corso statica e scienza i)ellf~...

$: CORSO STATICA E SCIENZA I)ELLF~ c()srl~I{UZI()NIpeople.unica.it/antoniocazzani/files/2008/05/SdC_SdA_20140916-p2-s.pdf)(sereizio ll. 2 (7 punti) . P~r. Ia struttura . isostatica.,$
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CORSO DI STATICA E SCIENZA I)ELLF~ c()srl~IUZI()NI

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Prova scritta in aula del 16092014 Parte II - (csto 1

ClS Edilizia D CdS Ad(~ I I (is SdA II

N(ta I risultati numerici vanno riporlali a penna su queslo slessofiJglio nei riqulldri Jrelli(sli i calcofi (informa ordinata) vanno allegati sui solifiJgli a quadrelli che sono SllllifiJrnili

IAllievo e-mail Matricola ~]

Esercizio D 1 (17 punti)

Ilisolvere mediante il Principio dei Lavori Virtuali (IgtLIV) la struttura iperstatica riportata in (4igllra~

assllmendo come incognita iperstatica la coppia in D Mn

()opo avere determinato Iiperstatica tenendo c(Jnlo solo della defiJrn1ahililitflessionllle calcolarc Ie reazioni vincolari Ie azioni interne e tracciare nello spazio predisposto nella pagina a fr()111c i corrispondenti grafici Calcolare infine riapplicando il PLV 10 spostamento orizzontale del punto A l(

~~fi rammenta che if diagramma del momenloflellenle va rijJorlalo dalla parle delletihre lese

Universita di CagJiari SdC_SdA 160914001

b

b

1M o

---+-~-~

qgtW 3b xuHp

Scritto 16092014 Parte II ---- Testo I pag 1

)(sereizio ll 2 (7 punti)

P~r Ia struttura isostatica indicata in Figura dcternlinare Ie reazioni villcolari c rcsprLssiollC dl1 h ltlioni interne nonche Ie condizioni al contorno inlposte dai vincoli nci punti 1111 e ( I Jtilizzare quindi lequazione della linea elastica pcr detcnninarc

1 La deformata della linea dasse v(z) == V(ZI) U V2(Z2)~

2 La sua derivata prima v(z) == VI(ZI) U V2(Z2)~

3 Lo spostalnento verticale del punto 13 VN~

4 Lo spostamento verticale del punto ( V(

Universita di Cagliari SdC_SdA 160914001

yvVq

bull c

cpW 2b 3b xuHp

HA (~) = Q MA (~) = 6~k VII (fr) = 6ampbull f 2

Q - 3a~ll M bOt _ ~ ozNAn = bullbull TAn = middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot1middotmiddotmiddotmiddot An = middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotrrJmiddotmiddotmiddotmiddotmiddotmiddotmiddot-flmiddotmiddotmiddotmiddotrmiddotmiddotmiddotiIebull ~

Nllc = bullbullbullbullbullbullbullbullbullbullt2 TI(= bullbullbullbullbullbullbullbullbullbullbullbullQ MI(middot= bullbullbullbullbullbullbullO

Cc in A = y~f~~Q)no Cc in B = ~LfF4 bull2b)J2tezmiddot~rl=Q I ~ (Ir- )

_ v (Cc=2b)middot J2 ~2 =0it cc In C - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~

_ lo~_ 3~~g2 + f ~4 _ --6 ~balt -f- ~ iJJ3 VI (ZI) - bullbullbullbullbullbullbullbullbulll abulltQ g m v I (z I) - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullliObullbullbullbull$

_ _ 8 ~gg-2- _ _ 8~- V2(Z2) - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~J V2 (Z2) - bullbull middotmiddotmiddotmiddotmiddotmiddotbmiddot~middotmiddotmiddotmiddotmiddotpoundt)middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot VII = bullbullbullQ v(middot = =2~ ~ ~ t)

Scritto 16092014 Parte II -l~esto I pag2

I~~ ercizio D 3 (9 punti)

l JrJ solido di forma cubica di spigolo lJ X5 111111 COil Ic f~lCCC p~lr~lllclc agl i assi del ~istCmiddot II I~ I di ri ferilnento ecostituito di un materialc caratlcrizzato dallc costal1ti eillst iche scgucnti Ij 1()( ) ( IP~l~

(i = 40 GPa E soggetto a uno stato di sforzo piano~ con1rlet~lnlente individu(lto dll ililsfl

comp()nenti ax == +40 MPa (jy == -80 MPa~ (J () MPu Si ricl1iede di

1 Determinare la lunghezza a defornlazionc avvcnuta dcllo spigolo originarianlcillc palnlhlo alIa direzione dell asse x Lx ~

2 Determinare la lunghezza a defornlaziol1c avvcl1uta dcllo spigolo originariall1cnte palullllo alia direzione dell asse ZJ 1J

3 Valutare la variazione di volume (rapportata al VOlulllciniialc) chc il cubo sllhis~l pll

effetto della deformazione 4 Determinare quale sarebbe in base al criterio di sicurczza di (ialilci-lUllkillC il 111aSsil110

sforzo piano ax == a(I) (Jy == -2(J(1) (valori csprcssi in MPa) sopportabilc dul 111atcrialc I1L1

caso in cui la tensione ammissibile in regime ITIonoassialc valga k = 160 MPa~

5 Determinare quale sarebbe in base al criterio di sicurezza di rresca il 111assilllo sf()rzo piano ax == a(2) a y == -2a(2) (valori espressi in MPa) sopportabilc dal 111atcrialc ncl caso ill cui la tensione ammissibile in regime monoassialc valga k == 160 MPa~

6 Determinare quale sarebbeJ in base al criterio di sicurczza di von Miscs il 111HSsil110 sf(umiddotlo piano ax == a(3) a y == -2(J(3) (valori sempre espressi in MPa) sopportuhilc dal 111utcrialc Ilel caso ill cui la tensione amnlissibile in regilTIe ITIonoassialc valga k ~ 160 MPu

- ac oSlo mmmiddot L - ~ FJCQC~ 111111middotIX - bullbullbullbullbullbullbullbullbullbullbullQlbulltt = - ~~MrlGlJ

V V)v- --OtrKlc1Qoo bull (I) - ampJ O(i~-) MI) middot (- - bullbullbullbullbullbullbullbullbullbull~~~bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull (1 - bullbullbullbullbullbullbullbullbullbullbullbullbullbull I bullbullbullbullbullbullbullbull-bullbullbullQbullbullbullbullbullbullbullbull bullbullbull bull bull bull 1

(1(2) = ~3J~~$3 MPa (1(3) = ~llt3 M Pa

Scritto 16092014 Parte II -lesto I pag3

S-_middotM__- 4fb -shy - -

q 2b

3 f 2-shy-102

NAil = middotmiddotmiddotmiddotl1b TAli = -~IltI MAII = ~i-4 N II( = ft~b 7~w = ~qb MII( = ftbf~tfb~~

N -)( shy Q 1 07 - -3~b 1gt( - bull bullbull bull bullbull middotf oM - )( shy ~ 3 ~2--tgh ~ -z middot middotmiddotmiddot2middot-Ttgtbull bull rmiddot~middotQow

UA = bullbull-bullbull~t clt) Scritto 16092014 Parte II-Testo 1 pag4

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CORSO DI STATICA E SCIENZA IlELLE (~()STI~lJZI()NI

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Prova scritta in aula del 16092014 Parte II - resto 2

( 1 S Edilizia D C~dS Ad( I I (lSSdAII

N()la I risultati numerici vanno riportati a penna su queslo slessofiJglio nei riquliliri JreliisOSI i i cllc()li (informa ordinata) vanno allegati sui solifiJKli a quadretti che sono sllllifhrnili

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I~s~rcizio D 1 (1 7 punti)

fisolvere mediante il Principio dei Lavori Virtuali (poundgt1-V) la struttura ipcrstatica riportata in FigllrH~

assumendo come incognita iperstatica la coppia in D Mj) IJopo avere detern1inato Iiperstatica tenendo conto solo della defhrmahililaflessionlile calenture Ie reazioni vincolari Ie azioni interne e tracciare nello spazio predisposto nella pagina a tr()l1tc i corrispondenti grafici C~alcolare infine riapplicando il PLY 10 spostamento orizzontale del punto A lJ

i rammenta che il diagramma del momentotlettente va riportato dalla parte dellefihre lese


tyvVq ~ 3o~f- ~2- c

B------------------~t-(~ LshyL_bull __ L-- t- shyLshyLshyLshyL-_b L- L- L-shyL__ _ L-shyLshy4 _ t- shyZ==5q L--- Lshy~_-shyL-shyL-_ L-_ LshyL--b L LshyLshyL-shyLshyLshyL-shyL-- L __

A

middot121

L--+~ _ - bull-____+_ --------------- -~-__t--- ---___+

3b xUHp

Scritto 16092014 Parte II - resto 2 pag1

I~sercizio D 2 (7 punti)

P~r la struttura isostatica indicata in l~igura detcrnlinarc Ie reaziolli vincolari e Iespressiolll- (klll azioni interne nonche Ie condizioni al contorno inlpostc dui vinco)i nei punti 11 1 e (Y

l Jtilizzare quindi lequazione della linea elastica per detenlliJlarc J La deformata della lil1ea dasse v(z) == VI(ZI) U V2(Z2)~

2 La sua derivata prima v(z) == VI (ZI) U V2(Z2)~

3 Lo spostamento verticale del punto B Vn~

4 Lo spostamento verticale del punto ( v(


4qyvVq

I A lllllllll~ c

I ~I iii r~ r

ltPW b 4b xuHp

HA (q) = t2 MA () = 2t VII (fr) = ft~

l)~b2 0middot ~ z _ = ~ b ~ ] = ~ --lt[ bullltI _NAil - bullbullbullbullbullbullbullbullbullQ TAli bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull MAli bullbullbullbullbullbullbullbullbullbullbullbullbull middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotfmiddot middot1middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot

NB( = bullQ TfU = bullbullbullbullbullbullbullbullbullbullQ middot Mu( = bullbullbullbullbullbullbullbullbullbullbullbullbullQ middotmiddotmiddotmiddotmiddotmiddotmiddotmiddot middotmiddot ~

I f I 1 r- ( )middot _ f f =-0 = middot - L 7 =1gt = UQ 2=O -0 CC In A - bullbullbullbullbullbullbullbullbull4l L J Qo ~~ c B - bullbullbullbullbull middotmiddot~middoti7i~6 j=middotmiddotmiddot~i~Jmiddotmiddotmiddot middotmiddotmiddot

cc 111 C - )

_ -li ~4 _ 9b~tl2 + 1 ~4 __ 2~b-( i- 2 lttr5

VI (ZI) - 6middotmiddotf0middotmiddotb~t)middotmiddotmiddotmiddotmiddotmiddotmiddot6middotmiddotmiddotmmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot VI (ZI) - bullbullbullbullbullbullbullbullmiddotmiddot-t~b5middotmiddotmiddotmiddotmiddotmiddot ~middotmiddotmiddotmiddott~F)middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot middotmiddotmiddotmiddotmiddot V2(Z2) = ~ ~ V2(Z2) = =tibull~~

It gb~ ~J _ rJ _ __ _ J=- 1+

VB - bullbullbullbullbullbullbullbullbull~bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull V(middot - bullbullbullbullbullbullbullbullbullbullbullbull~ bullbull ~- bull bullbullbullbullbullbullbullbull~lbullbull bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull

I~s ~rcizio n 3 (9 punti)

lJ rl so1ido di f0 rnla cubiea dispigolor ~ 65 111111 CO 11 Ic I~ ICC l p~ In til ell agIi ass ide J sistC Illcl di ri lerimel1to eeostituito di Ull lllatcrialc c~lrattcrizzato dalle costant i ll~lst iche segllllli i F I I ( ) ( J PI

( ~ 50 OPa E soggctto a UllO stahl di s(lrZO piano c0l11plctanllntc individllato d~l -Illesle

c0r11ponenti (Jx ==+-80 MPa~ Or ~ -160 MPa~ (f 0 MPa Si richiede di

1 Determinarc la lunghezza a defonnaziollc avvcnulu dcllo spigolo originHrialllclllLll pHItlllllo alIa direzione dell asse x lJx ~

2 Determinare la lunghezza a dcfornlazionc avvcl1uta dcllo spigolo originarial11cIltc rnunlhlo alIa direzionc dell asse z IJ ~

3 Valutare la variazione di VOlUlllC (rapportata al VOlUI11C inizialc) che il cuho sdhisLl per effetto della deformazione~

4 Determinare quale sarebbe in hase al critcrio di sicurczzH di (ialilci-Iallkillcl il 111assilllo sforzo piano (Jx == iT(I) iT) == -2iT(l) (valori esprcssi in MPa) sopportahilc Jul Illatcrialcl Ill~1 easo in cui la tensione ammissibile in regilne rTIolloassiale valga k 190 MPa~

5 Determinare quale sarebbe in base al criteria di sicurczza di rrcscal il 111aSsil110 sl()rzo piano (Jx == (J(2) (Jy == -2(J(2) (valori espressi in MPa) sopportuhilc dal nlutcrialc llcl caso in cui la tensione ammissibile in regime manoassiale valga k == 190 MPa~

6 Determil1are quale sarebbe in base al criteria di sicurczza di von Miscsl il nlHssinlo sf())zo piano (Jx == (1(3) (Jy == -2iT(3) (valori scmprc esprcssi in MPa) sopportahilc dnl nlatcrialcl ncl caso in cui la tensione amn1issibile in regime monoassialc valgu k _- 190 MPH

lJx = bullbullbull(J$ ~2~plusmn~ mm L= == bullbullbullbull(S Qf-2~ bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull 111111

V)v- ---0 ( ~JmiddotICmiddot ca2 bull ~(I) - 95(ucJpound~) MI)fl(V - - bullbullbullbullbullbullbullbullbullbullbullbullbull -r~ bull~~ bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull1( ( (2) = _ 3s ) (3) == c )t bull~ 1 l ~ 3Z65 3 3 MI a (1 bullbullbullbullbullbullbullbullbullbull bullbull bullbullbullbull 1 MI 1 I(1 bull bullbullbullbull bull bull 2 i

Scritto 16092014 Parte II - Testo 2 pag3

middot__------- middot---------cmiddotmiddot

+-B-~ 4th

poundb

I

I 4

f

- - - -

(f) c II

~bI -) II

I I

II

29 6 I I I

iG~

Hc (cent) = (g~ ~ (11) = q Ho (cent) = 2T 2~l- Mo (Jgt) = 1(q6~

- NfI = middotmiddotmiddotmiddotmiddotmiddotmiddot21middot~middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot TAli = middotmiddotmiddotmiddotmiddotmiddot5imiddot-Kimiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot M1II =bull~bull~amp

~ - - - middot - - -0 - NII - lJly Ilc - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull MII - bullbullbullbullbull1 f(b 2 ~ 4 h yen ~L ~

_ _ -r- bull _ _Lmiddot ~ +PJCmiddot ~~ No( - Q To( - bullbullbullbullbullbullbullbull~2gb Mol - bullbullbullbullbullbullbullbull(gt~b ~ bull~ I

l - ~middot--r cL~

Itt J1

- -~ ~ UA - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull ~ bullbullbullbull[ bullbullbullbullbullJ Scritto 16092014 Parte II - Testo 2 pag4

CORSO DI STATICA E SCIENZA IlELLE (()STI~lJZI()NI

AA 20 13-20 14

Prova scritta ill aula del 16()92() 14 Parte II - -resto J

I(~d S Edilizia D (~dS Ad( I I (IS SdA I

Nota I risultati numerici vanno riportali a penna su queslo slessofi)~io nei riqulJliri lJreliilr)()sli i (alcoli (informa ordinata) vanno alleKali sui solifhKli a quadrelli che sono SlllliiJrnili

IAllievo e-mail Matricola middot1

Escrcizio D 1 (1 7 punti)

isolvere mediante il Principio dei Lavori Virtuali (PL--IV) la struttura ipcrstatica ri portata in I~ igllra~

assumendo~ come incognita iperstatica la coppia in D Mo l)opo avere detern1inato liperstatica tenendo conto solo della defiJrmahililaflessi()n(lle~ calcnlure Ie reazioni vincolari Ie azioni interne e tracciare nello spazio predisposto nella pagina a fr()l1te i corrispondenti grafici Calcolare infine riapplicando il PLV 10 spostamento orizzontale del punto A ~ lJ

~Si rammenta che il diagramma del momentoflettente va riportalo dalla Jarle dellefihre lese


b

b

---I ~

ltpW 3b xuHp

Scritto 16092014 Parte 11- Testo 3 pagl

- -

Esercizio n 2 (7 pUl1ti)

Perla struttura isoslalica indicata in l~igura dctcrnlinare Ie reHzioni vincolari e resprcssiolll~ dvllv azior1i inter11e nonche Ie condizioni a] contorno inlposte di vincoli Ilei J1l1nti 1 11 c ( Utilizzare quindi lequazione della linea elastica per deienllilltlrc

1 La deformata della linea dassc v(z) == Vt(ZI) u V2(Z2)~

2 La sua derivata prilna v(z) == VI (ZI) U V2(Z2)~

3 Lo spostamento verticale del punia 3 Vn

4 Lo spostamento verticale del punio (l V(


2qYVVq

A lllllllllllllllllllllllllllllllll~ cI I iii

t-~4 2

t~~~ 1~ 3b 2b xuHp

HA (cent) = a MA (p) =~9b VII (11) = ca~

0fl g 2 ib2 - 0- middot - -- ~ - b -- ~ NAil - TAli - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull- Jbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull M1I1 - bullbullbullbullbullbullbullbullbullbullbullbull~ bullbullbullbullbullbullbullbullbullbullbull1 1bullbullbullbullbullbullbullbullbullbullbullbullbullbull

~ if r)Nu(middot = Q TI( = ~ Mj( = ltbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull ~

r-A

) ~11t =3b r-gt (12 =0) ~ltJ middot J =n - ~) middot lr- I raquo cc In A = lJ Qbullbullbullbull~~ bullbullbullbullbull~~ bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullc In B = ~i1i i~) =~oi7amp~

VH bullbullbullbullbullbullbullbullbullbullbullbull bullbullbull bullbullbullbullbullbullbull bull bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull ( bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~bullbullbullbullbullbullbullbullbullbullbullbulli Jbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull

VI (Z ) _-113S blr

J bull ~ shyoftbullbullbullbullbull6te

9 ccb ~~ - middotmiddotjmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot~middot1~middot 1 3 1 ~ +- An qlL ( ) _ - ~ 1s=-n i- ~ ~ t1J bullbull ~ bullbull~=3bullbullbullbullbullVI Z - bull_qbullbullbullbullbullbullbullbullbullbullbullbullbullbullongtbullbullbull_bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull

V2(Z2) _ -

_ JEt ~~3i2 ~

_ _ tB~gts V2 (Z2) - ~ bullbullbullbullbullbullbullb~17J middot middot middotmiddot _Q middot V - - 56 q~ ~ middot



lJI1 solido di forma cubica di spigolo r 95 n1111 COil Ic f~lCCC parallcle agli assi del sislc- In di riferimento ecostituito di un matcrialc caraltcrizzato dallc cosl~lllli claslichc scgucn1i ( I I ~ ( IP~I

( == 50 GPa I~ soggetto a uno sta10 di sforzo piano cOlnplclllllcnlc illdividullo ltItl ~Illsll cOlnponenti (Jx =--= +70 MPa~ ffy == -140 MPa~ (f 0 MPL Si richiede di

1 Determinare la lunghezza a deforn1azionc avvcnuta dcllo spigolo originarianlcillc palHlllln alIa direzione dell asse x Lx

2 Determinare la lunghezza a dcfornluzionc avvcnuta dcllo spigolo origil1arianlcntc pamiddot~d leln alIa direzione dell asse z 11 ~

3 Valutare la variaziol1e di VOIUlllC (rapportata al VOlUI11C il1izialc) chc il cliho suhis-l Pl)

effetto della deformazione 4 Detern1inare quale sarebbe in base al criterio di sicurczza di (ulilci-lankil1c il 111assilllo

sforzo piano (Jx == (J( I) (Jy == -2(J( I) (valori esprcssi in MPa) sopportabi Ie dal 111atcrial e Illl caso in cui la tensione ammissibile in regime Inonoassialc valga k == 180 MPa

5 Determinare quale sarebbe in base al criterio di sicurezza di rrrcscal il 111assinl0 slltuzo piano (Jx == (J(2) (Jy == -2(J(2) (valori espressi in MIla) sopportabilc dal ll1ateriale nel caso in cui la tensione amnlissibile in regime nl0noassialc valga k = ] 80 MPa~

6 Determinare quale sarebbe in base al criterio di sicurezza di von Misesl iI 111aSsinl() s()rzo

piano (Jx == (J(3) (Jy == -2(J(3) (valori sempre espressi in MPa) sopportahi Ie dal 111aterial e I1l~1 caso in cui la tensione ammissibile in regilne monoassiale valga k = 180 MPa

IJx == bullbullbullbullbullbullbullbull ~~lQ1S2 mm 11= == bullbullbullbullbullbullbull~Sro8t 111111

(V V)v- -0 nCiF-)At6 bull (1)- af()--~_)rc~ MI) - - bullbullbullbullbullbullbullbullbullbullbullbullbull Jbullbullrebullbullwt~bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull bullbull (J - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullmiddotJ~ middot~hmiddot bullbulli~bullbull I

1T(2) - ~~ rr--C10 MPa (J(3) - ~()2 MI)Imiddot( - Q~t~~ - JwltbullbullJL~ C I


C9 __---------~II------t-----------shy

~~~~~

5jhL-)

5lb2

~ (fr) = Hn (q) =JStff Mn (rlraquo = ~~~ ~

_ r~b 0 1~ - - 3qxJ oM _ _ ~cy ~z middot NAN - middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotJmiddot-rmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot AN - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbull bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull liN - middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotX~middotmiddotl~(middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotl

b 5 b c f2 c ~ - - ~~ 0 7~ - rid 0 M - - ~gt c~ + d--t 0NIU - bullbullbullbullbullbullbullbullbullbull middotmiddotmiddotf middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot N( - bullbullbullbullbullbullbullbullbull bull Imiddot - bullbullbullbullbullbullbullbull bullbull middot1middot l bull bullbullbullbullbullbullbullbullbullbullbullbullbullbull IN(

~ S b2 ~I Nn( = Q 7~)(middot = middot~1~ M)(middot = middotmiddotmiddotmiddotmiddot-middot~middot1middot -plusmnmiddot~~middot6D~

UA = =1 ~2 Scritto 160920 14 Parte II - Testo 3 pag4

DICAAR - l~acolta di Illgcgllcricl-Arcllitcttllr~l

CORSO DJ STATJCA E SCIENZA I)ELLI~ (()S~I~IlJZI()NI

AA 20 13-20 14

Prova scritta in aula del 16092() 14 Parte II - Icsto 4

(ltis Edilizia D (~dS Ad(~ I I (dS SdA I I

N(tcJ I risultati numerici vanno riporlali a penna su lueslo slesofiJKlio nei rilfuliliri Irellis~(si i calcoli (informa ordinala) vanno allegali sui solifiJ~li a luadrefli (he sono SllifiJrnii

IAllievo e-mail Matricola middot ]

[~sereizio D 1 (17 punti)

Iisolvere mediante il Principio dei Lavori Virtuali (IgtlJV) la struttura ipcrstatica riportata in 14igura~

assllmendo come incognita iperstatica la coppia in [J Mo [)opo avere determinato Iiperstatica lenendo conlo solo della defiJrmahilillflessionllle calcolarc Ie reazioni vincolari Ie azioni interne e tracciare nello spazio prcdisposto nella pagillu a fr()nle i corrispondenti grafici Calcolare infine riapplicando il PLV 10 spostamento orizzontale del punto A 1(

~~1i rammenta che il diagramma del momentofletlenfe va ri)orfafo dalla Jarfe dellefihre lese


B _

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LshyL-shyLshyi shy L-

b L--shyL___ L-shyL __ _ - Lshy

~4q LshyLshyLshyLshy L_ __ b L-_ L-shyLshy 1 L-shyI bull Lshy

_ middotmiddot--4-1-------------- __ ----+-----+

ltpW 3b xuHp

Scritto 16092014 Parte II _ Testo 4 pag1

I~~ercizio D 2 (7 punti)

PeIla struttura isotalica indicata in Figura deternlinarc Ie rcazioni vincohlri c IcsprcssiollL dllll aJlioni interne nonche Ie condizioni al contorno inlpostc thli vincoli lci pUllti if 1 c (1

tJtilizzare quindi lequazione della linea elastica pcr dctclIllinarc

] La deformata della linea dassc v(z) == VI(ZI) U V2(Z2)~

2 La sua derivata prilna v(z) == VI(ZI) U V2 (Z2)~

3 Lo spostamento verticale del puntoB Vn~

4 Lo spostamento verticale del punto (1 V(middot

Universitadi Cagliari SdC_SdA 160914004

yvVq

lilA llllllllllllllllllllllrlllllllllllllllllllll~ bull c

t1~~ 1Y1~~ ---------+----------------~I----------4-_--+

lpW 4b b xuHp

HA (cent) = 0 MA (amp-) = l-rgltEfhfbull VII (It) = ~11 _ ~ ~2 s ~2- 0 - - - -

NAil - 1AU - bullbullbullbullbullbullbullbullbullbullbullbullbull5~ ~tl MAli - bullbull bullQ middot 2~bullbull

Nul = 0 Tu( = () Mu( = 0 1

cc in A = 1Gltl~Q)~~Q~ c in B = f~~~~~~~J~~~~~ _ ~~4_ 20 c~~~~middot5tlmiddot~~middotmiddot~=middot~middotbZ~~pound $3

VI (z I) - middotbullbullbull3middotmiddotmiddotmiddotmiddotE3middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot~-rymiddotmiddotmiddotmiddotmiddotPtmiddotmiddotmmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot VI (z I) - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullm euro -m V2(Z2) = ~ ~~)~ V2(Z2) = ~-~~ l

- Q v - -- ~ ~~ 1--1-) VH - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull ( - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull$ 4Jt middotmiddotmiddotmiddotmiddot middotmiddotmiddot shy

ScnttoI6092014 Parte 11- Testo 4 pag2


lJn solido di forma cubica di spigoJo 11 == 75 111n1 con Ie Illcce parallele agli (Issi del siste 1111 di riterilnento ecostituito di un ma1criale caratt~rizzato (hlll~ costallti ~lltlstiche sl~glll~llli 1 11() (PL

( = 50 GPa E soggetto a uno sta1n di sforzo piano conlpl~tlI11ellt~ individultlto lt1lt1 t)tllsll

C0111p011enti ax == +90 MPa ay == -] 80 MPa (J ~ 0 MPa Si ric11iede di

] Deterlninare la lunghezza a deformazionc avvcIl1l1a dello spigolo originaritlll1cnte palallllo alia direzione dell asse x IJx ~

2 Determinare la lunghezza a dcrornlazion~ avv~lluta d~lIo spigolo origil1arialllcllt~ PHId Illn alIa direzione dellasse z 1-1~

3 Valutare la variazione di volulne (rapportata al volllllle iniziale) che il cuho sllhislt~l Pll

effetto della deformazione 4 Detern1inare quale sarebbe in base al criterio di siclIrczza di (ialilci-Iunkinc il 111assilllo

sforzo piano (Jx == (J(I) (Jy == -2a(l) (valori esprcssi in MPa) soppor1abilc dnl 1l1a1criall nel caso in cui la tensione an1missibile in regime monoassialc valga k ~ 230 MPa~

5 Detern1inare quale sarebbe in base al critcrio di sicurczza ltIi rrrcscul il 111USsil110 ~()Jzo

piano (Jx = (J(2) (Jy = -2(J(2) (valori espressi in MPa) sopportabitc dal tnatcrialcl net caso in cui la tensione ammissibile in regime monoassiale valga k = 230 MPa~

6 Determinare quale sarebbe in base al criteria di sicurezza di von Miscs il nlassin10 Si(lrZO

piano (Jx == (J(3) (Jy = -2(J(3) (valori sempre espressi in MIla) sopportabilc dal 111atcrialc Ilel caso in cui la tensione ammissibile in regime 111onoassiale valga k == 230 MPu

l-1x = ~151Q~l8B mm L= = x$QfJ3 111111

V)V= -([)I 000 -crJO bull (I) == (1 Sgt7 [)0 () MI) (V - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~~1-dbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull bullbullbull (1 bullbullbullbullbullbullbullbullbullbullbullbullbull bullbullbullbull1 1

(2) - 1( 61 t7_ MPa (1(3) - (Q~- 9ltJQ MI)(1 - bullbullbullbullbullbullbull ~1~bullbullbull1bullbullbullbull~~ - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~~~() I

Scritto 16092014 Parte II Testa 4 pag3

- ~b I L

--1

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He- (cent) = lI2~b ~- (D) = 2Iifb Hn (cent) = lujb Mn (~) = =tlJfb ~

N AJj = middotmiddotmiddotmiddot2middotmiddot1middotbmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot T AJj = 1114 MAJj = 2qtI~ - B b 2 ~j

NJj( = a~ TJj( = ~b MJj( = j ~4 6 d 3 t2 f 2 L 1s J 0

Nn( = amiddot8middotmiddotmiddotmiddotmiddotmiddotmiddotbmiddot4middotmiddotmiddotmiddotmiddotmiddotmiddot Tn( = middotmiddotmiddotmiddot-pound-middotmiddottful Mn( = middotmiddotmiddot-middot~middotmiddotrbmiddotmiddotmiddotmiddotmiddotmiddot~~~~bl~middotmiddotmiddotmiddotmiddot

_ -- - ~L 14-)U1 - bullbullbullbullbullbullbullbullbull abullbullbullbullbull~T) bullbullbullbullbullbull bullbullbullbullbullbull bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull

Scritto 16092014 Parte II - lesto 4 pag4

)(sereizio ll 2 (7 punti)

P~r Ia struttura isostatica indicata in Figura dcternlinare Ie reazioni villcolari c rcsprLssiollC dl1 h ltlioni interne nonche Ie condizioni al contorno inlposte dai vincoli nci punti 1111 e ( I Jtilizzare quindi lequazione della linea elastica pcr detcnninarc

1 La deformata della linea dasse v(z) == V(ZI) U V2(Z2)~

2 La sua derivata prima v(z) == VI(ZI) U V2(Z2)~

3 Lo spostalnento verticale del punto 13 VN~

4 Lo spostamento verticale del punto ( V(


yvVq

bull c

cpW 2b 3b xuHp

HA (~) = Q MA (~) = 6~k VII (fr) = 6ampbull f 2

Q - 3a~ll M bOt _ ~ ozNAn = bullbull TAn = middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot1middotmiddotmiddotmiddot An = middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotrrJmiddotmiddotmiddotmiddotmiddotmiddotmiddot-flmiddotmiddotmiddotmiddotrmiddotmiddotmiddotiIebull ~

Nllc = bullbullbullbullbullbullbullbullbullbullt2 TI(= bullbullbullbullbullbullbullbullbullbullbullbullQ MI(middot= bullbullbullbullbullbullbullO

Cc in A = y~f~~Q)no Cc in B = ~LfF4 bull2b)J2tezmiddot~rl=Q I ~ (Ir- )

_ v (Cc=2b)middot J2 ~2 =0it cc In C - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~

_ lo~_ 3~~g2 + f ~4 _ --6 ~balt -f- ~ iJJ3 VI (ZI) - bullbullbullbullbullbullbullbullbulll abulltQ g m v I (z I) - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullliObullbullbullbull$

_ _ 8 ~gg-2- _ _ 8~- V2(Z2) - bullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbullbull~J V2 (Z2) - bullbull middotmiddotmiddotmiddotmiddotmiddotbmiddot~middotmiddotmiddotmiddotmiddotpoundt)middotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddotmiddot VII = bullbullbullQ v(middot = =2~ ~ ~ t)

Scritto 16092014 Parte II -l~esto I pag2















(1(2) = ~3J~~$3 MPa (1(3) = ~llt3 M Pa



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_ middotmiddot--4-1-------------- __ ----+-----+

ltpW 3b xuHp










yvVq


t1~~ 1Y1~~ ---------+----------------~I----------4-_--+

lpW 4b b xuHp



Nul = 0 Tu( = () Mu( = 0 1


















l-1x = ~151Q~l8B mm L= = x$QfJ3 111111




- ~b I L

--1

e I I I

~















yvVq


t1~~ 1Y1~~ ---------+----------------~I----------4-_--+

lpW 4b b xuHp



Nul = 0 Tu( = () Mu( = 0 1


















l-1x = ~151Q~l8B mm L= = x$QfJ3 111111




- ~b I L

--1

e I I I

~




















l-1x = ~151Q~l8B mm L= = x$QfJ3 111111




- ~b I L

--1

e I I I

~







- ~b I L

--1

e I I I

~







corso statica e scienza i)ellf~...

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