Despre Rezervoare

N= 8

REZERVOR CIRCULAR PRECOMPRIMAT DIN BETON ARMAT PENTRU INMAGAZINAREA APEI POTABILE

Alcatuirea, calculul si dimensionarea structurii unui rezervor circular precomprimat pentru inmagazinarea apei. Se vor elabora calcule pentru determinarea starii de eforturi si a modului de armare a pricipalelor elemente structurale.

2.2 inaltime apa: Ha= 5.00 + 0.05*N = 5.4 m;2.3 nivelul terenului natural se gaseste la cota +1.50m fata de cota de fundare a rezervorului;

0.25 m;

3.1 predimensionare rezervor;3.2 stabilire incarcari;3.3.1.1 determinarea starii de eforturi pentru ansamblulul structural perete cilindric - inel din actiunea

greutatii proprii( GP);3.3.1.2 determinarea starii de eforturi pentru ansamblulul structural perete cilindric - inel din actiunea

presiunii hidrostatice( Ph);3.3.1.3 determinarea starii de eforturi pentru ansamblulul structural perete cilindric - inel din actiunea

precomprimarii;3.3.2.1 determinarea starii de eforturi pentru greutate proprie + presiune hidrostatica;(G.P.+P.H.)3.3.2.2 determinarea starii de eforturi pentru greutate proprie + precomprimare(rezervor gol);(G.P.+P.R.)3.3.2.3 determinarea starii de eforturi pentru greutate proprie + presiune hidrostatica + precomprimare

(proba de etanseitate);(G.P.+P.H.+P.R.)3.4 verificarea conditiei de stabilitate generala a ansamblului din actiunea seismica, a ansamblului talpa

inelara de fundare - perete cilindric;3.5 dimensionarea si armarea ansamblului structural perete cilindric - inel la starea limita de rezistenta

si starea limita de deschidere a fisurilor;3.6 plan de situatie;3.7 plansa armare si cofraj pentru ansamblul strutural perete cilindric - inel.

5000

5.4

35 m

0.25 m

17.5 ml = 0.3 m

3.41 m nr. chesoane: n =2*p*Ri' / l = 72 chesoane

L = 1.52 m

Re = 17.37 ml = 0.3 m

nr. strat h

1 placa cheson 5 2.5 0.132 sapa egalizare(mortar) 2 2.2 0.043 bariera vapori(1c+2b) 0.01 1.14 termoizolatie(polistiren) 0 1.2 I = 2…7

5 sapa protectie(1b+1c) 0.01 1.4

6 hidroizolatie(2p+1c+4b) 0.02 0.17 strat protectie hidroizolatie(nisip) 0

1o Obiectul temei

2o Date de tema

2.1 capacitate rezervor : V=5000m3;

2.4 grosimea peretelui cilindric : hpc

=

3o Se cere :

Vi = m3

Ha = m3

Di = ( 4*V

i / (π*H

a)) 0,5 =

hpc

=

Ri = D

i / 2 =

Ri =

2*π*Re/n=

Ri - 0.25 + 0.12 =

2*π*Ri'/n=

γ (tf/m3) qin (tf/m3)

n1 =

ni =

n8 =

q8 = tf/m3

q = 0.37

2.5

13.96 m 13.84 m

P1 = 0.4 tf/ml

0.12 m

0.35 m

P2 = 0.17 tf/ml

Q1 = 0.09 tf

Q2 = 0.02 tf

Q3 = 0.01 tf/ml

0.05 m

0.05 m

R1= 2.37 tfR2= 1.58 tf

M2 = 0 V1 = 2.33 tf

M1 = 0 V2 = 1.73 tf

v = 3.08 tf/ml

17.31 m

DETERMINAREA STARII DE EFORTURI A ANSAMBLULUI STRUCTURAL PERETE CILINDRIC - INEL DIN ACTIUNEA GREUTATII PROPRII (G.P.)

-facem o modelare a structurii reale:--modelare geometrica -reducerea placii la suprafata mediana a acesteia cupastrarea grosimii placii;--modelare fizica -comportare liniar-elastica a materialului;

Σ qi*n

i = tf/m2

γ m = tf/m3

l0 = l

0 ' =

q*L/ 2 + γ m*1.1* b

n* h

n =

bn =

hn =

q* l/ 2 + γ m*1.1* b

n* h

n =

γ m*1.1*b

n*h

n*L/2 =

γ m*1.1*b

n*h

n*l/2 =

γ m* 1.1*b

n' *h

n' *(L + l) / 2 =

bn

' =

hn

' =

V1*l0' - Q1*l

0' - R2*l

0' *2/3 - R1*l

0' *1/2 = 0

V2*l0' - Q2*l

0' - R2*l

0' *1/3 - R1*l

0' *1/2 = 0

n * V1 / (π * Rv ) =

Rv = R

i - 0.25 + 0.06 =

--modelarea incarcarilor.Placa cilindrica este incarcata in ipoteza de

incarcare G.P. de eforturile uniform distribuite urmatoare:-X1 , X2 , X3 , X4 (eforturi necunoscute)

Inelul este incarcat cu eforturile uniform distribuite X2 , X3 si X4 si efortul uniform distribuit v(provenit din greutatea acoperisului) si gi (greutatea pe m.l. a inelului).

OBS: In ipoteza de incarcare G.P. placa cilindrica este incarcata doar dupa directia generatoarei (in planul placii nu exista incarcare dupa directia normalei la planul median alplacii (directia razei))

a = 17.63 m

Ha + 0.75 + 0.10 = 6.25 mK = factor de amortizare al placii cilindrice

K= (3* (1 - m2))1/4/ (a* hpc)1/2 =K= 0.62

1/6= 0.17

3.37x4 = 3.37

0.34hi = 0.25 m

bi = 0.5 m

17.5 m

1.01 tfm/ml

0.13 m

0.19 m3.89

i Fi(k*l)1 598.37 02 599.29 03 599.83 04 598.83 05 0.46 06 598.83 07 -17.93 18 -16.67 09 1.23 0

10 -34.6 0

0A = b * h = 0.13

1548.36

1202.26

964.71

1548.36

473760

-59220

9870

473390.1

-59173.76

-gpx=g

m*h

pc*I*(1-x)*1,1 -incarcarea din greutatea peretelui

-gi=g

m*gi*bi*1,1 -incarcarea din greutatea inelului

R i + h

pc / 2 =

Hpc

=

µ =

x4 = (v* Rv+ g

i* R

Gi)/ a =

gi = γ

m* h

i* b

i* 1.1 =

hpc

+ 0.25 =

RGi

=

Mi = x

4 * b

x4 + v * b

v =

bx4

=

bv =

k*l=

F(k*l*ξ)

Ii = b*h3/12 = m4

Eδ11

c = 2 * a2 * k / h * ( F4 / F1) =Eδ

22c = 4 * a2* k3 / h

pc *( F3 / F1) =

Eδ23

c = 2 * a2* k2 / hpc

*( F2 / F1) =

Eδ33

c = 2 * a2 * k / h * ( F4 / F1) =Eδ

22i = a * R

Gii / I

i =

Eδ23

i = a* RGi

/ Ii* h

i/ 2 =

Eδ33

i = a* R Gi

i / Ii *( I

i/ A

i+ h

i2 /4) =

E∆2p

i = M i * R

GI2 / I

i =

E∆3p

i = M i * R

GI2 / I

i * h

i / 2 =

0 x1 = 0

0 ; 474962.26

0 ; 11418.36

-58255.29

474962.26 -58255.29 473390.1 = 0

-58255.29 11418.36 -59173.76 = 0

x3 = 0.26x2 = -0.96

Pentru a determina starea de eforturi din actiunea G.P. vom incarca placa cilindrica cu momentul X2= 1,291008768 tfm/ml , cu efortul uniform distribuit radial X3=0,26165208 tf/ml si cu incarcarile exterioare X4 =4,19866 tf/ml si gp=0,6875 tf/m2 .

Folosindu-ne de principiul suprapunerii efectelor gasim starea de eforturi pentru placa cilindrica pentru ipoteza de incarcarea greutatii proprii .

Pentru inelul circular:

0.26 tf/ml

0.97 tfm/ml

0.03 tfm/ml

1.01 tfm/mlmt=m2+m3-mi= 0 tfm/ml

-4.58 tf/ml

-0.03 tfm/mlverificare:

-528.07

-528.07

528.07 -528.07 =0 ==>O.K.

este torsorul fortei (rezulta din reducerea fortelor exterioare aplicate pe inel in centrul de greutate al inelului

mt -efortul uniform distribuit pe cercul de raza RgiValoarea momentului incovoietor intr-o sectiune transversala pe inel

centrul de greutate al inelului dat de fortele exterioare

DETERMINAREA STARII DE EFORTURI A ANSAMBLULUI STRUCTURAL PERETE CILINDRIC - INEL DIN ACTIUNEA PRESIUNII HIDROSTATICE (P.H.)

Datorita faptului ca cilindrul este lung rezulta ca detetminarea necunoscutelor X2 si X3 se face folosind urmatorul sistem de ecuatii de compatibilitate:

0 ;

0 ;

-1073.57

5.4 tf/m2

Eδ11

c*x1 + E∆

1pc =

Eδ22

"*x2+ Eδ23

"*x3+ E∆2p

" = δ22

" = δ22

c + δ22

i =

Eδ32

"*x2+ Eδ33

"*x3+ E∆3p

" = δ33

" = δ33

c + δ33

i =

δ32

" = δ23

" = δ23

c + δ23

i =

* x2 * x

3 +

* x2 + * x

3 +

P=X3*a/RGi

=

m2=X2*a/RGi

=

m3=X2*(a/RGi

)*(hi/2)=

mi=v*bv+X4*b

x4=

NQi=P*R

Gi=

Mi=mt*RGi

=

Edri= -P*RGi

2/A+mt*(RGi

2/I)*(hi/2)=

Edrc=Ed32

c*X2+Ed33

c*X3+ED3P

c=

Edri - Edrc==>

Edri - deplasarea radiala in punctul A pe inel

Edrc - deplasarea radiala in punctul A pe cilindru

P -efortul uniform distribuit pe cercul de raza RGi

Valoarea fortei axiale NQi

in inel va fi egala cu P*RGi

unde P

este calculata cu relatia Mi=mt*RGi

, unde mt este momentul incovoietor in

Eδ22

"*x2+ Eδ23

"*x3+ E∆2p

"PH =

Eδ32

"*x2+ Eδ33

"*x3+ E∆3p

"PH =

E∆2p

"PH = - po*a2/(h

pc*l) =

po=ga*Ha=

E∆3p

"PH = 0

474962.26

11418.36

-58255.29

474962.26 * x2 -58255.29 * x3 + -1073.57 = 0

-58255.29 * x2 + 11418.36 * x3 + 0 = 0

x3 = 0.03 tfm/mlx2 = 0.01 tf/m

4.34 tf/m


0.03 tf/mlmt=X2*a/Rgi-X3*(a/Rgi)*(hi/2)= 0 tfm/ml

-0.54 tf/ml

0.04 tfm/ml

DETERMINAREA STARII DE EFORTURI A ANSAMBLULUI STRUCTURAL PERETE CILINDRIC - INEL DIN ACTIUNEA PRECOMPRIMARII (P.R.)

1.42 tf/m2a= 17.63 m

0.25 cm

6.82 tf/m2

are urmatoarele valori :

1073.57

-282

1548.36

1202.26

964.71

1548.36

473760

-59220

9870

δ22

" = δ22

c + δ22

i =

δ33

" = δ33

c + δ33

i =

δ32

" = δ23

" = δ23

c + δ23

i =

X1= (p0/2*k)*(F4/F1) =

P=X3*a/RGi

=

NQi=P*R

Gi=

Mi=mt*RGi

=

Valoarea incarcarilor se determina pornind de la faptul ca in placa cilindrica sa avem un efort NQ de compresiune si un

efort normal remanent sQ

remanent =10 daN/cm2

p1

PR=100*hpr/a=

hpr=

p0

PR=p1PR+p

0

In cazul de incarcare al precomprimarii , din incarcarile exterioare , eforturile Nx , Qx si Mx pot fi neglijate , doar NQ

NQ =p1PR*a+po*(1-x)*a

ED2p

c,PR= po*a2/(h

pc*l) =

ED3p

c,PR=- p1PR*a2/(h

pc*l) =

ED2p

i,PR=0

ED3p

i,PR=0

Eδ11

c = 2 * a2 * k / h * ( F4 / F1) =Eδ

22c = 4 * a2* k3 / h

pc *( F3 / F1) =

Eδ23

c = 2 * a2* k2 / hpc

*( F2 / F1) =

Eδ33

c = 2 * a2 * k / h * ( F4 / F1) =Eδ

22i = a * R

Gii / I

i =

Eδ23

i = a* RGi

/ Ii* h

i/ 2 =

Eδ33

i = a* R Gi

i / Ii *( I

i/ A

i+ h

i2 /4) =

###

11418.36

δ22

" = δ22

c + δ22

i =

δ33

" = δ33

c + δ33

i =

-58255.29

474962.26 * x2 -58255.29 * x3 + 1073.57 = 0

-58255.29 * x2 + 11418.36 * x3 + -282 = 0

x3 = 0.04 tf/mlx2 = 0 tfm/ml


0.04 tf/mlmt=X2*a/Rgi-X3*(a/Rgi)*(hi/2)= 0 tfm/ml

-0.62 tf/ml

-0.04 tfm/ml

DIMENSIONARE CORDON ELASTIC Lc1= 6 cm = 0.06 mHc= 5 cm 0.05 mf=(Hc*P*1,2)/[2*(1+μc)*K1*Gc*Lc1]

4.12 tf/m2 (din diagrama G.P.)μc 0.47K1 1.1Gc= 95 tf/m2f= 0.01 m = 1.34 cm

k= 0.54Eb= 3500000 tf/m2 (BC30)hpc= 0.25 m a= 17.63 m

6.82 tf/m2Lc=2*Lc1= 12 cm = 0.12 m

0.67 tf/ml

ARMARE ANSAMBLU STRUCTURAL PERETE CILINDRIC - INELArmare inel

I- pentru GP :

-4.58 tf/ml

-0.03 tfm/ml

II- pentru GP+PR :

-5.2 tf/ml eforturile la care dimensionez inelul

-0.08 tfm/ml

III- pentru GP+PR+PH:

-5.75 tf/ml

-0.04 tfm/ml

-solocitarea in inel este de compresiune excentrica ==>schema de calcul:

-5.2 tf/ml 5203.61 N/ml - compresiuneMi= -0.08 tfm/ml 751128.02 N*mm = 0.75 kN*m

b= 500 mm 0.5 mh= 250 mm 2.5 m

a= 40 mm = 0.4 m

ho= 210 mm

0Rc= 15 N/mm2

δ32

" = δ23

" = δ23

c + δ23

i =

P=X3*a/RGi

=

NQi=P*R

Gi=

Mi=mt*RGi

=

P=Nxmax/2=

FL=p

0RP/[2*k+(Hc-f)*Eb*h

pc/Lc*Gc*a2]

p0

PR=

FL =X1

NQi=P*R

Gi=

Mi=mt*RGi

=

NQi=P*R

Gi=

Mi=mt*RGi

=

NQi=P*R

Gi=

Mi=mt*RGi

=

NQi=

1. Se apreciaza a

2. Sectiunea este intr-o zona plastica potemtiala la solicitari seismice? DA3. Ho=h-a

4.x=N/b*ho*Rc

x=

5. xb=0,55 -PC52

6. x<xb ? DA

8. x<0,4 ? DA

Nx

FL=X1

10. ea=h/30 =20mm

8.33 mm = 8.33 <=20 mm ==>

20 mm

Mc= ### N*mm

n=N/(b*h*Rc)= 0

0a/h= 0.16

0.01

Aa=A'a= 68.75 mm2

103.13 mm2Ra= 300 N/mm2

distantele maxime si minime intre bare.Aa= 314.16 mm2 (4Φ10)A'a= 314.16 mm2 (4Φ10)

Aa(total)/sectiune= 628.32 mm2

p(pe o latura)= 0.25 %p(total)=Aa(total)*100/(b*h)p(total)= 0.5

pmin= 0.2 %pmax= 2 %

Armare perete circular:a) stabilirea incarcarilor necesare din precomprimare

1.42 tf/m2a= 17.63 m

0.25 cm

6.82 tf/m2

b) stabilire numar nervuri de ancorare-se stabilesc din doua conditii: *numarul nervurilor trebuie sa fie minim 4, dar par *lungimea unui fascicol sa nu depaseasca 40---45 m pentru a nu creste exagerat de mult pierderile de tensiune din frecareAleg:numar nervuri= 8numar fascicole= 4==>lungime fascicol=2*π*a/4= 27.69 m <40---45mPentru a uniformiza incarcarile din precompreimare dealungul unui cerc paralel, ancorarea fascicolelor se decaleaza de la rand la

rand dupa urmatoarea regula: fascicolele dupa randurile fara sot se ancoreaza in nervurile cu sot

forta de tragere pe fascicol Npk Npk=σpk*Ap , unde:σpk - efortul unitar de control in armatura σpk= 0,9*Ra 11520 daN/cm2Ra= 12800 daN/cm2Ap -aria sarmelor din fascicolAp= 9.24 cm2Npk= ### daN = 106.402 tf

c) determinarea capacitatii portante a unui fascicol in faza finala:Daca facem media eforturilor unitare pe armatura dealungul unui cerc paralel rezulta forta pe care contez dupa ce s-au produs

ea=

ea=

11. Mc=M+N*ea

12. Se calculeaza :

m=M/(b*h2*Rc)=

13. Din tabelul 9 se scoate a

a=14. Aa=A'a=α*b*h*Rc/Ra

AaSLDF=A'aSLDF=1,5*Aa=

15. Se stabileste Aa si A' aefectiva si dispozitia lor cu respectarea regulilor constructive privind diametrele minime de bare si

16 Se calcleaza Aa(total)/sectiune

17. p(pe o latura)=Aaef*100/(b*h)

18. Din tabelul 6B se scot valorile :

19. p(total)<pmax ? DA20. Se verifica: p(pe o latura)>0,2% ? DAp(total)>pmin ? DA21. Armarea stabilita la 15. este buna

Valoarea incarcarilor se determina pornind de la faptul ca in placa cilindrica sa avem un efort NQ de compresiune si un

efort normal remanent sQ

remanent =10 daN/cm2

p0

PR=100*hpr/a=

hpr=

p1

PR=p1

PR+p0

-pentru precomprimare folosesc ancoraje inel si con 24Φ7

Aa

A'a

4 10

4 10

8064 daN/cm2

74481.38 daN = 74.48 tfd) determinarea numarului de randuri de fascicole necesare:

P=(po+p1)*l/2l= 6.25 mP= 25.74 tf/m

453.67 tf -numarul de randuri:

n'= 6.09 ==> n=10 randuri

e)distributia fascicolelor pe structura:

74.48 tf

4.23 tf/m

0.5 1.42 *x1 -4.23 = 0X1= 1.816 mpx1= 3.235 tf/m2Px1= 4.23 1.211 m fata de partea superioara

0.5 3.235 *x2 -4.23 = 0x2= 1.114 mpx2= 4.349 tf/m2Px2= 4.23 2.559 m fata de partea superioara









l'=x1+x2+x3+x4+x5+x6+x7+x8+x9+x10= 7.884 m

pierderile de tensiune in starea finala σpp

med,f

σpp

med,f=0,7*σpk

σpp

med,f=

Ncap

fasc=σpp

med,f*Ap

Ncap

fasc=

NQtotal=P*a -compresiunea totala ce trebuie introdusa pe intreaga structura

NQtotal=

n=NQ

total/Ncap

fasc

px1= po+γa*x1

Px1=[(p0+ga*x1+po)*x1]/2 =po*x1+0,5*ga*x12

Px1*a=Ncap

fasc =

Px1=Ncap

fasc/a=

*X12 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

*x22 +

tf/m la 2/3*x1'=

x1Px1

p1

px1

a=23.125

l=7.075

C.Gx1

P

Desfasurata peretelui cilindric:

Armare perete circular:

-2.57 tfm/ml 256650.6 daNcm/ml

7.26 tf/ml 7258.69 daN/mlh=hpc= 25 cmb= 100 cma=a'= 4 cmho=h-a= 21 cmha=ho-a'= 17 cmeo=M/N= 35.36 cmea=h/30= 0.83 cm==>ea= 2 cmeoc=eo+ea= 37.36 cmMc=N*ea= 14517.38 daNcmx=N/b*Rc= 0.47Rc= 155 daN/cm2

0.02Ma=Mc+N*ha/2= 76216.22

Aa=Aa'=Ma/(Ra*ha)-N/Ra= -0.93 cm2Ra= 3000 daN/cm2Amin=pmin*b*ho/100= 4.2 cm2pmin= 0.2 %

6.28 cm2

9.42 cm2

11.31 cm2

M=Mxmax(GP+PR+PH)

=

N=Nx(GP+PR+PH)

=

x=x/ho= <0,6 ? DA

x>2a' ? NU

Aaef

SLR=8Φ12/ml=

AaSLFD=1,5*Aaef

SLR=

Aaef

SLDF=10Φ12/ml=

N1

N8

N7

N6

N5

N4

N3

N2

N3

N2

N1

N8

N7

N6

N5

N4

fascicolele de pe fascicolele de pe

N1 N2 N3 N4 N5 N7 N6 N8

UNIVERSITATEA TEHNICA DE CONSTRUCTII BUCURESTI

Facultatea de Hidrotehnica

PROIECT

CALCULUL STRUCTURILOR EDILITARE

1.1. incarcarea cu x2= 0.96 tfm/ml

x x1= 00 0 -0.73 0 0 x2= 0.96

0.1 0 -0.12 0.01 -0.01 x3= 0.260.2 0 0.52 0.01 -0.01 x4= 3.370.3 0 1.21 0.01 0.02 gp= 0.690.4 0 1.94 -0.02 0.07 l= 7.080.5 0 2.57 -0.09 0.14 a= 23.130.6 0 2.82 -0.22 0.22 k= 0.540.7 0 2.18 -0.4 0.3 kl= 3.850.8 0 -0.11 -0.63 0.340.9 0 -4.95 -0.86 0.27

1 0 -13.2 -0.96 0 i x Fi(k.l.x) Fi1 1 546.07 F12 1 546.91 F2

1.2. incarcarea cu x3= 0.26 tf/ml 3 1 547.48 F3

x 4 1 546.5 F40 0 -0.03 0 0 5 1 0.42 F5

0.1 0 -0.17 0 0 6 1 546.49 F60.2 0 -0.31 0.01 -0.01 7 1 -17.84 F70.3 0 -0.42 0.02 -0.02 8 1 -15.13 F80.4 0 -0.46 0.04 -0.04 9 1 2.68 F90.5 0 -0.35 0.07 -0.05 10 1 -32.96 F100.6 0 0.03 0.1 -0.05 11 1 0.49 F110.7 0 0.83 0.14 -0.04 12 1 546.99 F120.8 0 2.17 0.15 0 13 1 -2.71 F130.9 0 4.13 0.12 0.1 14 1 -32.97 F141 0 6.54 0 0.26 15 1 -15.14 F15

16 1 -17.82 F16

1.3. incarcarea cu x4= 3.37 tf/ml si gp=0,6875 tf/ml

x

0 8.23 0 0 0 i x Fi0.1 7.75 0 0 0 1 0 00.2 7.26 0 0 0 2 0 00.3 6.77 0 0 0 3 0 00.4 6.29 0 0 0 4 0 00.5 5.8 0 0 0 5 0 00.6 5.31 0 0 0 6 0 00.7 4.83 0 0 0 7 0 10.8 4.34 0 0 0 8 0 00.9 3.85 0 0 0 9 0 01 3.37 0 0 0 10 0 0

11 0 012 0 013 0 1

1. STAREA DE EFORTURI DIN G.P. 14 0 1

x 15 0 00 8.23 -0.76 0 0 16 0 0

0.1 7.75 -0.29 0.01 -0.020.2 7.26 0.22 0.02 -0.020.3 6.77 0.8 0.03 00.4 6.29 1.48 0.02 0.030.5 5.8 2.22 -0.02 0.090.6 5.31 2.85 -0.11 0.170.7 4.83 3.01 -0.26 0.260.8 4.34 2.07 -0.48 0.340.9 3.85 -0.82 -0.73 0.371 3.37 -6.66 -0.96 0.26

NxG.P.

X2 NQG.P.

X2 MxG.P.

X2 QxG.P.

X2

NxG.P.

X3 NQG.P.

X3 MxG.P.

X3 QxG.P.

X3

NxG.P.

X4 NQG.P.

X4 MxG.P.

X4 QxG.P.

X4

Fi(k.l.x)

NxG.P.

NQG.P.

MxG.P.

QxG.P.

3 4 5 6 7 8 9-0.2

0

0.2

0.4

0.6

0.8

1

Nx din G.P.

Nx

ξ

-8 -6 -4 -2 0 2 4-0.2

0

0.2

0.4

0.6

0.8

1

NQ din G.P.

NQ

ξ

-1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2-0.2

0

0.2

0.4

0.6

0.8

1

Mx din G.P.

Mx

ξ

-0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-0.2

0

0.2

0.4

0.6

0.8

1

Qx din G.P.

Qx

ξ

-0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-0.2

0

0.2

0.4

0.6

0.8

1

Qx din G.P.

Qxξ

2.1. incarcarea cu x1= 4.34 tf/ml

x0 0 -125.06 0 4.34 x1= 4.34

0.1 0 -72.79 -1.91 1.33 x2= 0.010.2 0 -32.92 -2.23 -0.24 x3= 0.030.3 0 -8.11 -1.8 -0.84 l= 7.080.4 0 4.29 -1.18 -0.87 a= 23.130.5 0 8.45 -0.63 -0.66 k= 0.620.6 0 8.14 -0.26 -0.4 kl= 4.40.7 0 5.93 -0.06 -0.180.8 0 3.24 0.01 -0.040.9 0 0.59 0.01 0.02

1 0 -1.99 0 0

2.2. incarcarea cu x2= 0.01 tfm/ml i x Fi(k.l.x) Fi

x 1 1 1672.720 0 0.01 0 0 2 1 1674.54

0.1 0 0 0 0 3 1 1674.420.2 0 0 0 0 4 1 1673.840.3 0 -0.01 0 0 5 1 0.910.4 0 -0.01 0 0 6 1 1673.630.5 0 -0.02 0 0 7 1 -12.390.6 0 -0.02 0 0 8 1 -38.990.7 0 -0.02 0 0 9 1 -26.610.8 0 -0.01 0 0 10 1 -51.390.9 0 0.03 0.01 0 11 1 0.29

1 0 0.11 0.01 0 12 1 1674.1313 1 26.5914 1 -51.38

2.3. incarcarea cu x3= 0.03 tf/ml 15 1 -39

x 16 1 -12.390 0 0.01 0 0

0.1 0 0 0 00.2 0 -0.02 0 00.3 0 -0.04 0 0 i x Fi(k.l.x) Fi0.4 0 -0.06 0 0 1 0 00.5 0 -0.06 0 0 2 0 00.6 0 -0.03 0.01 -0.01 3 0 00.7 0 0.06 0.01 -0.01 4 0 00.8 0 0.23 0.02 0 5 0 00.9 0 0.52 0.01 0.01 6 0 0

1 0 0.89 0 0.03 7 0 18 0 0

2.4. Incarcarea cu fortele exterioare (membrana) 9 0 0

x 10 0 00 0 95.18 0 0 11 0 0

0.1 0 85.66 0 0 12 0 00.2 0 76.14 0 0 13 0 10.3 0 66.62 0 0 14 0 10.4 0 57.11 0 0 15 0 00.5 0 47.59 0 0 16 0 00.6 0 38.07 0 00.7 0 28.55 0 00.8 0 19.04 0 00.9 0 9.52 0 0

1 0 0 0 0

2. STAREA DE EFORTURI DIN P.H.

x0 0 -29.86 0 4.34

0.1 0 12.86 -1.91 1.330.2 0 43.2 -2.23 -0.240.3 0 58.47 -1.8 -0.840.4 0 61.32 -1.17 -0.870.5 0 55.96 -0.62 -0.660.6 0 46.16 -0.25 -0.40.7 0 34.52 -0.04 -0.190.8 0 22.5 0.03 -0.040.9 0 10.65 0.03 0.03

1 0 -0.99 0.01 0.03

NxP.H.X1 NQP.H.

X1 MxP.H.X1 QxP.H.

X1

NxP.H.

X2 NQP.H.

X2 MxP.H.

X2 QxP.H.

X2

NxP.H.

X3 NQP.H.

X3 MxP.H.

X3 QxP.H.

X3

NxP.H.

m NQP.H.

m MxP.H.

m QxP.H.

m

NxP.H. NQP.H. MxP.H. QxP.H.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.2

0

0.2

0.4

0.6

0.8

1

Nx din P.H.

Nx

ξ

-40 -20 0 20 40 60 80-0.2

0

0.2

0.4

0.6

0.8

1

NQ din P.H.

NQ

ξ

-2.5 -2 -1.5 -1 -0.5 0 0.5-0.2

0

0.2

0.4

0.6

0.8

1

Mx din P.H.

Mx

ξ

-2 -1 0 1 2 3 4 5-0.2

0

0.2

0.4

0.6

0.8

1

Qx din G.P.

Qx

ξ

F1F2F3F4F5F6F7F8F9F10F11F12F13F14F15

F16

po= 6.15 tf/m2

1.85 tf/m3a= 16.23 m

3.1. Incarcarea cu fortele exterioare (membrana)

x0 0 -129.78361 0 0

0.1 0 -119.80524 0 00.2 0 -109.82686 0 00.3 0 -99.84849 0 00.4 0 -89.87011 0 00.5 0 -79.89174 0 00.6 0 -69.91336 0 00.7 0 -59.93499 0 00.8 0 -49.95661 0 0 x1= 0.730.9 0 -39.97824 0 0 x2= 0.06

1 0 -29.99986 0 0 x3= 0.5po= 6.15

2.1. incarcarea cu x1= 0.73 tf/ml 1.85

x l= 7.020 0 -15.27608 0 0.72508 a= 16.23

0.1 0 -8.69705 0 0.21038 k= 0.650.2 0 -3.75914 0 -0.05190 kl= 4.560.3 0 -0.77758 0 -0.143440.4 0 0.63606 0 -0.141870.5 0 1.04912 0 -0.102780.6 0 0.94812 0 -0.05841

0.7 0 0.65361 0 -0.02345 i x Fi(k.l.x) Fi0.8 0 0.32837 0 -0.00224 1 1 2260.16490 F10.9 0 0.02352 0 0.00529 2 1 2262.11581 F2

1 0 -0.26698 0 0.00000 3 1 2261.79504 F34 1 2261.48557 F4

3.2. incarcarea cu x2= 0.06 tfm/ml 5 1 0.97545 F5

x 6 1 2261.14036 F60 0 0.03321 0 0 7 1 -7.45157 F7

0.1 0 0.01547 0 0 8 1 -46.96423 F80.2 0 -0.00485 0 0 9 1 -39.52469 F90.3 0 -0.03244 0 0 10 1 -54.42454 F100.4 0 -0.07093 0 0 11 1 0.15474 F110.5 0 -0.11808 0 -0.01 12 1 2261.64030 F120.6 0 -0.15917 0.01 -0.01 13 1 39.51266 F130.7 0 -0.15930 0.02 -0.02 14 1 -54.41580 F140.8 0 -0.05735 0.03 -0.02 15 1 -46.97461 F150.9 0 0.23231 0.05 -0.02 16 1 -7.44993 F161 0 0.79981 0.06 0

3.3. incarcarea cu x3= 0.5 tf/ml i x Fi

x 1 0 00 0 0.18280 0 0 2 0 0

0.1 0 -0.01611 -0.00177 0.00362 3 0 00.2 0 -0.22483 -0.00303 -0.00154 4 0 00.3 0 -0.44751 0.00258 -0.01606 5 0 00.4 0 -0.64916 0.02174 -0.04000 6 0 00.5 0 -0.71832 0.06029 -0.07037 7 0 10.6 0 -0.43550 0.11979 -0.09714 8 0 00.7 0 0.53240 0.19079 -0.09821 9 0 00.8 0 2.57382 0.24289 -0.03553 10 0 00.9 0 5.95472 0.21336 0.14405 11 0 01 0 10.45928 0 0.49645 12 0 0

13 0 114 0 1

3. STAREA DE EFORTURI DIN P.R.. 15 0 0

x 16 0 00 0 -144.84 0 0.73

0.1 0 -128.5 -0.31 0.220.2 0 -113.82 -0.36 -0.050.3 0 -101.11 -0.28 -0.160.4 0 -89.95 -0.15 -0.180.5 0 -79.68 -0.03 -0.180.6 0 -69.56 0.09 -0.170.7 0 -58.91 0.2 -0.14

p1PR=

NxP.H.m NQP.H.

m MxP.H.m QxP.H.

m

p1PR=

NxP.H.

X1 NQP.H.

X1 MxP.H.

X1 QxP.H.

X1

NxG.P.

X2 NQG.P.

X2 MxG.P.

X2 QxG.P.

X2

Fi(k.l.x)

NxG.P.

X3 NQG.P.

X3 MxG.P.

X3 QxG.P.

X3

NxG.P.

NQG.P.

MxG.P.

QxG.P.

0.8 0 -47.11 0.28 -0.060.9 0 -33.77 0.27 0.131 0 -19.01 0.06 0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.2

0

0.2

0.4

0.6

0.8

1

Nx din P.R.

Nx

ξ

-160 -140 -120 -100 -80 -60 -40 -20 0-0.2

0

0.2

0.4

0.6

0.8

1

NQ din P.R.

NQ

ξ

-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4-0.2

0

0.2

0.4

0.6

0.8

1

Mx din P.R.

Mx

ξ

-0.4 -0.2 0 0.2 0.4 0.6 0.8-0.2

0

0.2

0.4

0.6

0.8

1

Qx din P.R.

Qx

ξ

4. STAREA DE EFORTURI DIN GP+PH

x0 8.23 -30.62 0 4.34

0.1 7.75 12.57 -1.91 1.320.2 7.26 43.41 -2.21 -0.260.3 6.77 59.26 -1.77 -0.840.4 6.29 62.8 -1.16 -0.840.5 5.8 58.19 -0.65 -0.570.6 5.31 49.01 -0.36 -0.240.7 4.83 37.53 -0.31 0.070.8 4.34 24.57 -0.45 0.30.9 3.85 9.84 -0.7 0.391 3.37 -7.65 -0.96 0.29

NxG.P.

NQG.P.

MxG.P.

QxG.P.

3 4 5 6 7 8 9

-0.2

0

0.2

0.4

0.6

0.8

1

Nx

Nx

ξ

-40 -20 0 20 40 60 80-0.2

0

0.2

0.4

0.6

0.8

1

NQ

NQ

ξ

-2.5 -2 -1.5 -1 -0.5 0-0.2

0

0.2

0.4

0.6

0.8

1

Mx

Mx

ξ

-2 -1 0 1 2 3 4 5-0.2

0

0.2

0.4

0.6

0.8

1

Qx

Qx

ξ

4. STAREA DE EFORTURI DIN GP+PR

x0 8.23 -145.61 0 0.73

0.1 7.75 -128.79 -0.31 0.20.2 7.26 -113.6 -0.34 -0.070.3 6.77 -100.31 -0.25 -0.160.4 6.29 -88.47 -0.14 -0.150.5 5.8 -77.45 -0.05 -0.090.6 5.31 -66.71 -0.02 00.7 4.83 -55.9 -0.06 0.120.8 4.34 -45.05 -0.2 0.280.9 3.85 -34.59 -0.47 0.491 3.37 -25.67 -0.91 0.76

NxG.P. NQG.P. MxG.P. QxG.P.

3 4 5 6 7 8 9-0.2

0

0.2

0.4

0.6

0.8

1

Nx

Nx

ξ

-160 -140 -120 -100 -80 -60 -40 -20 0-0.2

0

0.2

0.4

0.6

0.8

1

NQ

NQ

ξ

-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1-0.2

0

0.2

0.4

0.6

0.8

1

Mx

Mx

ξ

-0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.2

0

0.2

0.4

0.6

0.8

1

Qx

Qx

ξ

4. STAREA DE EFORTURI DIN GP+PH+PR

x0 8.23 -175.47 0 5.07

0.1 7.75 -115.93 -2.22 1.530.2 7.26 -70.4 -2.57 -0.310.3 6.77 -41.84 -2.05 -10.4 6.29 -27.15 -1.31 -1.020.5 5.8 -21.49 -0.68 -0.750.6 5.31 -20.55 -0.27 -0.40.7 4.83 -21.38 -0.1 -0.070.8 4.34 -22.54 -0.17 0.240.9 3.85 -23.93 -0.44 0.521 3.37 -26.66 -0.9 0.79

NxG.P.

NQG.P.

MxG.P.

QxG.P.

3 4 5 6 7 8 9

-0.2

0

0.2

0.4

0.6

0.8

1

Nx

Nx

ξ

-200 -180 -160 -140 -120 -100 -80 -60 -40 -20 0-0.2

0

0.2

0.4

0.6

0.8

1

NQ

NQ

ξ

-3 -2.5 -2 -1.5 -1 -0.5 0-0.2

0

0.2

0.4

0.6

0.8

1

Mx

Mx

ξ

-2 -1 0 1 2 3 4 5 6-0.2

0

0.2

0.4

0.6

0.8

1

Qx

Qx

ξ

N1 N2 N3 N4 N5 N7 N6 N8

PLAN ARMARE PERETE CILINDRIC

N10

N9

N8

N7

N6

N5

N4N3N2N1

N1 N2 N3 N4 N5 N7 N6 N8

PLAN ARMARE PERETE CILINDRIC

N10

N9

N8

N7

N6

N5

N4N3N2N1

500

25

07

07

5

20

05

0

50

300

250

10 12/ml

500

25

07

07

5

20

05

0

50

300

250

10 12/ml

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