M r. d 2. R t a M. Structural Mechanics Section. Exam CT5141 Theory of Elasticity Friday 31 October 2003, 9:00 12:00 hours. Problem 1 (3 points)


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1 Delf Universiy of Technology Fculy of Civil Engineering nd Geosciences Srucurl echnics Secion Wrie your nme nd sudy numer he op righhnd of your work. Exm CT5 Theory of Elsiciy Fridy Ocoer 00, 9:00 :00 hours Prolem ( poins) Consider curved em of recngulr crosssecion (Figure ). The em is loded y momen. The norml sress σ θθ in crosssecion cn e derived s. where σ θθ Eε i fθθ, r r f θθ ln ln l r + n The momen in he crosssecion cn e derived s r R + R Figure. Curved em.0 d where E i εc, C ( ) + ln Derive he formul for he elsic secion modulus W (Duch: weersndsmomen). σ θθmx W W d d 6 d R Figure. Funcions W nd Derive he formul for he momen of ineri (Duch: rgheidsmomen). E κ (Noe h curvure κ is he chnge of roion of he crosssecions per rch lengh.). c The derived formuls W nd hve een ploed in Figure. Boh W nd cn e pproximed y he formuls for srigh ems. s his pproximion sfe or unsfe? Consider for his oh he serviceiliy limi se nd he ulime limi se of he srucurl componen. d The hypohesis of Bernoulli ses h plne crosssecions remin plne during ending. s his hypohesis correc for curved ems?
2 Prolem ( poins) Trdiionl imer floors consis of ems nd floorords (Duch: vloerdelen). n his prolem he deflecion of such floor is clculed. One of he ems crries line lod q [kn/m] (Figure ). The floorords rnsfer pr of his lod o he djcen ems. The conriuion of oher ems is negleced, so only hree ems re included in he srucurl model. The Young s modulus is E nd he momen of ineri of he ems is. The ords hve hickness. w q w w w floorord y em Figure. Crosssecion of imer floor We ssume deflecion funcion for he middle em, w deflecion funcion w for oh djcen ems nd deflecion funcion w for he ords. x x w wˆ + C + C l l x x w wˆ + C + C l l y y w w+ C5 + C 6 Figure. Srucurl model of imer floor Wh is he inerpreion of vriles w nd w ˆ. Formule he kinemic oundry ˆ condiions of he deflecion funcions. Which of hese oundry condiions re lredy sisfied? Show h he oundry condiions cn e used o solve he following coefficiens. C wˆ C C wˆ C C5 w w C6 Kinemic Equions d w d w κ κ dx dx d w κ dy Formule he poenil energy of he floor. Wrie i such h i cn e evlued y mhemicl compuer progrm. Consiuive Equions E κ E κ m E κ
3 c The resul of he compuer evluion is lrge funcion of he following vriles. Epo Epo(, l,, E,, q, wˆ, wˆ, C, C, C6 ) Wih respec o which vriles needs he poenil energy e miniml? How re he equions derived from which he deflecion cn e solved? d The compuer solves he deflecion s where 5 ql β+ 75β wˆ 8 E β+ 5β l β 000 The formul cn e simplified wih very lile loss of ccurcy o 5 ql 0 + 8β wˆ 8 E 0 + 8β Cn his resul e improved y using eer pproximions of he deflecion funcions? Prolem ( poins) A concree offshore plform hs cenrl shf wih fivecell crosssecion (Figure 5). The hickness of ech wll is, which is much smller hn he rdius R of he cells. The memrne nlogy will e used o deermine he orsion properies. Due o roion symmery jus wo cells need o e considered in he clculions. A nd A re he crosssecion res of hese cells. O nd O re he circumferences of hese cells. The following relions pply A A A R A A A A πr A ( π) R O πr O πr. Figure 5. Crosssecion of offshore plform shf Formule he equions from which he displcemens w nd w of he floing ples cn e solved. You do no need o solve hese equions.
4 The soluion of he equions is pr π+ w S 6π p R 8 w S π. Clcule he orsion siffness of he crosssecion. c Clcule he sher sresses in he wlls. ke simple wing of he crosssecion nd w he sher sresses in he correc direcions. Prolem ( poins) Consider n xil symmericl ple loded in ending. Derive he following equilirium equion. d ( r m rr ) dm + θθ rp
5 Exm CT5, Ocoer 00 Answers o Prolem Secion odulus n he lecure ook f θθ hs een ploed. From his we oserve h he exreme vlue occurs r. f θθ mx ln ln l + n ln 0 ln + ln 0+ ln ln 0 ln ln ( ) + ln Eε ic C W σ θθmx Eε ifθθmx fθθmx ln ( ) + ln W ln omen of neri Curvure is chnge of roion over rch lengh. ϕi ε κ i ( + ) ϕ + Eε ic C( + ) ( ) + ln ( E i + κ ε E + ( )( ) + ln ) c Approximion n he serviceiliy limi se he deformion is imporn. Since he pproximed is oo lrge he displcemens will e oo smll. Therefore he pproximion of is unsfe. n he ulime limi se he sresses re imporn. Since he pproximed W is oo lrge he clculed sresses will e oo smll. Therefore he pproximion of W is unsfe. 5
6 d Hypohesis The iniil srin ε i produces plne roions of he crosssecions. Therefore, he hypohesis is correc for curved ems. Encore (no n exm quesion) The firs wo non zero erms of he Tylor expnsion of W o d re W d 6 d 8. R The firs wo non zero erms of he Tylor expnsion of o d re 5 d d 65 R. These pproximions hve n error less hn % for d < R (Figure 6) W W d 6 d R + 5 d d 65 R d d d Figure 6. Funcions W, nd heir pproximions 6 d d R Answers o Prolem Kinemic Boundry Condiions The vriles w nd w ˆ re he mximum deflecions h occur in he middle of em nd ˆ respecively. The oundry condiions re x l w 0 x l w 0 x l w 0 This prolem hs een proposed y Dr. Linh Co Hong, COW Consul Denmrk 6
7 x l w 0 5 y 0 w w 6 y w w w 7 y 0 0 y The deflecion funcions re such h he oundry condiions,, 5 nd 7 re lredy sisfied. x l 0 wˆ + C + C C wˆ C Poenil Energy 6 x l 0 wˆ + C + C C 6 wˆ C 6 y w w+ C5 + C6 C5 w w C6 l l l l po κ + κ + κ x l x l x l y 0 x l E dx dx m dydx qw dx c iniml The poenil energy needs o e miniml wih respec o he prmeers h descrie he deformion,wˆ, w ˆ, C, C, C6. The equions re derived from Epo Epo Epo Epo Epo 0, 0, 0, 0, 0. ŵ ŵ C C C6 d mproved De inercion eween floorords nd em is no consn force over he em lengh. Therefore, fourh order polynomils re no sufficien o descrie he exc deflecion of he ems. So, he soluion is no exc nd in heory i cn e improved. Encore (no n exm prolem) The deflecion funcions cn e exended 6 x x x w wˆ + C + C + C 7 6 l l l 6 x x x w wˆ + C + C + C 8 6 l l l y y w w+ C5 + C 6 The clculed mximum deflecion is now 5 ql β β β wˆ 8 E β β β The pproximed formul 7
8 5 ql 0 + 8β wˆ 8 E 0 + 8β shows deviion eween 0. % nd 0. % for ll vlues of β. Answers o Prolem Consiuive Equions We ssume h ple hs higher liude hn ple. w w w Vericl equilirium of ple. pa S O + S O 0 w w Vericl equilirium of ple. pa S O 0 This cn e evlued o w w w pπr S π R + S π R 0 w w p( π) R S π R 0 The soluion of hese equions hs een provided. pr π+ w S 6π pr 8 w S π Torsion Siffness From he memrne he φ ule is oined using he following susiuions w φ p θ S. G π+ φ θgr 6π 8 φ θgr π The orsion momen is wo imes he volume of he φ ule. ( φ A+φA ) π+ 8 θgr π R + θgr ( π) R π π 6 8 GR θ 8π+ π For wire frme model of he shf we wrie G θ. 8
9 Therefore he orsion siffness is 8 G GR 8π+ π. c Sher Sresses The sher sress is he slope of he φ ule. Firs we rewrie he formul for he orsion momen π GRθ π + 8 R nd express φ nd φ in he orsion momen. π π+ π+ φ 6 π + 8 R π π + 8 R π 8 6 φ π + 8 R π π + 8 R The sher sresses re φ σ π+ σ π + 8 R 6 π+ φ φ π + 8 R π + 8 R π σ σ π + 8 R. 9
10 Answer o Prolem Vericl equilirium p ϕ r qr ϕr ( qr + dqr )( r + ) ϕ p r qr r qr( r + ) dqr( r + ) p r qr r qr r qr dqr r dqr p r qr dqr r dqr dq p r q r r r dqr dq pr q r r r d pr ( r q r ) () omen equilirium ϕ 0 ( mrr + dmrr ) ϕ ( r + ) mrr ϕ r + p ϕr qr ϕr mθθ 0 ( mrr + dmrr )( r + ) mrr r + p r qr r mθθ 0 mrr ( r + ) + dmrr ( r + ) mrr r + p r qr r mθθ 0 mrr r + mrr + dmrr r + dmrr mrr r + p r qr r mθθ 0 mrr + dmrr r + dmrr + p r qr r mθθ dm 0 rr mrr + r + dmrr + p r qr r mθθ dm 0 m rr rr + r qr r mθθ m ( ) 0 rr qr r mθθ () Susiuion of () in () gives pr d ( r mrr ) d + m θθ. Q. E. D. 0
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