CALCULE asf

5. Evaluarea incarcarilor din zapada

Localitate: VATRA DORNEIS=γi*µi*Ce*Ct*Sk

unde: i factorul de importanta-expunere pentru actiunea zapezii;i = 1 pentru clasa de importanta-expunere IIIμi coeficientul de forma al incarcarii din zapada pe acoperis;μi = 0.8 0°≤ °≤30°Sk valoarea caracteristica a incarcarii din zapada pe sol [kN/m2], in amplasament;Sk = 2 in functie de localitate - Vatra DorneiCe coeficientul de expunere al constructiei in amplasament;Ce = 1 pentru expunere normala Ct coeficientul termic;Ct = 1

S=1*0.8*1*1*2=1.6 [kN/m2]=160 [daN/m2]

6. Evaluarea incarcarii seismicea) Calculul greutatii cladirii

Gtotal = n x (Gpereti.exteriori + Gpereti interiori + Gstalpisori) + (n-1) x Gplanseu curent + Gplanseu terasa + Gatic

unde n- numarul de niveluri ( n = 5 )

1. Greutatea peretilor exterioriGp.exteriori = Gferestre + Gbuiandrugi + Gcentura + Gp.rezistenta

Aferestre = (1.50*1.80*8 + 1.50*1.50 + 1.50*1.20*2) = 27.45 [m2]Gferestre = 27.45 x 25 = 686.25[daN]Abuiandrungi = (0.30*2.40*8+0.30*2.10+0.30*1.8*2) = 7.47 [m2]Gbuiandrugi = 7.47 x 0.25 x 2500 = 4668.75 [daN]Acentura = 19.25 [m2]Gcentura = 19.25 x 0.13 x 2500 = 6256.25 [daN]Ap.rezistenta = 25.25*2.88*2 + 13.25*2.88*2 - 27.25 - 7.47 - 19.25 = 167.79 [m2]Gp.rezistenta = 167.79 x 210 = 35235.90 [daN]

Gp.exteriori = 686.25 + 4668.75 + 6256.25 + 35235.90 = 46847.15 [daN]

2. Greutatea peretilor interioriGp.interiori = Gusi + goluri + Gbuiandrugi + Gp.rezistenta + Gp.compartimentare

Ausi+goluri p.rez = (1.00*2.10*3 + 0.90*2.10*2) + (1.40*2.10 +0.90*2.10*4) = 20.58 [m2]Gusi+goluri p.rez = 20.58 x 25 = 514.5[daN]Ausi p.com = 0.90*2.10*9 + 0.80*2.10*10=33.81 [m2]Gusi p.com = 33.81 x 25 = 845.25[daN]Abuiandrungi = (2*0.3 + 1.6*0.3*3 + 1.5*0.3*6 ) = 4.74 [m2]Gbuiandrugi = 4.74 x 0.25 x 2500 = 2962.5 [daN]Ap.rezistenta = 2.88*(36.35+35.05) - 4.74 = 200.90 [m2]Gp.rezistenta = 200.90 x 210 = 42187.32 [daN] Ap.compartimentare = 2.88*42.26 = 121.71 [m2]Gp.compartimentare =121.71 x 82 = 9980.12 [daN]

Gp.interiori = 514.5 + 845.25 + 2962.5 + 42187.32 + 9980.12 = 56489.70 [daN]

3. Greutatea stalpisorilorGstalpisori= 31*0.25*0.25*2.88*2500 = 13950 [daN]

4. Greutatea planseu curentGplanseu.curent = Apard.calda x INpard.calda + Apard.rece in bai x INpard.rece in bai + Apard.rece in buc,hol,etc x INpard.rece in buc,hol,etc +

Acasa scarii x INcasa scarii + Ascara x INscara + 0.3 x [(Apard.calda + Apard.rece in bai + Apard.rece in buc,hol,etc) x INutila (incaperi) + (Acasa scarii + Ascara) x INutila (scari)]Apard.calda = 14.96+14.96+14.96+14.96+11.81+13.06+13.54+3.68 = 101.93 [m2]Apard.rece in bai = 4.50+6.74+2.18+5.22 = 18.64 [m2]Apard.rece in buc,hol,etc = 13.65+2.64+3.19+3.68+3.3+6.78+14.96+4.72+13.06+1.8+8.66+5.62+8.31

+13.543.15+5.22+3.15 = 112.28 [m2]Acasa scarii = 9.76*2.75 = 26.84 [m2]Ascara = 1.45*5.4 = 7.83 [m2]

Gplanseu.curent = 101.93 x 435.6 + 18.64 x 535.66 + 112.28 x 463 + 26.84 x 451.5 + 7.83 x 664 + 0.3 x [(101.93 + 18.64 + 112.28) x 175 + (26.84 + 7.83) x 300]

= 139033.36 [daN]

5. Greutatea planseului terasaGplanseu terasa= Aplanseu terasa x (incarcarea normata + 0.4 x max(zapada;incarcarea utila)

- Incarcarea caracteristica = 721.5 [daN/m2]- Zapada = 160 [daN/m2]- Incarcarea utila = 75 [daN/m2]

Gplanseu terasa = 304.5625*(721.5+0.4*160) = 239233.84 [daN]

6. Greutatea aticuluiGatic= 19.25 * 271 = 5216.75 [daN]

Gtotal = 5 x (46847.15 + 56489.70 + 13950) + (5-1) x 139033.36 + 239233.84 + 5216.75Gtotal = 1387018.28 [daN]

b) Calculul fortei seismice

Fb = 1,e * ❑0×agq

* m * * η

m = Gtotg

, g = 9.81 m/s2

m = 1407478.16

g

ag = 0.10g Vatra Dornei0 = 2.5 ordonata maxima a spectrului elastic

q = 2.50 x ❑u

❑1 = 2.50 x 1.25 = 3.125 factor de comportare

1,e = 1 factor de importanta = 0.85 pentru cladirile ≥P+2Eη = 0.88 factorul de reducere care tine seama de amortizarea zidariei = 8%

Fb = 1 x 2.5×0.10 g3.125

x 1407478.16

g x 0.85 x 0.88 = 84223.49 daN = 842.23 kN

7. Combinatii de incarcariEd = Σ Gk,j + P + AEd + Σ ψ2,i x Qk,i

AEd = I x AEk = Fb

8. Determinarea sectiunilor active ale diafragmelor transversale

a) Determinarea sectiunilor active pentru montanti

1. DIAFRAGMA TRANSVERSALA DT1

Determinarea sectiunilor active pentru montanti

- Latimea de conlucrare a talpii inferioare (ax A)

Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ;1.48) = 1.48 mLeff1 = 1.48+0.25 = 1.73 m

- Latimea de conlucrare a talpii intermediare (ax B)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 5.6) = 1.50 mLeff2 = 1.50+0.25 = 1.75 m

- Latimea de conlucrare a talpii intermediare (ax C)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 2.55) = 1.50 mLeff2 = 1.50+0.25 = 1.75 m

- Latimea de conlucrare a talpii superioare (ax E)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 1.48) = 1.48 mLeff2 = 1.48+0.25 = 1.73 m

Calculul caracteristicilor geometrice

- Aria sectiunii active la compresiune excentrica

Am1 = 0.25*(13.25+2*1.48+2*1.50) = 4.8025 m2

- Aria sectiunii active la forta taietoare

Av1 = 0.25*13.25 = 3.3125 m2

- Pozitia centrului de greutate

ZG = 0.25∗(2∗1.48∗0.99+2∗1.50∗1.0+13.25∗0.125)

4.8025 = 0.395 m

YG = 0.25∗(1.48∗0.125+1.50∗5.125+1.50∗8.125+1.48∗13.125+13.25∗6.625)

4.8025 = 6.625 m

- Momentul de inertie

I =2*( 1.48∗0.25³

12 + 1.48*0.25*6.52) + 2*(

1.50∗0.25³12

+ 1.50*0.25*1.52) + 0.25∗13.25³

12 = 81,4228 m4



- Latimea de conlucrare a talpii inferioare (ax A)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ;1.48) = 1.48 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;1.48) = 1.48 mLeff1 =2*1.48+0.25 = 3.21 m

- Latimea de conlucrare a talpii intermediare (ax B)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 1.92) = 1.50 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 0.80) = 0.80 mLeff2 = 1.50+0.25+0.80 = 2.55 m

- Latimea de conlucrare a talpii intermediare (ax C)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 0.60) = 0.60 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 5.6) = 1.50 mLeff2 = 1.50+0.25+0.60 = 2.35 m

- Latimea de conlucrare a talpii superioare (ax E)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ;1.48) = 1.48 mΔbst = min(6T ; distanta pana la primul gol)

Δbst = min(6*0.25 ;1.48) = 1.48 mLeff1 =2*1.48+0.25 = 3.21 m



Am1 = 1.615 m2

Am2 = 2.0625 m2

Am3 = 1.615 m2


Av1 = 0. 25*3.50 = 0.875 m2

Av2 = 0.25*3.85 = 0.9625 m2

Av3 = 0. 25*3.50 = 0.875 m2

- Pozitia centrelor de greutate

YG1 = 0.25∗(3.21∗0.125+3.25∗1.875)

1.615 = 1.005 m

ZG1 = 0.25∗(3.21∗1.605+3.25∗1.605)

1.615 = 1.605 m

YG2 = 0.25∗(2.55∗0.125+3.60∗2.05+1.50∗3.125+0.60∗3.125)

2.0625 = 1.729 m

ZG2 = 0.25∗(2.55∗1.975+3.60∗1.625+1.50∗0.75+0.60∗2.05)

2.0625 = 1.605 m

YG3 = 0.25∗(3.21∗0.125+3.25∗1.875)

1.615 = 1.005 m

ZG3 = 0.25∗(3.21∗1.605+3.25∗1.605)

1.615 = 1.605 m

- Momentele de inertie

I1 = 3.21∗0.25³

12 + 3.21*0.25*0.882 +

0.25∗3.25³12

+ 3.25*0.25*0.872 = 1.9558 m4

I2 = 2.55∗0.25³

12 + 2.55*0.25*1.6042 +

0.25∗3.60³12

+ 3.60*0.25*0.3212 + 1.50∗0.25³

12 +

1.50*0.25*1.3962 + 0.60∗0.25³

12 + 0.60*0.25*1.3962 = 3.7341 m4

I3 = 3.21∗0.25³

12 + 3.21*0.25*0.882 +

0.25∗3.25³12

+ 3.25*0.25*0.872 = 1.9558 m4

- Pozitia centrului de greutate al diafragmei

YG = 1.615∗1.005+2.0625∗(1.005+5.724)+1.615∗(13.25−1.005)

5.2925 = 6.6655 m

- Momentul de inertie al diafragmei

I0 = 1.9558 + 1.615*5.6605 + 3.7341 + 2.0625*0.0635 + 1.9558 + 1.615*5.5795 = 25.9292 m4



- Latimea de conlucrare a talpii inferioare (ax A)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ;0.77) = 0.77 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;1.48) = 1.48 mLeff1 =1.48+0.25+0.77 = 2.50 m


- Latimea de conlucrare a talpii intermediare (ax C)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 4.75) = 1.50 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 1.93) = 1.50 mLeff2 = 1.50+0.25+1.50 = 3.25 m

- Latimea de conlucrare a talpii superioare (ax E)Δbdr = min(6T ; distanta pana la primul gol)

Δbdr = min(6*0.25 ;1.48) = 1.48 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;1.48) = 1.48 mLeff1 =2*1.48+0.25 = 3.21 m


- Aria sectiunii active la compresiune excentricaAm1 = 1.615 m2

Am2 = 1.0313 m2

Am3 = 2.8438 m2

- Aria sectiunii active la forta taietoareAv1 = 0. 25*3.50 = 0.875 m2

Av2 = 0.25*1.125 = 0.28125 m2

Av3 = 0. 25*6.12 = 1.53 m2


YG1 = 0.25∗(3.21∗0.125+3.25∗1.875)

1.615 = 1.005 m

ZG1 = 0.25∗(3.21∗1.605+3.25∗1.605)

1.615 = 1.605 m

YG2 = 0.25∗(3.25∗0.125+0.875∗0.6875)

1.0313 = 0.244 m

ZG2 = 0.25∗(3.25∗1.625+0.875∗1.625)

1.0313 = 1.625 m

YG3 = 0.25∗(0.87∗0.435+3.25∗0.995+4.75∗3.495+2.5∗5.995)

2.8438 = 3.0945 m

ZG3 = 0.25∗(0.87∗1.625+3.25∗1.625+4.75∗1.625+2.5∗1.25)

2.8438 = 1.542 m


I1 = 3.21∗0.25³

12 + 3.21*0.25*0.882 +

0.25∗3.25³12

+ 3.25*0.25*0.872 = 1.9558 m4

I2 = 3.25∗0.25³

12 + 3.25*0.25*0.1192 +

0.25∗0.875³12

+ 0.875*0.25*0.44352 = 0.0727 m4

I3 = 0.25∗0.87³

12 + 0.87*0.25*0.4352 +

3.25∗0.25³12

+ 3.25*0.25*0.9952 + 0.25∗4.75³

12 +

4.75*0.25*3.4952 + 2.50∗0.25³

12 + 2.50*0.25*5.9952 = 40.0674 m4


YG = 1.615∗1.005+1.0313∗(1.005+5.244 )+2.8438∗(13.25−1.542)

5.4901 = 7.534 m


I0 = 1,9558 + 1,615 *6,529 + 0,0727 + 1,0313 *2,290 + 40,0674 + 2,8438 *2,6905 = 62.653 m4




Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ;0.98) = 0.98 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;0.77) = 0.77 m

Leff1 =0.98+0.25+0.77 = 2 m- Latimea de conlucrare a talpii superioare (ax B)

Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 1.38) = 1.38 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 2.75) = 1.50 mLeff2 = 1.50+0.25+1.38 = 3.13 m



Am2 = 1.3125 m2


Av2 = 0.25*3.50 = 0.875 m2


YG1 = 0.25∗(3.13∗0.125+0.60∗0.55)

0.9313 = 0.194 m

ZG1 = 0.25∗(3.13∗1.565+0.60∗1.565)

0.9313 = 1.565 m

YG2 = 0.25∗(3.25∗1.625+2∗3.375)

1.3125 = 2.292 m

ZG2 = 0.25∗(3.25∗0.895+2∗1.00)

1.3125 = 0.935 m


I1 = 3.13∗0.25³

12 + 3.13*0.25*0.0692 +

0.25∗0.60³12

+ 0.60*0.25*0.3562 = 0.0313 m4

I2 = 3.25∗0.25³

12 + 3.25*0.25*0.6672 +

0.25∗2.00³12

+ 2.00*0.25*1.0832 = 1.1188 m4


YG = 0.9313∗0.194+1.3125∗(5.25−2.292)

2.2438 = 1.8108 m


I0 = 0.0313 + 0.9313*3.2452 + 1.1188 + 1.3125*0.6028 = 4.9635 m4



- Latimea de conlucrare a talpii inferioare (ax C)Δbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;1.48) = 1.50 m

Leff1 =1.48+0.25= 1.73 m- Latimea de conlucrare a talpii intermediare (ax D)

Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 1.78) = 1.50 mLeff2 = 1.50+0.25= 1.75 m

- Latimea de conlucrare a talpii superioare (ax E)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 4.75) = 1.50 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 1.93) = 1.50 mLeff3 = 1.50+0.25+1.50 = 3.25 m





YG1 = 0.25∗(1.48∗0.125+5∗2.50+1.50∗3.125+3.25∗5.125)

2.8062 = 3.0315 m

ZG1 = 0.25∗(1.48∗0.75+5∗1.625+1.50∗2.50+3.25∗1.625)

2.8062 = 1.627 m


I1 = 1.48∗0.25³

12 + 1.48*0.25*2.09652 +

0.25∗5.00³12

+ 5.00*0.25*0.53152 + 1.50∗0.25³

12 +

1.50*0.25*0.09352 + 3.25∗0.25³

12 + 3.25*0.25*2.09352 = 8.1559 m4




Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 0.63) = 0.63 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;0.97) = 0.97 mLeff1 =0.97+0.25+0.63 = 1.85 m

- Latimea de conlucrare a talpii superioare (ax B)

Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 2.75) = 1.50 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 1.38) = 1.38 mLeff2 = 1.50+0.25+1.38 = 3.13 m



Am2 = 1.2750 m2


Av2 = 0.25*1.125 = 0.875 m2


YG1 = 0.25∗(3.13∗0.125+0.60∗0.55)

0.9312 = 0.194 m

ZG1 = 0.25∗(3.13∗1.565+0.60∗1.505)

0.9312 = 1.558 m

YG2 = 0.25∗(3.25∗1.625+1.85∗3.375)

1.2750 = 2.260 m

ZG2 = 0.25∗¿¿ = 0.925 m


I1 = 3.13∗0.25³

12 + 3.13*0.25*0.0692 +

0.25∗0.60³12

+ 0.60*0.25*0.3562 = 0.0313 m4

I2 = 3.25∗0.25³

12 + 3.25*0.25*0.6672 +

0.25∗1.85³12

+ 1.85*0.25*1.0832 = 1.1188 m4


YG = 0.9313∗0.194+1.3125∗(5.25−2.292)

2.2438 = 1.8108 m


I0 = 0.0313 + 0.9313*3.2452 + 1.1188 + 1.3125*0.6028 = 4.9635 m4



- Latimea de conlucrare a talpii inferioare (ax A)Δbdr = min(6T ; distanta pana la primul gol)Δbdr = min(6*0.25 ; 1.48) = 1.48 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 0.62) = 0.62 mLeff2 = 1.48+0.25+0.62 = 2.35 m


- Latimea de conlucrare a talpii superioare (ax D)Δbdr = min(6T ; distanta pana la primul gol)

Δbdr = min(6*0.25 ;1.48) = 1.48 mΔbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;1.77) = 1.50 mLeff1 =1.48+0.25+1.50 = 3.23 m



Am2 = 2.3453 m2


Av2 = 0.25*5.65 = 1.4125 m2


YG1 = 0.25∗(3.23∗0.125+3.35∗1.925)1.6437

= 1.042 m

ZG1 = 0.25∗(3.23∗1.615+3.35∗1.615)

1.6437 = 1.615 m

YG2 = 0.25∗(0.40∗0.20+2.18∗0.525+4.75∗3.025+2.35∗5.525)

2.3453 = 3.046 m

ZG2 = 0.25∗(0.40∗1.625+2.18∗1.09+4.75∗1.625+2.35∗2.05)

2.3453 = 1.659 m


I1 = 3.23∗0.25³

12 + 3.23*0.25*0.9172 +

0.25∗3.35³12

+ 3.35*0.25*0.8832 = 2.119 m4

I2 = 0,25∗0.40³

12 + 0,40*0,25*2,8462 +

2,18∗0.25³12

+ 2,18*0.25*2,52102 + 0.25∗4.75³

12 +

4,75*0.25*0.7212 + 2,35∗0.25³

12 + 2,35*0.25*2,47902 = 10,7412 m4


YG = 1.6437∗1.042+2.3453∗(10.25−3.046)

3,989 = 4,665 m


I0 = 2.119 + 1.6437 *4,5430 + 10,7412 + 2.3453 *2,061 = 25.1612 m4



- Latimea de conlucrare a talpii inferioare (ax A)Δbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 1.48) = 1.48 mLeff2 = 1.48+0.25= 1.73 m

- Latimea de conlucrare a talpii intermediare (ax B)Δbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ; 3.43) = 1.50 mLeff2 = 0.25+1.50 = 1.75 m

- Latimea de conlucrare a talpii superioare (ax D)Δbst = min(6T ; distanta pana la primul gol)Δbst = min(6*0.25 ;1.48) = 1.48 mLeff1 =1.48+0.25 = 1.73 m



Am1 = 3.7211 m2


Av1 = 0.25*10.25 = 2.5625 m2


YG = 0.25∗(1.48∗0.125+1.50∗5.125+10.25∗5.125+1.48∗10.125)

3.7211 = 5.125

m

ZG = 0.25∗(2∗1.48∗0.99+1.50∗1+10.25∗0.125)

3.7211 = 0.384 m

- Momentul de inertie

I =2*( 1.48∗0.25³

12 + 1.48*0.25*52) +

1.50∗0.25³12

+ 0.25∗10.25³

12 = 40.940 m4

b) Determinarea sectiunilor active pentru rigle

1. RIGLA 1 (L 0 = 0.90 m)

Δbrst = Δbr

dr = 0.15 * L0 ≤ 2hpl

Δbrst = Δbr

dr = 0.15 * 0.90 ≤ 2*0.13Δbr

st = Δbrdr = 0.135 m

Lp =0.135+0.25+0.135 = 0.52 mhp = 0.13 mhr = 0.91 m


- Aria rigleiAr = 0.2626 m2

- Aria sectiunii active la forta taietoare a rigleiAr1 = 0.2275 m2


Yr = 0.13∗0.52∗0.065+0.25∗0.78∗0.52

0.2626 = 0.403 m

- Momentul de inertie al riglei

Ir = 0.52∗0.13³

12 + 0.52*0.13*0.3382 +

0.25∗0.78³12

+ 0.78*0.25*0.1172 = 0.0204 m4

2. RIGLA 2 (L 0 = 1.00 m)

Δbrst = Δbr

dr = 0.15 * L0 ≤ 2hpl

Δbrst = Δbr

dr = 0.15 * 1.00 ≤ 2*0.13Δbr


Lp =0.15+0.25+0.15 = 0.55 mhp = 0.13 mhr = 0.91 m





Yr = 0.13∗0.55∗0.065+0.25∗0.78∗0.52

0.2665 = 0.398 m


Ir = 0.55∗0.13³

12 + 0.55*0.13*0.3332 +

0.25∗0.78³12

+ 0.78*0.25*0.1222 =

0.0217 m4

3. RIGLA 3 (L 0 = 1.50 m)

Δbrst = Δbr

dr = 0.15 * L0 ≤ 2hpl

Δbrst = Δbr

dr = 0.15 * 1.50 ≤ 2*0.13Δbr


Lp =0.225+0.25+0.225 = 0.7 mhp = 0.13 mhr = 0.91 m





Yr = 0.13∗0.7∗0.065+0.25∗0.78∗0.52

0.2860 = 0.375 m


Ir = 0.7∗0.13³12

+ 0.7*0.13*0.3102 + 0.25∗0.78³

12 + 0.78*0.25*0.1452 = 0.0226 m4

9. Calculul caracteristicilor de rigiditate pentru o diafragma cu 2 siruri de goluri (DT3)

a) CADRUL REAL: RIGIDITATEA MONTANTILOR LA DISTORSIUNE

km1 = (Im1 / He) x Em=(1.9558 x2.8) x Em=5.47624 x Em

km2 = (Im2 / He) x Em = (0.0727 x 2.8) x Em = 0.20356 x Em

km3 = (Im3 / He) x Em = (40.0674 x 2.8) x Em = 112.18872 x Em

RIGIDITATEA RIGLEI LA DISTORSIUNEa = 0.35 x hr = 0.35 x 0.91= 0.3185 m

l = l1-2 + 2 x a =1.5 + 2 x 0.3185 = 2.137 mEr = 2 x Em

µ1 = 1 / (1 + (30 x Ir / Av x l2)) =1

1+30×0.0226

0.2275×2.137❑2

=¿0.605

kr1-2 = ((6 x Er x Ir)/L1-2) x (L1-2/l)2 x µ =((6 x 2 x Em x 0.0226)/ 5) x (5 / 2.137)3 x 0.605 =40 x Em

a = 0.35 x hr = 0.35 x 0.91 = 0.3185 m l = l2-3+ 2 x a = 3 + 2 x 0.3185 = 3.637 m

Er = 2 x Em

µ2 = 1 / (1 + (30 x Ir / Av x l2)) = 1

1+30×0.0217

0.2275×3.637❑2

=¿ 0.216

kr2-3 = ((6 x Er x Ir)/L2-3) x (L2-3/l)2 x µ =((6 x 2 x Em x 0.0217)/ 3) x (3 / 3.637)3 x 0.216 =0.01x Em

b) CADRUL ECHIVALENT: RIGIDITATEA MONTANTILOR LA DISTORSIUNE

KME=(5.47624 +0.20356 +112.18872)/2 x Em=58.93 x Em

RIGIDITATEA RIGLEI LA DISTORSIUNEKRE=40 x Em+ 0.01x Em=40.01 x Em

c) MARIMILE λ, γ, α, I 0, Ies

λ = kre / kme = 40.01 x Em / 58.93 x Em = 0.679

γ = 1 + ((Im1 + Im2+ Im3) / L1-52) x (1/Am1 + 1 / Am2+1/ Am3) = 1 + ((1.9558 + 0.0727 + 40.0674) / 13.252) x

(1/1.615 + 1 /1.0313 +1/2.8438) = 1.480

α = n x √ λ x γ = 4 x √0.679 x1.480 = 4.01 =>DIAFRAGMA CU GOLURI MIJLOCII

I0 = (γ / (γ-1)) x (Im1 + Im2+ Im3) = (1.480 / (1.480-1)) x (1.9558 + 0.0727 + 40.0674) = 129.79 m4

νs = 11.25 – 4 niveluri

0.0 ......... 0.1310.022 ... x => x= 0.1498 0.05 .......... 0.173

0.0......... 0.1670.022 ... y => y = 0.190320.05 .......... 0.22

1.15 ..... 0.1498 1.188 ... ηs => ηs = 0.18059521.20 ...... 0.19032

Ies = (ηs x I0) / (1+νs x (ηs x I0 / Amt x H2)) = (0.1805952x 129.79) / (1+11.25 x ((0.1805952x 129.79 / 5.4901 x (2.8x4)2)) = 16.95 m4

10.Distributia incarcarii orizontale totale la diafragmele transversale

1. DETERMINAREA AXEI CENTRULUI MASELOR (CM)

Gbuiandrug,centura,stalpi,ext=0.005*1900+0.1*20+0.25*2500+0.015*1900=665 daN/m2

Gbuiandrug,centura,stalpi,int=0.015*1900+0.25*2500+0.015*1900=682 daN/m2

ELEMENT Gi [kN] yi [m] Gi * yi [kNm]

DT1

Trama 1

Trama 2

Trama 3

DT2

Trama 4

Trama 5

Trama 6

DT3

Trama 7

Trama 8

Trama 9

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