FINITE DEFORMATION OF ELASTIC CABLES UNDER 3-D LOADING. ANALYTIC SOLUTION

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1 4 th National Conference on Steel Structures Ma 4-5, 00, Patras-Greece FINIE DEFORMAION OF ELASIC CABLES UNDER 3-D LOADING. ANALYIC SOLUION J.. Katsikadelis School of Civil Engineering National echnical Universit of Athens, GR-5773 Athens, Greece. SUMMARY In this aer an analtic solution is resented for the finite deformation analsis, i.e. large dislacements and large strains, of stretched (taut) elastic cables under 3-D loading. he cable ma be restressed b aial tension and the in-service loads act in the three directions. hus, deformed shae of the cable is a sace curve. he material is linearl elastic or incomressible herelastic. he influence of the contraction of the cross section on the deformation is also taken into account. In the resented here solution the roblem is formulated in terms of the three dislacements comonents, which are obtained b direct integration of the equilibrium equations. he introduction of a constitutive relation, that adequatel aroimates the herelastic incomressible materials, makes the analtic integration ossible. Besides, an analtic solution for moderate large deflections is resented, which is used to establish the range of validit of various cable theories as well as to anale cables with small load-to-stiffness ratios, where eisting the numerical methods, e.g. FEM, fail to give reliable results. Eamle roblems are studied, which illustrate the alication of the method and demonstrate its efficienc.. DERIVAION OF HE DIFFERENIAL EQUAIONS OF HE CABLE FOR 3- D FINIE DEFORMAION. Equilibrium equations Consider a fleible cable having length L and uniform cross section A and consisting of elastic or herelastic material ling on the -ais in the undeformed state. he cable ma be restressed b aial tension or end dislacements and the in-service loads act in the three directions, namel in the aial and two transverse directions. hus, the deformed shae of the cable is a sace curve. Using the notation: d, ds : Undeformed and Deformed cable elements u u( ), v( ), w( ) : Dislacement comonents (,, ), (,, ) : Load vectors on the undeformed and deformed elements,,, : he comonents of the tension force and magnitude,, 3: Direction cosines of the tension force and referring to Figure, the equilibrium of the deformed cable element ields 56

2 ( ),, ( ), s s, ( ), s (a,b,c) or taking into account that d, d ds, d ds (a,b,c) ds we can write, ( ), ( 3), (3a,b,c) ( ), d B ' + d d d A' ds (,, ) i 3 i i A w u v d (,, ) w dw u du B v dv Figure. Deformed cable element and notation. Eression of the direction cosines in terms of the dislacements he osition vector of a oint,, after deformation is written as r=( u)i vi wi (4) 3 which ields ds dr ( u, ) v, w, (5) d d where reresents the etension ratio. he tangential vector is dr dr d t ( u, )i v, i w, i 3/ (6) ds d ds and the direction cosines are obtained as ( u, ) v,, w,, 3 (7a,b,c).3 Constitutive laws he material used in cable fabrication ehibit nonlinear tension-etension, that is ( ) (8) If this law is not is not given from eeriments for the considered cable, we can aroimate it using its material constants as follows: 57

3 (i). Linearl elastic material without contraction of the cable cross section In this case we have EA EA( ) (9) where ds d/ d ( ) is the engineering strain and E is the modulus of elasticit. (ii). Linearl elastic material with contraction of the cable cross section aking into account that 0, we have. hus the area of the deformed cross section becomes A ( )( ) ( ) A (0) hus we have EA EA EA () (iii). Linearl elastic material with constant element volume In this case we have A ds Ad, hence A A / and EA EA( )/ () (iv). Herelastic material. Neo-Hookean Assuming herelastic material we have for homogeneous simle tension W ii i i ( i,,3 no summation on i ) (3) where i are the etension ratio along the i, is a hdrostatic ressure and W W( I, I, I ) is the strain energ function, where the strain invariants are given as 3 3 I, 3 3 I, For the Neo-Hookean material it is W c0( I 3) []. Hence I c c ii 0 i 0 i i 3 I (4a,b,c) For simle tension in direction, we have 33 0 and because of smmetr 3. Moreover, assuming incomressible material we have 3 and setting we find /. hen eqn (5) for i or i 3 gives c c (6) which is substituted in eqn (3) for i to ield c ( ) (7) 0 he area of the deformed cross section is obtained from the relation ields A A/. Hence 0 (5) A A, which A Ac ( ) (8) 58

4 (v). Herelastic material. Moone-Rivlin For the Moone-Rivlin material the strain energ function is W c0( I 3) c0( I 3) []. Substituting this eression in eqn (4) and following the revious rocedure we find ension force /A Neo-Hookean Moone-Rivlin Linear Elastic-Constant Element Volume Linear Elastic-Constant Cross Section Linear elastic_cross section contraction 0 c0 c0 3 ( 0 0)( ) Aial Strain ε Figure. Constitutive laws ( ) ( ) (9) A A c c (0) In Figure the variation of / A versus the strain for the above five constitutive laws is deicted. he material constants are c kN/m, c kN/m. he modulus of elasticit has been chosen E 6c0 c0, that is equal to the sloe of the Moone- Rivlin material at ( 0 ). From this figure we ma conclude that the linear material with constant cross-section (case i) ields accetable results for relativel small etension ratios (.), while the linear material with varing cross-section (case ii) can be alied for greater etension ratios (.5). Finall the constant element volume assumtion can adequatel aroimate the herelastic materials. 3. ANALYIC SOLUION 3. Finite deformations A first integration of eqns (3) gives, where, 3 (a,b,c) ( ) d, ( ) da, ( ) d A3 (a,b,c) Solving eqn (8) for, ( ) and using eqns (7), we can write eqns () as u, ( ),, v ( ),, w ( ) (3a,b,c) 59

5 Eqns (3) can be further integrated to ield ( ), v ( ) d B u d B, w ( ) d B3 (4a,b,c) A i, B i arbitrar constants, which are established from the boundar conditions u(0) (0) 3 ul ( ) ( L) 3 (5a,b) v(0) (0) 3 vl ( ) ( L) 3 (6a,b) w, (0) (0) 3 w, ( L) ( L) 3 (7a,b) where ai, i, i are secified constants. 3. Moderate large deflections deformations In this case it is u,, v,, w,, hence, ds d,, v,, 3 w,. Using the Lagrangian strain and noting that u, u,, we can write EA( u, v, w, ) (8) and the equilibrium equations (3) become, ( v, ),, ( w, ), (9a,b,c), Without restricting the generalit, we assume uniform loading. hen eqns (9) ields A (30) A B v ln B (3) A C w ln C (3) Combining eqns (8) and (30) we obtain EA( u, v, w, ) (33) Substituting eqns (3) and (3) in (33) and integrating give B A u ln 3 3 A B A EA A C ln 3 3 A C A A, A, B, B, C, C are arbitrar constants, which can be established using the boundar conditions. heir evaluation is simlified, if we introduce the new arbitrar constants d A B e C (35a,b) (34) 530

6 hus, eqns (3), (3) and (34) are written as d v ln A B, e w ln C (36a,b) A d u dln 3 3 A EA e eln 3 3 A A (36c) he boundar conditions for restressed cable are ua ( ) ua, va ( ) va, wa ( ) wa, and ub ( ) ub, vb ( ) vb, wb ( ) wb. Aling eqns (36a,b,c) at the end oints and subtracting ield b a L u L EA Ld e d e b ln 3 a b A a A b dln F v L, eln F w L, F (38a,b) a A Eqns (37) and (38a,b) are solved numericall to ield A, d, e. hen B, C are comuted from eqns (35a,b). Finall, B, C and A are comuted from eqns (36a,b,c), if the are alied at either of the end oints a or b. (37) 4. EXAMPLES 4. Finite deformations: Solution for linear and constant volume constitutive law he solution is obtained b rogramming the rocedure in the smbolic language Male. ABLE,, ma u min u 3, 0, 5 5, 50, 40 0, 00, 00 Linear Const. Vol Linear Const. Vol Linear Const. Vol ma v ma w ma min ma min (0)

7 he emloed data are: E kn / m, A 0.4m, L 5m. Results for various loadings are given in able. Moreover, the variation of the quantities ( ), ( ), w ( ) and A / A / ( ) for 0 kn / m, 00 kn / m, 00 kn / m for both constitutive laws are shown in Figure 3 through Figure Constant Volume Linear 4 Constant Volume Linear (kn) 00 w (m) (m) Figure 3. ension force (m) Figure 4. Deflection w Constant Volume Linear 0.6 (kn) 0 A /A Constant Volume (m) Figure 5. ension comonent (m) Figure 6. A/A 4. Moderate large deflections: Inclined cable under its own weight. he cable shown in Figure 7 is analed under its own weight (0,0) sin cos (00,45) Figure 7. Inclined cable he emloed data are: Diameter D 39.5mm, E N / mm, 3, 6 kn / m. he cable is ver stiff and unrestessed, the ratio / EA is ver small and the numerical solutions, e.g. FEM, fail to converge. Onl the analtical solution can give reliable results. he comuted arbitrar constants are: , A , B = , B , C C 0. he obtained solution is 53

8 u= e /( ) ln( ) (39) v= ln( ) (40) = e5/[( ) ] e/( ) (4) [ /( )] he relation between the load and the midoint vertical deflection is deicted in Figure Vertical dislacement at midoint Figure 8. Load versus midoint dislacement 5. CONCLUSIONS An analtic solution to the roblem of finite deformations of cables under 3-D loading is resented. he formulation is in terms of the cable dislacements. he integration of the resulting differential equations is simle when the contraction of the cross-section is ignored, which, however, is not realistic for structural cables undergoing large etension ratios. Constitutive laws, taking into account the contraction of the cross section are derived for linear elastic as well as for incomressible herelastic materials. Nevertheless, these laws can not be emloed, because their inversion roduces eressions ( ) that can not racticall be integrated. his drawback is circumvented b introducing a constitutive law, which on one hand aroimates adequatel the incomressible Neo-Hookean and Moone-Rivlin herelastic materials and on the other hand its inversion roduces eressions that can be readil integrated. he solution rocedure can be rogrammed in smbolic languages. Besides, an analtic solution is derived for moderate large deflections, which is useful to establish the range of validit of the various cable theories as well as to anale cables with small load-to-stiffness ratios, where the eisting numerical methods, e.g. FEM, fail to give reliable results. 6. REFERENCES [] Oden, J.., Finite Elements of Nonlinear Continua, McGraw-Hill Inc., (97). 533

9 ΠΕΠΕΡΑΣΜΕΝΕΣ ΠΑΡΑΜΟΡΦΩΣΕΙΣ ΚΑΛΩΔΙΩΝ ΥΠΟ ΤΡΙΣΔΙAΣΤΑΤΗ ΦΟΡΤΙΣΗ. ΑΚΡΙΒΗΣ ΑΝΑΛΥΤΙΚΗ ΛΥΣΗ Ι.Θ. Κατσικαδέλης Σχολή Πολιτικών Μηχανικών Εθνικό Μετσόβιο Πολυτεχνείο, ΤΚ-5773 Αθήνα, Ελλάδα ΠΕΡΙΛΗΨΗ Στην εργασία αυτή παρουσιάζεται ακριβής επίλυση του προβλήματος πεπερασμένων παραμορφώσεων ελαστικών καλωδίων υπό τρισδιάστατη φόρτιση. Το καλώδιο μπορεί να προενταθεί, ενώ τα φορτία λειτουργίας επενεργούν κατά τις τρεις διευθύνσεις, αξονική και δύο εγκάρσιες. Το προκύπτον παραμορφωμένο σχήμα είναι χωρική καμπύλη. Η επιρροή της συστολής της διατομής του καλωδίου λαμβάνεται ωσαύτως υπόψη. Οι εξισώσεις ισορροπίας διατυπώνονται κατ αρχάς ως προς τις τρεις μετατοπίσεις, οι οποίες ακολούθως αποσυμπλέκονται και μετασχηματίζονται σε μορφή που επιδέχεται απ ευθείας αναλυτική ολοκλήρωση. Η ολοκλήρωση είναι εύκολη όταν ο καταστατικός νόμος δεν λαμβάνει υπόψη τη συστολή της διατομής, EA( ) όπου ο λόγος επιμηκύνσεως. Η περίπτωση αυτή δεν ανταποκρίνεται στη συμπεριφορά των πραγματικών καλωδίων, στα οποία ο καταστατικός νόμος για τα χρησιμοποιούμενα υλικά είναι μη γραμμικός, ( ). Εισάγονται και εξετάζονται μη γραμμικοί καταστατικοί νόμοι για γραμμικά ελαστικά και υπερελαστικά υλικά που λαμβάνουν υπόψη τη συστολή της διατομής. Εντούτοις οι νόμοι αυτοί δεν μπορούν να χρησιμοποιηθούν, διότι η αντιστροφή τους δίδει εκφράσεις ( ), που δεν μπορούν να ολοκληρωθούν αναλυτικά. Το μειονέκτημα αυτό αίρεται εισάγοντας καταστατικό νόμο, ο οποίος αφενός μεν προσεγγίζει επαρκώς τα ασυμπίεστα υπερελαστικά υλικά τύπου Neo-Hookean και Moone-Rivlin, αφ ετέρου δε η αντιστροφή του δίδει έκφραση είναι απλή που μπορεί να ολοκληρωθεί αναλυτικά. Επιπλέον παρουσιάζεται αναλυτική μέθοδος επιλύσεως για την περίπτωση των μετρίως μεγάλων παραμορφώσεων του καλωδίου. Η λύση αυτή είναι χρήσιμη για το καθορισμό των ορίων ισχύος των διαφόρων θεωριών αναφορικά με το μέγεθος των παραμορφώσεων, αλλά και για τη λήψη αριθμητικών αποτελεσμάτων στη περίπτωση που οι αριθμητικές μέθοδοι αστοχούν, όπως π.χ. η FEM όταν ο λόγος του φορτίου προς την ακαμψία είναι πολύ μικρός. Από τα παραδείγματα που αναλύονται προκύπτει ότι η παραμέληση της συστολής της διατομής δίδει αποκλίσεις από την ακριβή λύση. 534