despre rezervoare
Post on 29-Dec-2015
126 Views
Preview:
DESCRIPTION
TRANSCRIPT
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
Pentru inelul circular:
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) =
-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
Pentru inelul circular:
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
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
0.5 4.349 *x3 -4.23 = 0x3= 0.882 mpx3= 5.231 tf/m2Px3= 4.23 3.519 m fata de partea superioara
0.5 5.231 *x4 -4.23 = 0x4= 0.754 mpx4= 5.985 tf/m2Px4= 4.23 4.315 m fata de partea superioara
0.5 5.985 *x5 -4.23 = 0x5= 0.669 mpx5= 6.654 tf/m2Px5= 4.23 5.012 m fata de partea superioara
0.5 6.654 *x6 -4.23 = 0x6= 0.607 mpx6= 7.261 tf/m2Px6= 4.23 5.640 m fata de partea superioara
0.5 7.261 *x7 -4.23 = 0x7= 0.560 mpx7= 7.821 tf/m2Px7= 4.23 6.216 m fata de partea superioara
0.5 7.821 *x8 -4.23 = 0x8= 0.523 mpx8= 8.344 tf/m2Px8= 4.23 6.752 m fata de partea superioara
0.5 8.344 *x9 -4.23 = 0x9= 0.492 mpx9= 8.836 tf/m2Px9= 4.23 7.254 m fata de partea superioara
0.5 8.836 *x10 -4.23 = 0x10= 0.466 mpx10= 9.302 tf/m2Px10= 4.23 7.728 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
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
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
top related