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CALCULUL STRUCTURII - PREDIMENSIONARE

Constructii

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CALCULUL STRUCTURII

PREDIMENSIONARE

  1. Predimensionare stilpilor

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

  1. Predimensionarea grinzilor

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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