Beton i Paranderur dk

PROJEKT KURSI Te llogaritet soleta e paranderur kur jepet:

A. Seksioni terthor faktik eshte si ne figure:

Skema statike eshte:

PROJEKT KURSI, B/A

1

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Te dhenat: LK=900 cm

h’f =3cm

B=200 cm

a sp =a’sp=1.5 cm

hf =3 cm

=30 cm B’f =Bf =30

A’sp = 0.52 cm²

h=45 cm

t=3 cm

tm =13 cm B. Ngarkesat: p n=75daN/m2

 t=1.4

C. Betoni: Marka 400 Ro=320 daN / cm²

Rb tsh=21.4 daN / cm²

Rb t=14.3 daN / cm²

Rb=210 daN / cm²

Eb=3.5*10ˆ5 daN / cm²

D. Armatura: n

Rsp =7000 daN / cm²

Rsp =3600 daN / cm²

Rsw =0.8 Rs =2880 daN / cm²

Es=1.8*10ˆ6 daN / cm²

E. Shtresat: -granil

1cm;

 =1500 kg/m3;

 t=1.2

-hidroizolim

1cm;

 =1100 kg/m3;

 t=1.2

-lluster

2cm;

 =2200 kg/m3;

 t=1.2





KERKESAT:

1.Llogaritja ne aftesine mbajtese kundrejt momentit perkules dhe forces prerese ( llog. sipas gjendjes kufitare te gruit te pare ).

2.Llogaritja sipas formimit te te plasurave ku kontrollohet nese lindin te plasurat ne element.





ZHVILLIM 1. Llogaritja ne aftesine mbajtese kundrejt momentit perkules dhe forces prerese . Skema llogaritese eshte:

Seksioni llogarites eshte:

b

t sin 

*2 

3 0.54

* 2  11.1cm

sin  

h x



45 83.22

 0.54





Ngarkesat e seksionit: 

E perhershme

Granil ( 1cm ) 0.01( 2* x + B’f + Bf ) ρ = = 0.01( 2*0.8322 + 2*0.3 2*0.3 ) 1500 = 34 daN / ml * 1.2 = 40.8 daN / ml Hidroizolim ( 1cm ) 0.01( 2*x + B’f + Bf ) ρ = = 0.01( 2*0.8322 + 2*0.3 ) 1100 = 25 daN / ml * 1.2 = 30 daN / ml Lluster çimento ( 2cm ) 0.02( 2* x + B’f + Bf ) ρ = = 0.02( 2*0.8322 + 2*0.3 ) 2200 = 97 daN daN / ml * 1.2 = 116.4 116.4 daN / ml Betoni ( h’f * B’f + hf * Bf + 2 * x * t ) ρ = = ( 0.03*0.3 + 0.03*0.3 + 2*0.8322*0.3 ) 2500 = 165 daN / ml * 1.1 = 181.5 daN / ml Monolitizim

*0.03* ρ = 0.3 * 0.03 * 2400 = 21.6 daN / ml * 1.2 = 25.82 daN / ml B’f *0.03*

ngarkesa e normuar g n  342.6 daN

ml

ngarkesa llogaritese g ll  394.5 daN



ml

E perkoheshme n P = 75 * 2 = 150 daN m l

P = 150 * 1.4 = 210 daN m l 

E plote

n n n q = g + p = 342.6 + 150 = 492.6 daN m l

q = g + p = 394.5 + 210 = 604.5 daN m l







Percaktojme momentet dhe forcat prerese: q *l 2

M 

8

M  n

Q

qn *l 2 8

q *l

Q  n





2

qn *l 2

604.5 * 9 8



 6120.6daNm

492.6 * 9 8

604.5 * 9 2



2

 4987.6daNm

 2720.3daN

492.6 * 9 2

2

 2216.7daN

σ’o = 0.71*Rⁿsp = 4970 daN / cm² Pranoj σ’h = ( 1000 – 2500 ) daN / cm² Humbjet pranoj σ’h = 1500 daN / cm² σ’sp = Rsp – γt* σo1 Sforcimi σo1 = σo - σ’h = 4970 – 1500 = 3470 daN / cm² γt = 1 (koefiçient) σ’sp = 3600 – 3470 = 130 daN / cm² Llogaritja e armatures:









Llogarisim Momentin e pllakes





M pll  B f ' * h f ' * Rb h0  0.5 * h f '  A' sp * ' sp (ho  a' sp ) 

 30 * 3 * 210 * 43.5  0.5 * 3  0.52 *130 * (43.5  1.5)  796639 daNcm  M  612060 daNcm

Elementi konsiderohet katerkendor me permasa( B' f xh ) do te kemi: A0 

M   ' sp * A' sp (ho  a sp ) B ' f * h02 * Rb



612060  130 * 0.52 * 42 2

30 * 42 * 210

 0.055

  1  1  2 * A0  1  1  2 * 0.055  0.057 A' ' sp  ( * B f * h0 * Rb  A' sp * ' sp )

1

Rso

 (0.057 * 30 * 43.5 * 210  0.52 *130)

1 3600

 4.34cm 2

Per te rritur aftesine mbajtese kundrejt plasaritjeve A sp e llogarisim duke e zmadhuar me 15 % te vleres se gjetur. Asp  1.15 * 4.34  5.0cm 2

Percaktojme numrin e shufrave n

5.0 0.125

 40shufra

Asp e vendosur do te jete: Asp  40 * 0.125  5.0cm 2

Llogaritja e aftesise mbajtese kundrejt forcave prerese: Q

 * b * h * R

0 6 *11 1* 43 5 *14 3

4142 85daN  Q

2720 3daN





2. Llogaritja sipas formimit formimit te te plasurave plasurave ku kontrollohet kontrollohet nese lindin te te plasurat ne element. Krakteristikat gjeometrike dhe inerciale te seksionit s eksionit

A

h =3 ys 0

0 A

yp

h =3

Ared  Ab   Asp   A ' sp   As  B f * h f  B' f *h' f (h  h f  h' f ) * b   ( Asp  A' sp  As ) 

 3 * 30  3 * 30  (45  3  3) *11.1  5.14 * (5.0  0.52  1.04)  646.6cm 2

Percaktojme momentin statik: 11

S red

 B' f *h' f (h  0.5h' f )  B f * h f * 0.5h f  (h  h' f h f )b * 0.5h   * A' sp (h  a' sp ) 

  * Asp * 0.5h f   * As * 0.5h   30 * 3(45  1.5)  30 * 3 *1.5  (45  3  3)11.1 * 22.5  5.14 * 0.52(45  1.5)   5.14 * 5.0 *1.5  5.14 *1.04 * 22.5  14065.3cm 3 11

Y p 

S

red

Ared



14065.3 646.6

 21.75cm





00

W

s

p

W

s

r

r

bt

red



146536



146536

y s

red

bt

p



red

I





I 00 red y p

W

P

red



s

red

21.75



6737.3



6302.6

646.6

Ared W

23.25

646.6

Ared

 6302.6cm 3 ;

 6737.3cm 3 .

 10.42cm ;

 9.75cm .

Llogaritja e humbjeve Humbjet e para  o  0.71* R n sp  0.71* 7000  4970daN / cm 2  h1   3   4   7   3   4

 3 (0.27 *  4 

 0 Rsp

 1   2 L

n

 0.1) o  (0.27 *

* E sp 

0.1  0.1 10000



  

7000

 0.1) * 4970  456daN / cm 2

* 1.8 * 10  36daN / cm

 h1  456  36  492dan / cm 2

Humbjet e dyta

4970

6

2





 b 

24718.6 646.6



24718.6 * 20.4 * (21.75  1.5) 146536



165 * 9 8

2

*

(21.75  1.5) 146536

 107.7daN / cm 2

Beton i Paranderur dk

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