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CALCULUL STRUCTURII
PREDIMENSIONARE
Stilpii cadrelor sunt elemente supuse la solicitari compuse:M,N,T in proportii diferite,dupa pozitia lor in cadru.
Sectiunea transversala a stilpilor se alege de forma dreptunghiulara si circulara.
Dimensiunile sectiunii transversale se iau multiplu de 5 cm, pina la 80 cm si multiplu de 10 cm peste 80 cm.
Dimensiunea minima admisa este de 30x30 cm in situatii justificate dimensiunea minima se poate lua 25x25 cm.
Pentru asigurarea unei ductilitati corespunzatoare trebuie ca : Ab
unde : Rc-rezistenta la compresiune a betonului ;
Ab=bxh -suprafata sectiunii transversale a stilpului ;
n-coeficient functie de pozitia stilpului in constructie :
n=0.35 stilp intermediar
n=0.30 stilp marginal
n=0.25 stilp de colt
N-efortul axial in stilp
Materiale folosite: B200 Rc=95 daN/cm2
PC52 Ra=2900 daN/cm2
Nn=Saf*nniv *qn unde:
Saf-suprafata aferenta de descarcare a stilpului
nniv-numarul de niveluri
qn-incarcarea utila
Saf=5.4*5.4=29.16 m2
nniv=3 T Nn=29.16*3*835.6=73098.28 daN
Ab= cm2
b=h= T b=h=50 cm
Inaltimea "h" a grinzii fata de deschiderea de calcul "l" este de obicei multiplu de 5 cm pentru hI cm si de 10 cm pentru h 80 cm.
hmin -pentru rigle si grinzi principale
hmin -pentru grinzi secundare
l=5040 cm T cm T hmin=40 cm
Se recomanda ca raportul dintre inaltime si latimea sectiunii transversale sa se ia I pentru elementele cu sectiune dreptunghiulara.
Inaltimea grinzii "b" are in mod obisnuit urmatoarele valori :
-pentru grinzi secundare 18,20,22 si max 25 cm
-pentru grinzi principale 25.40 cm
In final vom alege : hg=60 cm si bg=30 cm.
CALCULUL INCARCARILOR
CADRUL 3-3
5.4 5.4 5.4 4.5
E=300000 daN/m2
grupari incarc. | ||||
|
Stilpi : A=050*050=0.250 m2
Grinzi : A=0.3*0.6=0.18 m2
Calculul suprafetelor aferente cadrului 3-3
S1 S2
S3 S4
5.4 5.4 5.4 4.5
S1=S2= m2
S3= m2
S4= m2
Ipoteza 1: Incarcari permanente
A.Din greutatea planseului
pngr.1= daN/m
pngr.2== daN/m
pngr.3== daN/m
pngr.4== daN/m
pngr.5== daN/m
pngr.6== daN/m
pngr.7== daN/m
pngr.8== daN/m
pngr.9= daN/m
pngr.10= daN/m
pngr.11= daN/m
pngr.12= daN/m
B.Din greutatea grinzilor
pngr.1=0.3*0.6*2500=450 daN/m
pngr.2=0.3*0.6*2500=450 daN/m
pngr.3=0.3*0.6*2500=450 daN/m
pngr.4=0.3*0.6*2500=450 daN/m
pngr.5=0.3*0.6*2500=450 daN/m
pngr.6=0.3*0.6*2500=450 daN/m
pngr.7= 0.3*0.6*2500=450 daN/m
pngr.8=0.3*0.6*2500=450 daN/m
pngr.9=0.3*0.6*2500=450 daN/m
pngr.10=0.3*0.6*2500=450 daN/m
pngr.11=0.3*0.6*2500=450 daN/m
pngr.12= daN/m
C. Din greutatea stilpilor
pngr.1=0
pngr.2=0
pngr.3=0
pngr.4=0
pngr.5= daN/m
pngr.6= daN/m
pngr.7= daN/m
pngr.8= daN/m
pngr.9= daN/m
pngr.10= daN/m
pngr.11= daN/m
pngr.12= daN/m
D. Din pereti
pngr.1=0
pngr.2=0
pngr.3=0
pngr.4=0
pngr.5= daN/m
pngr.6=0
pngr.7= daN/m
pngr.8= daN/m
pngr.9= daN/m
pngr.10=0
pngr.11= daN/m
pngr.12= daN/m
E. Din atic
pngr.1=5.4*467=2521.8 daN/m
pngr.4=2.7*467=1260.9
daN/m
Total permanente pentru cadrul 3-3
qngr1=1750.2+450+0+0+2521.8=4722 daN/m
qngr2=1750.2+450+0+0=2200.2 daN/m
qngr3=875.1+450+0+0=1325.1 daN/m
qngr4=744.5+450+0+0+1260.9=2455.4 daN/m
qngr5=1356.8+450+434.02+675=2915.82 daN/m
qngr6=1356.8+450+434.02+0=2240.80 daN/m
qngr7=678.5+450+434.02+85.25=1647.77 daN/m
qngr8=577.125+450+520.83+511.5=2059.45 daN/m
qngr9=1356.8+450+434.02+800=3040.82 daN/m
qngr10=1356.8+450+434.02+0=2240.82 daN/m
qngr11=678.5+450+434.02+85.25=1647.77 daN/m
qngr12=577.125+450+520.83+252.87=1800.82 daN/m
Ipoteza 2 :Incarcari variabile-incarcari din zapada (STAS 10101/21-92)
pnz=ce*czi*qz unde :
ce-coeficient care tine seama de conditiile de expunere a constructie
cz-coeficient prin care se tine seama de aglomerarile de zapada de pe suprafata
constructie
a Tczi=1,00
qz-greutatea de referinta a stratului de zapada
-pentru DOROHOI care este in zona C Tqz=150 daN/m2
pnz=1*0.8*150=120 daN/m2
qngr1= daN/m
qngr2= daN/m
qngr3= daN/m
qngr4= daN/m
Ipoteza 3 :Incarcari variabile-incarcari utile
qngr1= daN/m
qngr2= daN/m
qngr3= daN/m
qngr4= daN/m
qngr5= daN/m
qngr6= daN/m
qngr7= daN/m
qngr8= daN/m
qngr9= daN/m
qngr10= daN/m
qngr11= daN/m
qngr12= daN/m
Ipoteza 4 : Incarcari variabile -incarcari din vint
pnv=b*Cn*Ch(z)*gv daN/m2
Ch(z)=0.65*0.7
gv=0.55 KN/m2 =55 daN/m2
b
Cn=0.8
Ch(z)1=0.65*0.7=0.32 =0.65
Ch(z)2=0.65*0.7=0.32=0.65
Ch(z)3=0.65*0.7=0.48=0.65
Ch(z)4=0.65* 0.7=063=065
pnn1,2,3,4=1.6*0.8*55*0.65=45.76 daN/m2
Pnv=pnv*het*ld=45.76*3*2.7=370.65 daN
Ipoteza 5: Ipoteza seismica
GLD1
GLD2
GLD3
5.4 5.4 5.4 4.5
m1=182*2.7*3+293*6.4*3+0.3*0.6*2500*(2.7+2.7+2.7)+0.5*0.5*2500*(1.8+
+1.5)+456*5.4*2.7=19455.78 daN
m2=0.3*0.6*2500*(2.7+2.7+2.7+2.7)+0.5*0.5*2500*(1.8+1.5)+456*5.4*5.4=
=20219.46
m3=182*3*2.2+0.3*0.6*2500*(2.7+2.7+2.7+2.7)+0.5*0.5*2500*(1.8+1.5)+456
*(2.7*5.4+2.7*2.7)=18096.42
m4=182*3*3+0.3*0.6*2500*(2.7+2.7+2.7)+0.5*0.5*2500*(1.8+1.5)+456*
*4.925*2.7=13409.16
m5=182*3*5+0.3*0.6*2500*(2.7+2.225)+0.5*0.5*2500*(1.8+1.5)+456*2.7*
*2.225=9748.17
m6=182*3*2.2+293*3*5.1+0.3*0.6*2500*(2.7+2.7+2.7)+0.5*0.5*2500*(1.5+1.5)
+456*5.4*2.7=17852.58
m7=293*3*6.78+0.3*0.6*2500*(2.7+2.7+2.7+2.7)+0.5*0.5*2500*(1.5+1.5)+
+456*5.4*5.4=25991.58
m8=182*3*2.2+0.3*0.6*2500*(2.7+2.7+2.7+2.7)+0.5*0.5*2500*(1.5+1.5)+
+456*(2.7*5.4+2.7*2.7)=17908.92
m9=182*3*3.6+0.3*0.6*2500*(2.7+2.7+2.7)+0.5*0.5*2500*(1.5+1.5)+
+456*4.925*2.7=13549.26
m10=182*3*5+0.3*0.6*2500*(2.7+2.225)+0.5*0.5*2500*(1.5+1.5)+456*2.7*
*2.225=9560.67
m11=636.2*5.4*2.7+293*6.4*1.5+182*2.7*1.5+0.3*0.6*2500*(2.7+2.7+2.7)+0.5*
*0.5*2500*1.5+467*5.4=19929.99
m12=636.2*5.4*5.4+293*1.5*6.78+0.3*0.6*2500*(2.7+2.7+2.7+2.7)+0.5*0.5*
*2500*1.5=27328.902
m13=636.2*(2.7*5.4+2.7*2.7)+182*1.5*2.2+0.3*0.6*2500*(2.7+2.7+2.7+2.7)+
+0.5*0.5*2500*1.5=20311.794
m14=636.2*4.925*2.7+182*1.5*3+0.3*0.6*2500*(2.7+2.7+2.7)+0.5*0.5*
*2500*1.5=13861.36
m15=636.2*2.7*2.225+182*5*1.5+0.3*0.6*
+467*2.7=9601.62
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