Themele

1

PROJEKT KURSI

Kerkesat e projektit

1. Planvendosja e projektit dhe sonda.

2. Profilet gjeologjike dhe karakteristikat e shtresave.

3. Planimetrite dhe prerjet e objektit.

4. Planet e strukturave.

5. Llogaritja e ngarkesave.

6. Llogaritja e plintave elastike e te shtanget.

7. Llogaritja e themeleve te podrumit.

8. Llogaritja ne bazament elastik (trare ose pllake).

9. Llogaritja e themeleve me pilota.

10. Krahasimi ekonomik.

Vizatimet.

11. Planimetria e themeleve me plinta, perpunimi i plintave, dhe e pjeses

se podrumit.

12. Planimetria e trareve me bazament elastik.

13. Perpunimi i nje trau.

14. Planimetria e themeleve me pilota dhe jasteku i tyre.

2

PLLAKA 0.7 cm

LLAÇ ÇIM. 2 cm

RERE 2 cm

BETON 5 cm

QERAMIKE 20 cm

SUVA 2 cm

5. Llogaritja e ngarkesave.

Ngarkesa e normuar e soletes per nje meter katror.

Soleta e nderkatit

1.Pllaka 0.007 x 2000 = 14 daN/m2

2.Llaç çim. 0.02 x 2200 = 44 daN/m2

3.Rere. 0.02 x 1600 = 32 daN/m2

4.Suva. 0.02 x 1900 = 38 daN/m2

5.SOLETA

a-Betoni. (0.05+2 x 0.13 x 0.2) x 2500 = 255 daN/m2

b-Qeramika 2 x 0.37 x 0.2 x 700 = 104 daN/m2

6.MURE TE BRENDSHEM

a.Mur tulle 20cm [(0.2 x 800 x 2.75) x 16.6] / 367.67 = 20 daN/m2

a.Mur tulle 10cm [(0.1 x 800 x 2.75) x 45.7] / 367.67 = 28 daN/m2

2.Suva. [(2 x 0.015 x 1900 x 2.75) x 62.3] / 367.67= 27 daN/m2

g = 565 daN/m2

Soleta e taraces.

1.Granil 0.02 x 1700 = 34 daN/m2

2.Hidroizolim 0.01 x 1200 = 12 daN/m2

3.Llaç çim. 0.02 x 2200 = 44 daN/m2

4.Rere. 0.02 x 1600 = 32 daN/m2

5.Suva. 0.02 x 1900 = 38 daN/m2

6.SOLETA

a-Betoni. (0.05+2 x 0.13 x 0.2) x 2500 = 255 daN/m2

b-Qeramika 2 x 0.37 x 0.2 x 700 = 104 daN/m2

g = 520 daN/m2

3

BETON 8 cm

HIDROIZOLIM 1 cm

PLLAKA 0.7 cm

LLAÇ ÇIM. 2 cm

RERE 2 cm

LLAÇ ÇIM. 2 cm

LLAÇ ÇIM. 2 cm

BETON 4 cm

Soleta e taraces.

1.Pllaka 0.007 x 2000 = 14 daN/m2

2.Llaç çim. 0.02 x 2200 = 44 daN/m2

3.Rere. 0.02 x 1600 = 32 daN/m2

4.Beton 0.08 x 2500 = 200 daN/m2

2.Llaç çim. 0.02 x 2200 = 44 daN/m2

2.Hidroizolim 0.01 x 1200 = 12 daN/m2

3.Llaç çim. 0.02 x 2200 = 44 daN/m2

4.Beton 0.05 x 2500 = 125 daN/m2

g = 515 daN/m2

Ngarkesat e perkohshme mbi:

Soleten e nderkatit dhe podrumit

150 daN/m2 x 1.4 = 210 daN/m2

Soleten e taraces

70 daN/m2 x 1.4 = 105 daN/m2

Ngarkesa e mureve per nje meter linear.

Per katin tip mur 20 cm

1.Mur tulle 0.2 x 800 x 2.66 = 426 daN/ml

2.Suva. 2 x 0.02 x 1900 x 3.06 = 233 daN/ml

g = 660 daN/ml

Per katin perdhe mur 20 cm


2.Suva. 2 x 0.02 x 1900 x 3.74 = 285 daN/ml

g = 820 daN/ml

Ngarkesa e parapetit per nje meter linear.


2.Suva. (2 x 0.9+0.25) x 0.015 x 1900=58 daN/ml

3.Kapuç. 0.18 x 0.05 x 2500 = 23 daN/ml

g = 153 daN/ml

4

GJEJ NGARKESAT QE TRANSMETOHEN NE THEMEL NGA KOLLONA.

Per kollonen K1

Ngarkesa nga:

Muret 5.05 x 660 x 1.2 x 5 + 5.05 x 820 x 1.2 = 24968 daN

Soleta e nderkatit 6.75 x (565 x 1.1+ 210) x 6 = 33676 daN

Soleta e taraces 6.75 x (520 x 1.1+ 105) x 1= 5242 daN

Parapeti 5.25 x 153 x 1.2 x 1 = 964 daN

Traret 5.25 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 20213 daN

Kollonat 0.4 x 0.4 x 2500 x 1.1 x (3.06 x 5+1 x 3.75) = 8382 daN

Ngarkesa totale ne K1 N = 93405 daN

Per kollonen K2

Ngarkesa nga:

Muret 4.55 x 660 x 1.2 x 5 + 4.55 x 820 x 1.2 = 22496 daN

Soleta e nderkatit 6.75 x (565 x 1.1+ 210) x 6 = 33676 daN

Soleta e taraces 11.14 x (520 x 1.1 + 105) x 1= 7542 daN

Shkallet 1.95 x 2.25 x 1000 x 6 x 1.1 = 28958 daN

Parapeti 4.95 x 153 x 1.2 x 1 = 909 daN

Traret 5.25 x 0.4 x 0.5 x 2500 x 1.1 x 7 + 1.95 x 0.4 x 0.5 x

2500 x 1.1 x 2 + 1.95 x 0.4 x 0.3 x 2500 x 5= 25283 daN



Per kollonen K3

Ngarkesa nga:

Muret 5.1 x 660 x 1.2 x 5 + 5.1 x 820 x 1.2 = 25215 daN

Soleta e nderkatit 7.5 x (565 x 1.1 + 210) x 6 = 37418 daN


Parapeti 5.5 x 153 x 1.2 x 1 = 1010 daN

Traret 5.5 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 21175 daN



5

Per kollonen K4

Ngarkesa nga:

Muret 4.55 x 660 x 1.2 x 5 + 4.55 x 820 x 1.2 = 22496 daN



Parapeti 4.55 x 153 x 1.2 x 1 = 836 daN

Traret 10.75 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 41388 daN



Per kollonen K5

Ngarkesa nga:



Shkallet 1.95 x 2.25 x 1000 x 6 x 1.1 = 28958 daN

Traret 9.7 x 0.4 x 0.5 x 2500 x 1.1 x 7= 37345 daN



Per kollonen K6

Ngarkesa nga:

Muret 4.6 x 660 x 1.2 x 5 + 4.6 x 820 x 1.2 = 22743 daN

Soleta e nderkatit 15 x (565 x 1.1 + 210) x 6 = 74835 daN

Soleta e taraces 15 x (520 x 1.1 + 105) x 1= 10155 daN

Parapeti 5 x 153 x 1.2 x 1 = 918 daN

Traret 8 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 30800 daN



Per kollonen K7

Ngarkesa nga:



Traret 11 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 42350 daN



6

Per kollonen K8

Ngarkesa nga:



Traret 9.95 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 38108 daN



Per kollonen K9

Ngarkesa nga:

Muret 5.6 x 660 x 1.2 x 5 + 5.6 x 820 x 1.2 = 27687 daN



Parapeti 6 x 153 x 1.2 x 1 = 1102 daN

Traret 8.5 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 32725 daN



Per kollonen K10

Ngarkesa nga:

Muret 4.55 x 660 x 1.2 x 5 + 4.55 x 820 x 1.2 = 22496 daN



Parapeti 4.95 x 153 x 1.2 x 1 = 909 daN

Traret 7.45 x 0.4 x 0.5 x 2500 x 1.1 x 7 = 28683 daN



7

LLOGARITJA E PLINTAVE TE SHTANGET

GJENDJA FIZIKE E SHTRESAVE.

Nr i

shtreses

γ0

(kN/m3)

γ

(kN/m3)

W

(%)

Ws

(%)

Wd

(%)

E·104

(kPa) φ

C

(kPa) k Emertimi

Shtresa 1 27.3 19.8 30.4 48 20 1.8 18 44 10-5 Shtresa 2 26.1 17 2.6 17.5 14.3 0.5 24 5 10-4 Shtresa 3 27.1 19.65 29.2 33.88 23.8 0.45 21 10 10-6

Treguesi I Perberjes Se Tokes.

Per shtresen e pare:

Ip=Ws1 – Wp1= 48 – 20= 28 %

Ne baze te Ip percaktoj llojin e shtreses:

ARGJILE

Per shtresen e dyte:

Ip=Ws2 – Wp2= 17.5 – 14.3= 3.2 %

Ne baze te Ip percaktoj llojin e shtreses:

SURERE

Per shtresen e trete:

Ip=Ws3 – Wp3= 33.88 – 23.8 = 10.08 %

Ne baze te Ip percaktoj llojin e shterses:

SUARGJILE

Treguesi I Gjendjes Se Porozitetit.


)W01.01(γ

γε 1

1

01+=

8.19

3.27ε = (1+0.01·30.04)=1.79

TOKE ME POROZITET TE LARTE


)W01.01(γ

γε 2

2

02+=

17

1.26ε = =(1+0.01·17.5)=1.803


8


)W01.01(γ

γε 3

3

03+=

65.19

1.27ε = (1+0.01·29.2)=1.78


Treguesi I Gjendjes Se Lageshtise.


371.0=28

204.30=

I

WW=I

1P

1P1

1K PLASTIKE


=2.3

12=

I

WW=I

2P

2P2

2K

4.3.6-3.66 E NGURTE


54.0=08.10

22=

I

WW=I

3P

3P3

3K

3.89.2 PLASTIKE E BUTE

FAKTORI HIDROLOGJIK.

Mbi NUNmax=0.5 m Tabani i ujrave nentokesore vihet → Hzh=1.8 m

Mbi NUNmin=1.5 m

Kjo behet per te ruajtur themelin nga rreziku i sufozionit dhe i gerryerjes se

bazamentit.

FAKTORI HIDROLOGJIK.

Hzh>Hngritjes se tokes=60 cm

Thellesia e zhytjes se themelit duhet te jete Hzh>Hçarjeve te tokes argjilore=80 cm

FAKTORI SIZMIK.

Hzh≥Hgodines/10≥1.8 m→Hzh=2.2 m

9

PLINTI P1

Nga llogaritjet e ngarkesave ne marrim ngarkesen e pergjithshme qe realizohet

ne koken e themelit.

N1=935 kN

M1= 227 kN·m

Q1= 309 kN

Gjej aftesine mbajtese te shtreses ku do te mbeshtet tabani i themelit duke

mare paraprakisht b=2m dhe Hzh=5.5m

Gjej siperfaqen e tabanit te themelit duke marre parasysh permasat A=b2.

Pranoj b=1.7 m

Gjej jashteqendersine e ngarkesave ne tabanin e themelit.

L = b+e=1.7+0.4=2.1 m Pranoj e = 0.4 m

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.9

3

5.31 44

R 382.245

b 1.673

Mp M Q Hzh =226.6+309 2.4 =430

Np N Ath Hzh m =935+2.8 2.4 20 =1069

eMp

Np =

430

1069=0.402

AthN

R Hzh m =

935

383 2.4 20

Ath 2.797

b Ath = 2.797

b 1.673

10

Meqe e > [e] themeli meret asimetrik dhe lartesia e themelit te shtanget do te

jete:

Kemi :αkuf=31o ak=0.4 m

Pranoj hth=1.45 m

Bej kontrollin e sforcimeve ne tabanin e themelit.

hthL ak

2 tan kuf =

2.1 0.4

2 tan 31( )o

hth 1.415

maxNp

b L

Mp 6

b L2

Np e 6

b L2

max=1069

1.7 2.1

6 430

1.7 2.12

6 1069 0.4

1.7 2.12

max 299.516

11

PLINTI P2



N1=1273 kN

M1= 127.3 kN·m

Q1= 190.5 kN




Pranoj b=2 m


Pranoj e = 0.4 m

L = b+e=2+0.4=2.4 m

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Mp M Q Hzh =127.3+190.95 2.4 =586

Np N Ath Hzh m =1273+2.67 2.4 20 =1456

eMp

Np =

586

1456=0.402

e 0.402

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

AthN

R Hzh m =

1273

383 2.4 20

Ath 3.809

b Ath = 3.8

b 1.952

12


jete:


Pranoj hth=1.7 m Hth+0.15<Hzh


σmax<1.2R

maxNp

b L

Mp 6

b L2

Np e 6

b L2

max=1456

2 2.4

6 586

2 2.42

6 1456 0.4

2 2.42

max 303.294

hthL ak

2 tan kuf =

2.4 0.4

2 tan 31( )o

hth 1.664

13

PLINTI P3



N1=983 kN

M1= 98.3 kN·m

Q1= 140 kN




Pranoj b = 1.8 m


Pranoj e = 0.4 m

L = b+e=1.75+0.4=2.2 m

AthN

R Hzh m =

983

382 2.4 20

Ath 2.941

b Ath = 2.941

b 1.715

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Mp M Q Hzh =98.3+148 20 =453

Np N Ath Hzh m =935+2.941 2.9 20 =1124

eMp

Np =

453

1124=0.402

e 0.402

14


jete:


Pranoj hth=1.5 m


max

Np

b L

Mp 6

b L2

Np 6 e

b L2

max=1124

1.8 2.2

453 6

1.8 2.22

1124 6 0.4

1.8 2.22

max 289.357

hthL ak

2 tan kuf =

2.2 0.4

2 tan 31( )o

hth 1.498

15

PLINTI P4



N1=1962 kN

M1= 196.2 kN·m

Q1= 295 kN




Pranoj b = 2.5 m


Pranoj e = 0.4 m

L = b+e=2.5+0.4=2.9 m

AthN

R Hzh m =

1962

383 2.4 20

Ath 5.87

b Ath = 5.87

b 2.423

Mp M Q Hzh =196.2+295 2.4 =903

Np N Ath Hzh m =1962+2.4 5.87 20 =2244

eMp

Np =

1050

2287=0.402

e 0.402

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

16


jete:


Pranoj hth=2 m


σmax<1.2R

hthL ak

2 tan kuf =

2.9 0.4

2 tan 31( )o

hth 2.08

max

Np

b L

Mp 6

b L2

Np 6 e

b L2

max=2244

2.5 2.9

903 6

2.5 2.92

2244 6 0.4

2.5 2.92

max 309.484

17

PLINTI P5



N1=1830 kN

M1= 183 kN·m

Q1= 295 kN




Pranoj b = 2.4 m


Pranoj e = 0.4 m

L = b+e=2.4+0.4=2.8 m

AthN

R Hzh m =

1830

382 2.4 20

Ath 5.475

b Ath = 5.475

b 2.34

Mp M Q Hzh =183+274.5 2.4 =842

Np N Ath Hzh m =1830+2.4 5.475 20 =2093Mp 841.8

eMp

Np =

842

2093=0.402

Np 2.093 103

e 0.402

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

18


jete:


Pranoj hth=2 m


σmax<1.2R

hthL ak

2 tan kuf =

2.8 0.4

2 tan 31( )o

hth 1.997

19

PLINTI P6



N1=1479 kN

M1= 148 kN·m

Q1= 222 kN




Pranoj b = 2.2 m


Pranoj e = 0.4 m

L = b+e=2.2+0.4=2.6 m

AthN

R Hzh m =

1479

382 2.4 20

Ath 4.425

b Ath = 4.425

b 2.104

b 2.104

Mp M Q Hzh =148+222 2.4 =680

Np N Ath Hzh m =1479+2.4 20 4.425 =1691

eMp

Np =

680

1691=0.402

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

20


jete:


Pranoj hth=1.85 m


hthL ak

2 tan kuf =

2.6 0.4

2 tan 31( )o

hth 1.831

max

Np

b L

Mp 6

b L2

Np 6 e

b L2

max=1691

2.2 2.6

680 6

2.2 2.62

1691 6 0.4

2.2 2.62

max 295.698

21

PLINTI P7



N1=2208 kN

M1= 221 kN·m

Q1= 331.2 kN




Pranoj b = 2.6 m


Pranoj e = 0.4 m

L = b+e=2.6+0.4=3 m

AthN

R Hzh m =

2208

383 2.4 20

Ath 6.606

b Ath = 6.606

b 2.57

b 2.57

Mp M Q Hzh =221+331.2 2.4 =1016

Np N Ath Hzh m =2208+2.4 20 6.6 =2525

eMp

Np =

1016

2525=0.402

e 0.402

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

22


jete:


Pranoj hth=2.2 m


σmax<1.2 R

hthL ak

2 tan kuf =

3 0.4

2 tan 31( )o

hth 2.164

max

Np

b L

Mp 6

b L2

Np 6 e

b L2

max=2525

2.6 3

1016 6

2.6 32

2525 6 0.4

2.6 32

max 323.729

23

PLINTI P8



N1=1868 kN

M1= 187 kN·m

Q1= 280 kN




Pranoj b = 2.4 m


Pranoj e = 0.4 m

L = b+e=2.4+0.46=2.8 m

AthN

R Hzh m =

1868

383 2.4 20

Ath 5.589

b Ath = 5.589

b 2.364

Q 280.2Mp M Q Hzh =187+280 2.4 =859

Np N Ath Hzh m =1868+2.9 5.589 20 =2136Mp 859.28

eMp

Np =

859

2136=0.402

Np 2.136 103

e 0.402

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

24


jete:


Pranoj hth=2 m Hth+0.15<Hzh


hthL ak

2 tan kuf =

2.8 0.4

2 tan 31( )o

hth 1.997

max

Np

b L

Mp 6

b L2

Np 6 e

b L2

max=2136

2.4 2.8

859 6

2.4 2.82

2136 6 0.4

2.4 2.82

max 317.896

25

PLINTI P9



N1=1549 kN

M1= 155 kN·m

Q1= 232 kN




Pranoj b = 2.2 m


Pranoj e = 0.4 m

L = b+e=2.2+0.4=2.6 m

AthN

R Hzh m =

1549

383 2.4 20

Ath 4.631

b Ath = 4.631

b 2.152

b 2.153

Mp M Q Hzh =280+233 2.4 =713

Np N Ath Hzh m =1549+2.4 4.634 20 =1771

eMp

Np =

713

1711=0.402

e 0.402

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

26


jete:




hthL ak

2 tan kuf =

2.6 0.4

2 tan 31( )o

hth 1.831

max

Np

b L

Mp 6

b L2

Np 6 e

b L2

max=1711

2.2 2.6

713 6

2.2 2.62

1711 6 0.4

2.2 2.62

max 309.694

27

PLINTI P10



N1=1306 kN

M1= 131 kN·m

Q1= 196 kN




Pranoj b = 2 m


Pranoj e = 0.4 m

L = b+e=2+0.4=2.4 m

AthN

R Hzh m =

1306

383 2.4 20

Ath 3.907

b Ath = 3.907

b 1.977

Mp M Q Hzh =131+196 2.4 =600

Np N Ath Hzh m =1306+3.907 2.4 20 =1494

eMp

Np =

600

1494=0.402

e 0.402

m 20 A 0.43 D 5.31 b 2 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2.4

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.4

3

5.31 44

R 382.245

28


jete:




hthL ak

2 tan kuf =

2.4 0.4

2 tan 31( )o

hth 1.664

maxNp

b L

Mp 6

b L2

Np e 6

b L2

max=1494

2 2.4

6 600

2 2.42

6 1494 0.4

2 2.42

max 311.157

29

LLOGARITJA E PLINTAVE ELASTIKE

Llogaritja e plintave elastike eshte e njejte me perjashtim te llogaritjes se lartesise

se themelit e cila percaktohet nga tre kondita.

Kondita ne prerje e betonit te plintit nga kollona

Kondita e sforcimeve kryesore

Kondita e inkastrimit te fitilave te kollones ne plint.

PLINTI P1

b=1.7 N1= 935 KN

L=2.1 M=93.5

Ath=3.57 m2 R=383 Kpa

Gjejme lartesine e themelit duke u nisur nga kondita ne prerje.

Pranoj hth=0.6 m

Gjej lartesine e dobishme duke u nisur nga kondita e sforcimeve kryesore.

Gjej sforcimet qe lindin ne skajet e tabanit te themelit.

σmax=σmin=300 KPa

Gjatesia e konsolit qe inkastrohet ne kollone eshte:

Sforcimi kryesor qe lind ne taban eshte:

Forca prerese qe lind eshte: Qmax=σ1·C=300·1.25=75 KN

hthN

2 ak bk Rpr m =

935

2 0.4 0.4( ) 1040 1

hth 0.562

min 300

CL ak

2e =

2.1 0.4

20.4

C 1.25

1 min

max min

L

LC

2

= 300300 300

2.1

2.11.25

2

1 300

30

Lartesia e dobishme e themelit eshte:

Pranoj hth=0.7 m

Kondita e inkastrimit te fitilave

hth > 30·φ=30·1.6=48 cm

Nga tre lartesite e gjetura pranojme me te madhen.

hth=70 cm

Llogaritja e siperfaqeve te armimit me ngarkesen llogaritese.

b=170 cm e=40 cm Rb=115 kg/cm2

L=210 cm ak=40 cm Rç=2100 kg/cm2

hth=80 cm a=5 cm σ=3 kg/cm2

ho=hth-a Llogarisim gjatesite e lira te konsolave te themelit

Gjej momentet perkulese qe veprojne ne tabanin e themelit.

MxCy

2

2L My

Cx2

2b

Mx 34728.75 kg m My 39843.75 kg m

Gjej siperfaqet e armimit sipas te dy drejtimeve

AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.036 Aoy 0.02557

x 1 1 2Aox y 1 1 2Aoy

x 0.037 y 0.0259

h0Qmax

Rt m =

935

540 1

h0 0.694

CxL ak

2e Cy

b ak

2

Cx210 0.4

20.4 =125 cm

Cy170 0.4

2 =105 cm

31

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.00258 m2

Asy 0.00223 m2

Bejme armimin e plintit sipas drejtimit x me:

11 x Φ18

Bejme armimin e plintit sipas drejtimit y me:

20 x Φ12

A

A

1

1

-2.50

-1.80

-2.50

-1.80

210

170

210

70

170

70

3 Ø5

6 Ø16

6 Ø16

11 Ø18 20 Ø12

32

PLINTI P2

b=2 N1= 1273 KN

L=2.4 M=93.5

Ath=3.57 m2 R=383 KPa


Pranoj hth=0.8 m



σmax=σmin=304 KPa





Pranoj hth=0.8 m


hth > 30·φ=30·1.6=48 cm


hth=80 cm

hthN

2 ak bk Rpr m =

1273

2 0.4 0.4( ) 1040 1

hth 0.765

min 300

CL ak

2e =

2.4 0.4

20.4

C 1.4

1 min

max min

L

LC

2

= 300304 304

2.1

2.41.4

2

1 304

h0Qmax

Rt m =

426

540 1

h0 0.788

33




hth=80 cm a=5 cm σ=3.06 kg/cm2

ho=hth-a

Llogarisim gjatesite e lira te konsolave te themelit

CxL ak

2e Cy

b ak

2

Cx 1.4 m Cy 0.8 m


MxCy

2

2L My

Cx2

2b

Mx 23500.8 kg m My 59976 kg m


AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.046 Aoy 0.01514

x 1 1 2Aox y 1 1 2Aoy

x 0.047 y 0.01525

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.0039 m2

Asy 0.0015 m2


16 x Φ18


14 x Φ12

34

PLINTI P3

b=1.8 N1= 983 KN

L=2.2 M=98.3 Ath=3.57 m2 R=383 KPa


Pranoj hth=0.6 m

Gjej lartesine e dobishme duke u nisur nga kondita e sforcimeve kryesore. Gjej sforcimet qe lindin ne skajet e tabanit te themelit.

σmax=σmin=290 Kpa


hthN

2 ak bk Rpr m =

983

2 0.4 0.4( ) 1040 1

hth 0.591

min 300

CL ak

2e =

2.2 0.4

20.4

C 1.3

1

-2.40

-1.60

-2.40

-1.60

240

200

240

80

200

80

Ø5 - 4

6 Ø16

6 Ø16

16 Ø18 14 Ø12

A

A 1

35




Pranoj hth=0.7 m


hth > 30·φ=30·1.6=48 cm


hth=70 cm





ho=hth-a


2.9kg

cm2

CxL ak

2e Cy

b ak

2

Cx 1.3 m Cy 0.7 m


MxCy

2

2L My

Cx2

2b

Mx 15631 kg m My 44109 kg m Gjej siperfaqet e armimit sipas te dy drejtimeve

AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.05 Aoy 0.01462

x 1 1 2Aox y 1 1 2Aoy

x 0.052 y 0.01473

C 1.3

1 min

max min

L

LC

2

= 290290 290

2.1

2.21.3

2

1 290

Qmax 377

h0Qmax

Rt m =

377

540 1

h0 0.698

36

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.0039 m2

Asy 0.0015 m2


14 x Φ18


11 x Φ12

A

A

1

1

-2.50

-1.80

-2.50

-1.80

220

180

220

70

180

70

3 Ø5

6 Ø16

6 Ø16

14 Ø18 11 Ø12

37

PLINTI P4

b=2.5 N1= 1962 KN

L=2.9 M=197

Ath=3.57 m2 R=383 KPa


Pranoj hth=1.2 m



σmax=σmin=310 KPa





Pranoj hth=0.8 m


hth > 30·φ=30·1.6=48 cm


hth=120 cm

hthN

2 ak bk Rpr m =

1962

2 0.4 0.4( ) 1040 1

hth 1.179

min 300

CL ak

2e =

2.9 0.4

20.4

C 1.65

C 1.3

1 min

max min

L

LC

2

= 310310 310

2.1

2.91.65

2

1 310

Qmax 403

h0Qmax

Rt m =

403

540 1

h0 0.746

38





ho=hth-a


CxL ak

2e Cy

b ak

2

Cx 1.65 m Cy 1.05 m


MxCy

2

2L My

Cx2

2b

Mx 49557.375kg m My 105496.875kg m


AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.028 Aoy 0.01124

x 1 1 2Aox y 1 1 2Aoy

x 0.028 y 0.0113 Asx

x b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.00443 m2

Asy 0.00206 m2


18 x Φ18


19 x Φ12

39

PLINTI P5

b=2.4 N1= 1830 KN

L=2.8 M=183 Ath=3.57 m2 R=383 KPa


Pranoj hth=1.1 m

Gjej lartesine e dobishme duke u nisur nga kondita e sforcimeve kryesore. Gjej sforcimet qe lindin ne skajet e tabanit te themelit.

σmax=σmin=305 KPa


hthN

2 ak bk Rpr m =

1830

2 0.4 0.4( ) 1040 1

hth 1.1

min 300

CL ak

2e =

2.2 0.4

20.4

C 1.6

A

A

1

1

-2.50

-1.30

-2.50

-1.30

290

250

290

120

250

120

6 Ø5

6 Ø16

6 Ø16

18 Ø18 19 Ø12

40




Pranoj hth=1 m


hth > 30·φ=30·1.6=48 cm


hth=110 cm





ho=hth-a


2.9kg

cm2

CxL ak

2e Cy

b ak

2

Cx 1.6 m Cy 1 m Gjej momentet perkulese qe veprojne ne tabanin e themelit.

MxCy

2

2L My

Cx2

2b

Mx 42700 kg m My 93696 kg m


AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.03079168 Aoy 0.01203

x 1 1 2Aox y 1 1 2Aoy

x 0.03128093 y 0.0121

C 1.3

1 min

max min

L

LC

2

= 306306 306

2.1

2.41.6

2

1 306

Qmax 489.6

h0Qmax

Rt m =

490

540 1

h0 0.907

41

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.00432 m2

Asy 0.00195 m2


17 x Φ18


18 x Φ12

A

A

1

1

-2.50

-1.40

-2.50

-1.40

280

240

280

110

240

110

5 Ø5 - 4

6 Ø16 - 3

6 Ø16

17 Ø18 18 Ø12

42

PLINTI P6

b=2.2 N1= 1479 KN

L=2.6 M=148

Ath=3.57 m2 R=383 KPa


Pranoj hth=0.9 m



σmax=σmin=296 KPa





Pranoj hth=0.9 m


hth > 30·φ=30·1.6=48 cm


hth=90 cm

hthN

2 ak bk Rpr m =

1479

2 0.4 0.4( ) 1040 1

hth 0.889

min 300

CL ak

2e =

2.6 0.4

20.4

C 1.5

C 1.5

1 min

max min

L

LC

2

= 296296 296

2.1

2.61.5

2

1 296

Qmax 444

h0Qmax

Rt m =

490

540 1

h0 0.822

43





ho=hth-a


CxL ak

2e Cy

b ak

2

Cx 1.5 m Cy 0.9 m


MxCy

2

2L My

Cx2

2b

Mx 31168.8 kg m My 73260 kg m


AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.04007823 Aoy 0.01443

x 1 1 2Aox y 1 1 2Aoy

x 0.04091526 y 0.01453

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.00419 m2

Asy 0.00176 m2


17 x Φ18


16 x Φ12

44

PLINTI P7

b=2.6 N1= 2208 KN

L=3 M=221 Ath=3.57 m2 R=383 KPa


Pranoj hth=1.4 m



σmax=σmin=324 KPa


hthN

2 ak bk Rpr m =

2208

2 0.4 0.4( ) 1040 1

hth 1.327

min 300

CL ak

2e =

3 0.4

20.4

C 1.7

A

A

1

1

-2.50

-1.60

-2.50

-1.60

260

220

260

90

220

90

4 Ø5

6 Ø16

6 Ø16

17 Ø18 16 Ø12

45




Pranoj hth=1.1 m


hth > 30·φ=30·1.6=48 cm


hth=140 cm





ho=hth-a


2.9kg

cm2

CxL ak

2e Cy

b ak

2

Cx 1.7 m Cy 1.1 m Gjej momentet perkulese qe veprojne ne tabanin e themelit.

MxCy

2

2L My

Cx2

2b

Mx 58806 kg m My 121726.8 kg m Gjej siperfaqet e armimit sipas te dy drejtimeve

AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.02233816 Aoy 0.00935

x 1 1 2Aox y 1 1 2Aoy

x 0.0225934 y 0.0094

C 1.5

1 min

max min

L

LC

2

= 324324 324

3

31.7

2

1 324

Qmax 550.8

h0Qmax

Rt m =

490

540 1

h0 1.02

46

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.00434 m2

Asy 0.00208 m2


18 x Φ18


19 x Φ12

A

A

1

1

-2.50

-1.10

-2.50

-1.10

300

260

300

140

260

140

7 Ø5

6 Ø16

6 Ø16

18 Ø5 19 Ø5

47

PLINTI P9

b=2.2 N1= 1594 KN

L=2.6 M=160

Ath=3.57 m2 R=383 KPa


Pranoj hth=1 m



σmax=σmin=310 KPa





Pranoj hth=0.9 m


hth > 30·φ=30·1.6=48 cm


hth=100 cm

hthN

2 ak bk Rpr m =

1594

2 0.4 0.4( ) 1040 1

hth 0.958

min 300

CL ak

2e =

2.6 0.4

20.4

C 1.5

C 1.5

1 min

max min

L

LC

2

= 310310 310

2.2

2.61.5

2

1 310

h0Qmax

Rt m =

465

540 1

h0 0.861

48




hth=100 cm a=5 cm σ=310 KPa

ho=hth-a


CxL ak

2e Cy

b ak

2

Cx 1.5 m Cy 0.9 m


MxCy

2

2L My

Cx2

2b

Mx 32643 kg m My 76725 kg m


AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.04197382 Aoy 0.01511

x 1 1 2Aox y 1 1 2Aoy

x 0.04289376 y 0.01523

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.00439 m2

Asy 0.00184 m2


18 x Φ18


17 x Φ12

49

PLINTI P10

b=2 N1= 1306 KN

L=2.4 M=221 Ath=3.57 m2 R=383 KPa


Pranoj hth=0.8 m



σmax=σmin=312 KPa


hthN

2 ak bk Rpr m =

1306

2 0.4 0.4( ) 1040 1

hth 0.785

min 300

CL ak

2e =

2.4 0.4

20.4

C 1.4

A

A

1

1

-2.50

-1.50

-2.50

-1.50

260

22

0

260

10

0

220

10

0

5 Ø5

6 Ø16

6 Ø16

18 Ø18 17 Ø12

50




Pranoj hth=0.9 m


hth > 30·φ=30·1.6=48 cm


hth=90 cm




hth=90 cm a=5 cm σ=312 KPa

ho=hth-a Llogarisim gjatesite e lira te konsolave te themelit

2.9kg

cm2

CxL ak

2e Cy

b ak

2

Cx 1.4 m Cy 0.8 m


MxCy

2

2L My

Cx2

2b

Mx 23961.6 kg m My 61152 kg m Gjej siperfaqet e armimit sipas te dy drejtimeve

AoxMy

b ho2

Rb

AoyMx

L ho2

Rb

Aox 0.03679976 Aoy 0.01202

x 1 1 2Aox y 1 1 2Aoy

x 0.037503 y 0.01209

1 min

max min

L

LC

2

= 312312 312

2.4

2.41.4

2

1 312

Qmax 436.8

h0Qmax

Rt m =

440

540 1

h0 0.809

51

Asxx b ho Rb

Rs Asy

y L ho Rb

Rs

Asx 0.00351 m2

Asy 0.00136 m2


14 x Φ18


12 x Φ12

A

A

1

1

-2.50

-1.60

-2.50

-1.60

240

200

240

90

200

90

6 Ø5

14 Ø18 12 Ø12

4 Ø5

6 Ø16

52

SHTRESAT z L b β=z/b α=L/b ko σz σtok σzmes

0 2.1 1.7 0.000 1.235 1.000 217.000 42.500 0.000

0.4 2.1 1.7 0.235 1.235 0.907 196.745 50.420 206.872

0.8 2.1 1.7 0.471 1.235 0.728 158.063 58.340 177.404

1.2 2.1 1.7 0.706 1.235 0.572 124.077 66.260 141.070

1.6 2.1 1.7 0.941 1.235 0.419 90.937 74.180 107.507

2 2.1 1.7 1.176 1.235 0.325 70.609 82.100 80.773

2.4 2.1 1.7 1.412 1.235 0.251 54.553 90.020 62.581

2.8 2.1 1.7 1.647 1.235 0.197 42.825 97.940 48.689

3.2 2.1 1.7 1.882 1.235 0.155 33.694 105.860 38.260

3.6 2.1 1.7 2.118 1.235 0.126 27.399 113.780 30.547

4 2.1 1.7 2.353 1.235 0.110 23.939 121.700 25.669

SHTRESA 2 0.008

LLOGARITJA E ULJEVE TE PLINTAVE ELASTIKE

Plinti P1

b=1.7 N1= 935 KN hsh1=2 m

L=2.1 R=383 Kpa hsh1=5 m

Ath=3.57 m2 Hzh=2.5 m hsh1=10 m

Gjej presionin llogarites:

kPa 2175.21857.3

935P

HγA

NPPP

ll

zh

th

1matll

Percaktoj hi=1/4 - 1/8 e b dhe kryej llogaritjet si ne tabelen e meposhtme

h1=0.2 m

Llogaritjet vazhdojne deri sa σz< σT/5

E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a

Ulja e plitit P1 eshte 8 mm

53


0 2.4 2 0.000 1.200 1.000 220.000 42.500 0.000

0.4 2.4 2 0.200 1.200 0.920 202.470 50.420 211.235

0.8 2.4 2 0.400 1.200 0.783 172.163 58.340 187.317

1.2 2.4 2 0.600 1.200 0.638 140.386 66.260 156.275

1.6 2.4 2 0.800 1.200 0.506 111.399 74.180 125.893

2 2.4 2 1.000 1.200 0.375 82.412 82.100 96.906

2.4 2.4 2 1.200 1.200 0.312 68.737 90.020 75.574

2.8 2.4 2 1.400 1.200 0.250 55.062 97.940 61.899

3.2 2.4 2 1.600 1.200 0.202 44.352 105.860 49.707

3.6 2.4 2 1.800 1.200 0.166 36.608 113.780 40.480

4 2.4 2 2.000 1.200 0.131 28.864 121.700 32.736

SHTRESA 2 0.009

Plinti P2

b=2 N1= 1273 KN hsh1=2 m

L=2.4 R=383 Kpa hsh1=5 m

Ath=4.8 m2 Hzh=2.5 m hsh1=10 m


kPa 2.2205.2188.4

1273P

HγA

NPPP

ll

zh

th

1matll

-

--


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


54


0 2.2 1.8 0.000 1.222 1.000 204.000 42.500 0.000

0.4 2.2 1.8 0.222 1.222 0.912 185.988 50.420 194.994

0.8 2.2 1.8 0.444 1.222 0.748 152.673 58.340 169.330

1.2 2.2 1.8 0.667 1.222 0.596 121.607 66.260 137.140

1.6 2.2 1.8 0.889 1.222 0.451 92.019 74.180 106.813

2 2.2 1.8 1.111 1.222 0.344 70.128 82.100 81.074

2.4 2.2 1.8 1.333 1.222 0.274 55.934 90.020 63.031

2.8 2.2 1.8 1.556 1.222 0.212 43.273 97.940 49.603

3.2 2.2 1.8 1.778 1.222 0.173 35.214 105.860 39.244

3.6 2.2 1.8 2.000 1.222 0.133 27.155 113.780 31.184

4 2.2 1.8 2.222 1.222 0.118 24.112 121.700 25.633

SHTRESA 2 0.008

Plinti P3

b=1.8 N1= 983 KN hsh1=2 m

L=2.2 R=383 Kpa hsh1=5 m



kPa 2045.21896.3

983P

HγA

NPPP

ll

zh

th

1matll

-

--


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


55


0 2.9 2.5 0.000 1.160 1.000 226.000 42.500 0.000

0.4 2.9 2.5 0.160 1.160 0.936 211.524 50.420 218.762

0.8 2.9 2.5 0.320 1.160 0.845 190.883 58.340 201.204

1.2 2.9 2.5 0.480 1.160 0.718 162.314 66.260 176.598

1.6 2.9 2.5 0.640 1.160 0.609 137.549 74.180 149.931

2 2.9 2.5 0.800 1.160 0.501 113.327 82.100 125.438

2.4 2.9 2.5 0.960 1.160 0.394 89.106 90.020 101.217

2.8 2.9 2.5 1.120 1.160 0.331 74.734 97.940 81.920

3.2 2.9 2.5 1.280 1.160 0.282 63.646 105.860 69.190

3.6 2.9 2.5 1.440 1.160 0.233 52.558 113.780 58.102

4 2.9 2.5 1.600 1.160 0.197 44.495 121.700 48.527

4.4 2.9 2.5 1.760 1.160 0.169 38.246 130.020 41.371

4.8 2.9 2.5 1.920 1.160 0.142 32.140 138.740 35.193

SHTRESA 2 0.012

Plinti P4

b=2.5 N4= 1962 KN hsh1=2 m

L=2.9 R=383 Kpa hsh1=5 m

Ath=7.25 m2 Hzh=2.5 m hsh1=10 m


kPa 2265.218-25.7

1962P

Hγ-A

NP-PP

ll

zh

th

1matll


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


56


0 2.8 2.4 0.000 1.167 1.000 228.000 42.500 0.000

0.4 2.8 2.4 0.167 1.167 0.933 212.800 50.420 220.400

0.8 2.8 2.4 0.333 1.167 0.834 190.203 58.340 201.501

1.2 2.8 2.4 0.500 1.167 0.703 160.208 66.260 175.205

1.6 2.8 2.4 0.667 1.167 0.591 134.824 74.180 147.516

2 2.8 2.4 0.833 1.167 0.480 109.440 82.100 122.132

2.4 2.8 2.4 1.000 1.167 0.369 84.056 90.020 96.748

2.8 2.8 2.4 1.167 1.167 0.317 72.377 97.940 78.217

3.2 2.8 2.4 1.333 1.167 0.266 60.699 105.860 66.538

3.6 2.8 2.4 1.500 1.167 0.215 49.020 113.780 54.859

4 2.8 2.4 1.667 1.167 0.186 42.433 121.700 45.727

4.4 2.8 2.4 1.833 1.167 0.157 35.847 130.020 39.140

4.8 2.8 2.4 2.000 1.167 0.128 29.260 138.740 32.553

SHTRESA 2 0.012

Plinti P5

b=2.4 N5= 1830 KN hsh1=2 m

L=2.8 R=383 Kpa hsh1=5 m

Ath=7.25 m2 Hzh=2.5 m hsh1=10 m


kPa 2285.21872.6

1830P

HγA

NPPP

ll

zh

th

1matll

-

-


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


57


0 2.6 2.2 0.000 1.182 1.000 214.000 42.500 0.000

0.4 2.6 2.2 0.182 1.182 0.927 198.465 50.420 206.232

0.8 2.6 2.2 0.364 1.182 0.811 173.485 58.340 185.975

1.2 2.6 2.2 0.545 1.182 0.673 144.043 66.260 158.764

1.6 2.6 2.2 0.727 1.182 0.552 118.215 74.180 131.129

2 2.6 2.2 0.909 1.182 0.432 92.386 82.100 105.300

2.4 2.6 2.2 1.091 1.182 0.343 73.462 90.020 82.924

2.8 2.6 2.2 1.273 1.182 0.287 61.443 97.940 67.452

3.2 2.6 2.2 1.455 1.182 0.231 49.423 105.860 55.433

3.6 2.6 2.2 1.636 1.182 0.193 41.325 113.780 45.374

4 2.6 2.2 1.818 1.182 0.161 34.534 121.700 37.929

4.4 2.6 2.2 2.000 1.182 0.130 27.742 130.020 31.138

4.8 2.6 2.2 2.182 1.182 0.118 25.210 138.740 26.476

SHTRESA 2 0.010

Plinti P6

b=2.2 N6= 1479 KN hsh1=2 m

L=2.6 R=383 Kpa hsh1=5 m

Ath=5.72 m2 Hzh=2.5 m hsh1=10 m


kPa 2145.21872.5

1479P

HγA

NPPP

ll

zh

th

1matll

-

--


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a

Ulja e plintit P6 eshte 10 mm

58


0 3 2.6 0.000 1.154 1.000 238.000 42.500 0.000

0.4 3 2.6 0.154 1.154 0.938 223.331 50.420 230.666

0.8 3 2.6 0.308 1.154 0.854 203.306 58.340 213.318

1.2 3 2.6 0.462 1.154 0.733 174.351 66.260 188.828

1.6 3 2.6 0.615 1.154 0.625 148.671 74.180 161.511

2 3 2.6 0.769 1.154 0.521 124.082 82.100 136.377

2.4 3 2.6 0.923 1.154 0.418 99.494 90.020 111.788

2.8 3 2.6 1.077 1.154 0.343 81.597 97.940 90.546

3.2 3 2.6 1.231 1.154 0.296 70.393 105.860 75.995

3.6 3 2.6 1.385 1.154 0.249 59.189 113.780 64.791

4 3 2.6 1.538 1.154 0.207 49.208 121.700 54.199

4.4 3 2.6 1.692 1.154 0.180 42.899 130.020 46.054

4.8 3 2.6 1.846 1.154 0.154 36.590 138.740 39.745

SHTRESA 2 0.013

Plinti P7

b=2.6 N6= 2208 KN hsh1=2 m

L=3 R=383 Kpa hsh1=5 m



kPa 2385.2188.7

2208P

HγA

NPPP

ll

zh

th

1matll

-

-


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


59


0 2.8 2.4 0.000 1.167 1.000 233.000 42.500 0.000

0.4 2.8 2.4 0.167 1.167 0.933 217.467 50.420 225.233

0.8 2.8 2.4 0.333 1.167 0.834 194.374 58.340 205.920

1.2 2.8 2.4 0.500 1.167 0.703 163.721 66.260 179.048

1.6 2.8 2.4 0.667 1.167 0.591 137.781 74.180 150.751

2 2.8 2.4 0.833 1.167 0.480 111.840 82.100 124.810

2.4 2.8 2.4 1.000 1.167 0.369 85.899 90.020 98.870

2.8 2.8 2.4 1.167 1.167 0.317 73.965 97.940 79.932

3.2 2.8 2.4 1.333 1.167 0.266 62.030 105.860 67.997

3.6 2.8 2.4 1.500 1.167 0.215 50.095 113.780 56.062

4 2.8 2.4 1.667 1.167 0.186 43.364 121.700 46.729

4.4 2.8 2.4 1.833 1.167 0.157 36.633 130.020 39.998

4.8 2.8 2.4 2.000 1.167 0.128 29.902 138.740 33.267

SHTRESA 2 0.012

Plinti P8

b=2.4 N8= 1868 KN hsh1=2 m

L=2.8 R=383 Kpa hsh1=5 m

Ath=6.72 m2 Hzh=2.5 m hsh1=10 m


kPa 2335.21872.6

1868P

HγA

NPPP

ll

zh

th

1matll

-

-


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


60


0 2.6 2.2 0.000 1.182 1.000 234.000 42.500 0.000

0.4 2.6 2.2 0.182 1.182 0.927 217.013 50.420 225.506

0.8 2.6 2.2 0.364 1.182 0.811 189.699 58.340 203.356

1.2 2.6 2.2 0.545 1.182 0.673 157.505 66.260 173.602

1.6 2.6 2.2 0.727 1.182 0.552 129.263 74.180 143.384

2 2.6 2.2 0.909 1.182 0.432 101.020 82.100 115.142

2.4 2.6 2.2 1.091 1.182 0.343 80.328 90.020 90.674

2.8 2.6 2.2 1.273 1.182 0.287 67.185 97.940 73.756

3.2 2.6 2.2 1.455 1.182 0.231 54.042 105.860 60.614

3.6 2.6 2.2 1.636 1.182 0.193 45.187 113.780 49.615

4 2.6 2.2 1.818 1.182 0.161 37.761 121.700 41.474

4.4 2.6 2.2 2.000 1.182 0.130 30.335 130.020 34.048

4.8 2.6 2.2 2.182 1.182 0.118 27.566 138.740 28.950

SHTRESA 2 0.011

Plinti P9

b=2.2 N8= 1594 KN hsh1=2 m

L=2.6 R=383 Kpa hsh1=5 m

Ath=5.72 m2 Hzh=2.5 m hsh1=10 m


kPa 2345.21872.5

1594P

HγA

NPPP

ll

zh

th

1matll

-

-


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


61


0 2.4 2 0.000 1.200 1.000 228.000 42.500 0.000

0.4 2.4 2 0.200 1.200 0.920 209.833 50.420 218.916

0.8 2.4 2 0.400 1.200 0.783 178.424 58.340 194.128

1.2 2.4 2 0.600 1.200 0.638 145.491 66.260 161.958

1.6 2.4 2 0.800 1.200 0.506 115.450 74.180 130.471

2 2.4 2 1.000 1.200 0.375 85.409 82.100 100.429

2.4 2.4 2 1.200 1.200 0.312 71.236 90.020 78.323

2.8 2.4 2 1.400 1.200 0.250 57.064 97.940 64.150

3.2 2.4 2 1.600 1.200 0.202 45.965 105.860 51.514

3.6 2.4 2 1.800 1.200 0.166 37.939 113.780 41.952

4 2.4 2 2.000 1.200 0.131 29.914 121.700 33.926

SHTRESA 2 0.010

Plinti P10

b=2 N1= 1306 KN hsh1=2 m

L=2.4 R=383 Kpa hsh1=5 m



kPa 2285.2188.4

1306P

HγA

NPPP

ll

zh

th

1matll


h1=0.2 m


E1=5000 β1=0.5

E2=18000 β2=0.4

E3=4500 β3=0.4

E

β=a


62

LLOGARITJA E THEMELIT TE PODRUMIT.

Llogarisim themelin e vazhduar nen murin e podrumit per kollonen me te

ngarkuar

Llogarisim aftesine mbajtese te shtreses ne te cilen do te mbeshtet tabani i

themelit te podrumit.

t1=1 m t2=0.3 m

383.1=18

233.0+1=

γ

γt+t=t

t

b2

1

Ngarkesen nga kollona e shperndajme ne menyre uniforme ne murin e

podrumit dhe bashke me peshen e murit marim ngarkesen e shperndare qe

vepron mbi themelin e vazhduar.

N’=N4/L=1962/5.25=374 kN/ml

Nm=bm·Hm·γ·n=0.3·2.81·23·1.1=22 kN/ml

Np=N’+Nm=374+22=396 kN/ml

Llogarisim gjeresine e tabanit te themelit te vazhduar

973.0120437

396

hγR

NF

thzhmes

p

m2

973.0=m1

A=b m pranoj b=1 m

Gjej raprtin bm/b dhe ne baze te tij percaktoj skemen.

3.0=1

3.0=

b

bm <0.7 marim skemen si me poshte

t 1.383

Rm1 m2

k A b

2B Hzh t 3

D C

R=1.21.2

1 0.43 2 2 2.72

2 2.06 1.383( )

3

19.8 5.31 44

R 436.187

63

Ea

Gjej presionin aktiv te dheut qe vepron mbi murin e podrumit

Gjej γmes dhe φmes

228.233.02

3.01822

hh

hφhφφ

88.173.02

3.08.19217

hh

hγhγγ

21

2211mes

21

2211mes

4

Presioni aktiv do te jete:

1.311.1598.02

3.288.17E

598.0)2

228.2845(tg)

2

φ45(tgk

ηk2

HγE

2

a

mesa

a

2mes

a

Gjej momentin e shkaktuar nga presioni aktiv.

Mo=Ea·L/3

Mo=31.1·4.36/3=45.2 kN·m

Gjej momentin e reduktuar

3m

z

oy

)b

b(

C

E31

MM

ku: Cz eshte funksion i aftesise mbajtese R Cz=76000

E=0.8·α·R R rezistenca ne shtypje e murit R=1200 kPa

E=0.8·1000·1200=960000

64

6.36=

)1

3.0(

36.476000

9600003+1

2.45=M

3y kN·m

6.615=

6

1

6.36+

1

396=

6

bl

M+

A

N=σ

32

yp

max kPa

4.176=

6

1

6.36

1

396=

6

bl

M

A

N=σ

32

yp

min kPa

σmax>1.2·R

Zmadhojme siperfaqen e themelit

Pranojme b=1.2 m

457=

6

2.1

6.36+

2.1

396=

6

bl

M+

A

N=σ

32

yp

max kPa

σmax<1.2·R

65

LLOGARITJA E THEMELIT ME BAZAMENT ELASTIK.

Gjejme aftesine mbajtese te tokes per Hzh=2.4 m

Gjej momentet e inertesise se pllakes sipas te dy akseve

Ix=31044 m4

Iy=9777 m4

Gjej forcen normale dhe momentet qe veprojne ne qender te pllakes,

∑N=30800 kN

My=N·ex ex=x’-L/2

Mx=N·ey ey=Y’-B/2

m 15.14'x

30800

)14789832(1.28)154922071963934((1.22

30800

)1306186718301272(1.16)1306186718301272(2.12

30800

)154922071963934(2.6)14789832(2.0'x

30800

)NN2(1.28)NNNN(1.22)NNNN(1.16

30800

)NNNN(2.12)NNNN(2.6)NN2(2.0'x

63974110852

10852974163

Rm1 m2

k A b

2

3BHzh

D C

R=1.2 1.2

10.43 2

2 2.72 2.9

3

5.31 44

R 544.24

m 20 A 0.43 D 5.31 b 16.9 m1 1.2

k 1 m2 1.2C 44 B 2.72 Hzh 2

66

m 22.7'y

30800

)127229342(7.14)18302196329832(2.10

30800

)186722207214782(2.5)13062154929832(2.0'x

30800

)N2N2(7.14

30800

)N2N2N2(2.10)N2N2N2(2.5)N2N2N2(2.0'y

21

5438761093

02

3.2815.14

2

L'xex

m 23.02

9.1422.7

2

B'yey

Mx = N·ey = 30800 · 0.23 = 7084

My = 0

A

K3

K6

B

E D

C

F

G

67

Gjejme sforcimet ne pikat skajore te rripave ne te cilat e ndajme pllaken.

X

X

Y

Y

I

yM

I

xM

F

Nσ

Pika A

2Am

KN 55.84

31044

75.37084

368

30800σ

Pika C

2Am

KN 55.84

31044

75.37084

368

30800σ

Pika D

2Am

KN 12.85

31044

25.67084

368

30800σ

Pika F

2Am

KN 12.85

31044

25.67084

368

30800σ

Pika G

2Am

KN 84

31044

25.17084

368

30800σ

Gjeresia e rripit ABEF eshte B1=3.2 m

2

FAmes

m

kN84.84

2

12.8555.84

2

σσσ

Gjejme reaksionin total te tokes i cili eshte

Ntokes=σmes· B1· L=84.84 · 3.2 · 28.3= 7684 kN

Nrrip=2·N3+N6=2 ·983+1478=3444 kN

55642

34447684

2

NNN

rriptokes

mes

kN

Modifikojme σmes

43.617684

556484.84

N

Nσσ

tokes

mesmes

modmes kPa

Modifikojme ngarkesat nga kolonat.

615.13444

5564

N

NF

rrip

mes

68

Kolona 3

Nmod=N3 · F= 983 · 1.615= 1588 kN

Kolona 6

Nmod=N6 · F= 1478 · 1.615= 2387 kN

1972.343.61Bσq 1modmes kN/m

Lartesia e pllakes gjendet:

33.11040)4.04.0(2

2207

R)ba(2

Nh

pr

bkk

maxpll

m

Gjejme momentet qe lindin ne kete pjese te pllakes duke e modeluar ate si nje tra

te thjeshte te permbysur ne te cilin vepron ngarkesa e shperndare e presionit te

tokes.

Duke pasur lartesine e pllakes dhe vlerat e momenteve bejme armimin e

pllakes sipas drejtimit te momentit qe kemi gjetur.

Per konstruimin e pllakes vendosim qe armimi do te behet ne brza qe jane sa 2°k

dhe 2bk pra ne breza prej 80 cm.

Per brezin e siperm te armimit

AoM

b ho2

Rb

Ao 0.085

1 1 2Ao

0.089

As

b ho Rb

Rs

As 0.00517 m2

69

Bejme armimin e pllakes ne brezin e siperm me:

21 x Φ18

Per brezin e poshtem te armimit

AoM

b ho2

Rb

Ao 0.151

1 1 2Ao

0.165

As

b ho Rb

Rs

As 0.00961 m2

Bejme armimin e pllakes ne brezin e siperm me:

21 x Φ24

70

LLOGARITJA E THEMELEVE TE THELLA ( PILOTAVE )

Pilota P6

Kollana me permasa 40 x 40 cm transmeton ne toke ngarkesat

kN222=Q

kNm148=M

kN1479=N

Projektojme themelin me pilota te parapergatitura me diameter d=40cm dhe me

jastek te ulet betonarme

1. Llogaritja e karakteristikave gjeometrike te pilotes se

parapergatitur

2

22

P m 1255.04

4.014.3

4

dπA

UP=π·d=3.14·0.4=1.256 m

2. Gjejme lartesine e jastekur betonarme duke u nisur nga kushti:

hj ≥ 2·dpi ≥ 2·40 = 80 Marrim hj=1m

3. E llogarisim piloten si pilote e varur

Lpil = 10m

m9=5+1+2+1=l+l+2

l+hj=z

m5.4=1+5.2+1=l+2

l+hj=z

m5.1=5.0+1=2

l+hj=z

213

3

12

2

11

15

45

90

71

Nga tabelat e librit marrim ferkimin kufitar ne funksion te gjendjes fizike te

shtresave dhe z = 11m

Ik1 = -3.66 → f1= 38.5

Ik2 = 0.371 → f2= 30.42

Ik3 = 0.54 → f3= 23.5

Llogarisim aftesine mbajtese te pilotes se varur

P = Pf + Pσ = m ·k·[Rmes · Apil + Upil ·(l1 · f1 + l2 · f2 + l3 · f3)]

P = 1 · 0.7 ·[2400·0.1255+1.256·(1·38.5+5·30.42+4·23.5)

P = 461 kPa

k = 0.7 per pilota qe punojne ne ngulje

Rmes = 2.2MPa – rezistenca ne majen e pilotes f (z = 11m dhe llojit te tokes )

Aftesia mbajtese sipas materialit

dpil = 40 > 30 ( m = 1 ) → φ = 1

Mb =200daN/ cm2

Rsh = 7000kPa

c – 3 → Rs = 2100daN/ cm2 = 210000kPa

As = 8φ20 = 22

sh

2

m002512.084

02.014.3A

4

dπ

Pm = mφ[Ab · Rb + As · Rs] = 1 · 1[0.2156 · 7000 + 0.002512 · 210000]

Pm = Pmat = 1456kPa

Perfundimisht aftesi mbajtese marrim:

Pmin = Pdheut = 461 kPa

4. Llogarisim numrin e pilotave qe vihen nen jastek

ηP

Nnpil η – koeficient qe merr parasysh veprimin e momentit

η =1.1 – 1.3

85.32.1461

1479npil pranojme n = 4 pilota

5. I veme pilotat ne planin e jastekut duke ruajtur distancat minimale

ndermjet tyre

amin ≥ 3dpil ( per toka te lidhura )

72

amin = 3 · 0.4 = 1.2

C = 30cm

bj = amin + 2dpil + 2C

bj = 1.2 + 2 · 0.4 +2 · 0.3 = 2.6m

aj = amin + dpil + 2C

aj = 1.2 + 2 · 0.4 +2 · 0.3 = 2.6m

6. Llogarisim ngarkesen e cdo pilote nga konstruksioni dhe i

krahasojme ato me vleren e lejuar

2i

iy

pil

j

0x

xM

n

NNP

Nj = aj · bj · hj · γb · η

Nj = 1.2 · 2.2 · 1 · 1.1 = 72.6 kN

∑ =x2+x2=x 22

21

2i 1.52 + 1.52 = 4.5

4365.4

45.1148

4

6.721479Pmax

0

kN

3405.4

45.1148

4

6.721479Pmin

0

kN

7. Kontrolli i presioneve qe merr cdo pilote ne lidhje me aftesine

mbajtese sipas dheut

min0P ≤ Pdh · 1.05

amin

aj

bj

C

x

73

340 ≤ 484

min0P ≤ Pdh · 1.2

436 ≤ 553

min0

max0

P

P ≤ 3

340

436 ≤ 3

nP

T

H

≤ m

PH – aftesia mbajtese horizontale f ( permasat e pilotes, llojit te tokes )

m – koeficient qe gjendet net abele f (nr pilotave )

PH = 35 kN

m = 0.8

61.0=480

222≤ m

8. Kontrolli i tufes se pilotave

18.20452

421518224

h

hφφ

i

ii

045.54

18.20

4

φmes

m37.3=045.5tg20+3=4

tgl2+6.1=B

m37.3=045.5tg20+3=4

tgl2+6.1=A

mesP

mesP

W

M±

A

N=σ tmax

min

Nt = Nkons + Nj + Npil + Ndheut

Npil = 4 · 0.1256 · 10 · 25 · 1.1 = 139 kN

Ndheut = 1.1γl2

)BA()16.1(mesP

.6

22.19452

46.1958.19217

h

hγγ

i

ii

mes

kN/m3

Ndheut = 17411.122.19102

)37.337.3()16.1(

.6 kN

74

Nt = 1479 + 72.6 + 139 + 1741 = 3432 kN

32637.3

6148

37.3

3432σ

32max

kPa ≤ 1.2R

27937.3

6148

37.3

3432σ

32min kPa ≤ 1.2R

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