# 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 righ-hnd of your work. Exm CT5 Theory of Elsiciy Fridy Ocoer 00, 9:00 :00 hours Prolem ( poins) Consider curved em of recngulr cross-secion (Figure ). The em is loded y momen. The norml sress σ θθ in cross-secion cn e derived s. where σ θθ Eε i fθθ, r r f θθ ln ln l r + n The momen in he cross-secion 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 cross-secions 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 cross-secions 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. Cross-secion 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 five-cell cross-secion (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. Cross-secion 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 cross-secion. c Clcule he sher sresses in he wlls. ke simple wing of he cross-secion 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 cross-secions. 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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