisic equaion (afer division b r e EA) can be wrien as: r λ + = 0 From which i follows: r = λ The wo roos herefore are: r = ; r = λ λ The oal soluion for φ including he paricular soluion becomes: φ( ) = Ae + Ae + f / λ / λ This soluion is ofen wrien is a somewha differen form. The firs erm of he righ-hand side is hen epressed in he coordinae opposie o he coordinae and saring a he free end (see Fig..7). Beween and he following relaion holds = ( l ). ( ) λ l The firs erm hen becomes Ae or Ae λ e λ. Afer inroducion of he new consan l A = Ae λ he soluion also can be wrien as: 35

37 φ( ) = Ae + Ae + f / λ / λ The homogeneous par consiss of a conribuion ha damps ou from he end = 0 and of a conribuion ha damps ou from he end = 0. The consans A and A can be obained from he boundar condiions. The elaboraion is simplified b he assumpion ha he bar has such a lengh ha boh damping erms do no reach he oher end. This is he case if he lengh l of he bar is hree o four imes is characerisic lengh λ. I hen is found: λ e = 0 = 0 φ = A + f = 0 A = f e λ 0 = e = λ 0 e = e = dφ λ λ = l A e A e A 0 A 0 λ = = = = e = 0 d λ λ λ Therefore he soluion is: / λ ( ) φ( ) = e f From he equilibrium equaion (.5) i direcl follows: N( ) = λ f e / λ / λ ( ) s ( ) = f e and he displacemen becomes: f u ( ) = k / λ ( e ) Fig..8 displas he normal force and displacemen disribuions. N u s Fig..8: Normal force and displacemen as a funcion of. Displacemen mehod In he displacemen mehod he boundar condiions have o be epressed in he displacemen u. For = 0 his is alread he case, because a ha posiion he kinemaic boundar condiion u = 0 applies. A he end = l he dnamic boundar condiion N = 0 holds, which can be reformulaed as du d = 0. Therefore, he following problem has o be solved: 36

38 du EA + k u = f (differenial equaion) d = 0 u = 0 = l du = 0 d (boundar condiions) Now, a paricular soluion is: u = f k As homogeneous soluion i is seleced: u = Ae r This delivers he characerisic equaion: ( EA r ) r + k e = 0 r Afer division b ke and inroducion of he characerisic lengh λ as previousl defined, he same characerisic equaion as in he force mehod is obained: r λ 0 + = From his equaion he same roos can be solved. In a similar wa as found for φ in he force mehod, he soluion for u becomes: u ( ) = Ae + Ae + k λ λ f Again he resricion is inroduced ha l λ. For he values of he consans i hen can be derived: = 0 f f u = A + A = k k = l du λ λ = A e A e = A = 0 A = 0 d λ λ λ The final soluion for u hen becomes: f u ( ) = k / λ ( e ) Subsequenl, from his relaion he quaniies N and s can be deermined. Naurall, his soluion is in agreemen wih he soluion found wih he force mehod. 37

39 .3 Saicall deerminae beam subjeced o bending In his secion a sraigh beam is considered in which bending momens are generaed caused b a disribued load f perpendicular o he beam-ais. Fig..9 shows he srucure and he smbols o be used during he derivaions. f ( ) l w ( ) M κ M z M V d f d V dm M + d dv + d d d Fig..9: Saicall deerminae beam subjeced o bending wih relevan quaniies. For he analsis of he problem i is assumed ha he reader knows he classical beam heor. In his case i is paricularl imporan ha he deformaions generaed b he shear force V can be negleced. In ha case he displacemen field can be described b onl one degree of freedom, which will be indicaed b w ( ). Therefore, also onl one eernal load is possible, he disribued load f ( ) per uni of beam lengh in he direcion of w. Ne o hese wo eernal quaniies, also inernal quaniies pla a role. The are he momen M and he curvaure κ, which is caused b he momen. In Fig..9 posiive M and κ are depiced. Remark I is rue ha he shear force V is appearing as well, bu no as a generalised sress ha plas a role in he followed soluion sraeg. Generalised sresses are quaniies ha are coupled wih deformaion energ. For he shear force his is no he case because he corresponding deformaion is no considered. Remark The deformaion caused b he momen is indicaed b κ. This parameer definiion is used frequenl in he engineering pracice in concree and seel. Think abou he use of M-κ diagrams. This means ha κ is defined posiive if i is caused b a posiive momen M. The quaniies, which are essenial for his discussion and he relaions beween hem are indicaed in he scheme of Fig..0. From he hree basic equaions, wo can be wrien down immediael: dw κ = (kinemaic equaion) (.9) d 38