Cladire administrativa

CLADIRE ADMINISTRATIVA

I. INCARCARI PERMANENTE

- placa beton armat:

0.15 x 2500 x 1.35 = 506.25

- tencuiala 2 cm:

0.02 x 2000 x 1.35 = 54

- gresie:

100 x 1.35 = 135

p = 695.25 695

INCARCARI UTILE

- hol (1):

300 x 1.5 = 450

- birou (2):

200 x 1.5 = 300

- depozit (3 + 4):

400 x 1.5 = 600

II. CALCUL STATIC

Ochiul 1

= = = 1.10 = 0.0384 β= 0.7854

= 0.0210 β= 0.2146

q= β q = 0.7854 1245 = 977.82 daN/m

q= β q = 0.2146 1245 = 267.17 daN/m

M= q l = 0.0384 1245 5 = 1195.20

M= q l = 0.0210 1245 5.5 = 790.88

M= - = = - 3055.69

M= - = = - 1010.24

Ochiul 2

= = = 1.10 = 0.0293 β= 0.8798

= 0.0133 β= 0.1202

q= β q = 0.8798 995 = 875.40 daN/m

q= β q = 0.1202 995 = 119.60 daN/m

M= q l = 0.0293 995 5 = 728.84

M= q l = 0.0133 995 5.5 = 400.31

M= - = = - 1823.75

M= - = = - 301.49

Ochiul 3

= = = 1.83 = 0.05621 β= 0.9177

= 0.00503 β= 0.0822

q= β q = 0.9177 1295 = 1188.42 daN/m

q= β q = 0.0822 1295 = 106.45 daN/m

M= q l = 0.0562 1295 3 = 655.01

M= q l = 0.00503 1295 5.5 = 197.04

M= - = = - 1336.97

M= - = = - 402.51

Ochiul 4

= = = 0.55 = 0.0093 β= 0.1862

= 0.0808 β= 0.8138

q= β q = 0.1862 1295 = 241.13 daN/m

q= β q = 0.8138 1295 = 1053.87 daN/m

M= q l = 0.0093 1295 3 = 108.39

M= q l = 0.0808 1295 5.5 = 3165.24

M= - = = - 271.27

M= - = = - 3984.95

III CALCULUL ARMATURILOR

1x - 1x

M = 1195.20 daN m

m = = = 0.086 = 0.955

h= h - a - = 15 - 2.5 - 0.8 = 12.10 cm

A= = = 3.45 cm

M = 728.84 daN m

m = = = 0.052 = 0.9733

A= = = 2.06 cm

M = 655.01 daN m

m = = = 0.047 = 0.976

A= = = 1.85 cm

M = 3055.69 daN m

m = = = 0.220 = 0.8742

A= = = 9.63 cm

M = 1823.75 daN m

m = = = 0.130 = 0.930

A= = = 5.40 cm

2x - 2x

M = 108.39 daN m

m = = = 0.007 = 0.995

A= = = 5.30 cm

1y - 1y

M = 790.88 daN m

m = = = 0.057 = 0.9706

A= = = 2.24 cm

2y - 2y

M = 400.31 daN m

m = = = 0.029 = 0.9885

A= = = 1.12 cm

3y - 3y

M = 197.04 daN m

m = = = 0.014 = 0.993

A= = = 0.55 cm

M = 3165.24 daN m

m = = = 0.23 = 0.8681

A= = = 10.04 cm

M = 3984.95 daN m

m = = = 0.29 = 0.8242

A= = = 13.32 cm

IN CAMP

REAZEME

GRINDA

A. Incarcari permanente

- greutate proprie

b h 2500 1.35 0.25 0.50 2500 1.35 = 421.88

- placa beton armat

S = S1 + S2

S1 = S2 = 7.5

S = 7.5 + 7.5 = 15.00

= = 1895.45

- zidarie

b h 1450 1.35 0.25 2.6 1450 1.35 = 1272.38

p= 421.88 + 1895.45 + 1272.38 = 3589.71 daNm

B. Incarcare utila

- placa beton armat

+ = 613.64 + 409.10 = 1022.73 daNm

q= p+ u= 3589.71 + 1022.73 = 4612.44 daNm

I. Dimensionarea in sectiuni normale

M= = = 17440.79

T= = = 12684.21

M = 17440.79

m = = 0.33

m = 0.33 = 0.7917

A= = 15.65 cm

Aleg 5 20 cu A15.71

- 3 bare intinse

- 2 bare inclinate

II. Dimensionarea in sectiuni inclinate

1) Alegerea etrierilor inchisi

- etrieri simpli

- distanta intre etrieri a

b = 25 cm a h = 37.5 cm

a b a15 0.6 = 9

a 30 cm 9 a 25 a= 20 cm

- diametrul etrierilor

6 pentru h 80 cm

- aria etrierilor din tabel

A= 0.283 cm

2) Verificarea nivelului de solicitare la forta taietoare

0.5 = 4

0.5 = 1.35 4

3) Forta taietoare preluata de beton si armatura

q= = = 67.92

p = 100 = 100 = 1.34

T= 2 = 2 = 11744.10

Daca T< T se va dispune armatura inclinata

4) Aria armaturii inclinate

A= = = 0.56 cm

FUNDATII

CALCUL FUNDATII EXTERIOARE

- planseu acoperis - terasa:

= = 655.5

- planseu intermediar (2 buc.):

= = 971.25

- zidarie 30 cm:

0.30 8.2 1450 1.35 = 4815.45

- greutate proprie

0.60 2.00 2500 1.35 = 4050

q = 11463.45

= 284.30 K

P = = = : 100 = 191.05 K

C= K (B - 1) = 284.30 0.05 (0.60 - 1) = - 5.69

C= = 284.30 = 0

P > P

Fundatia este cea calculata