Elaborati Ne Konstruksionet e Betonit III(Ultimate Limit State & Serviceability Limit State)

UNIVERSITETI I PRISHTINËS "HASAN PRISHTINA"

Për objektin e paraqitur më poshtë të bëhet : 1. Analiza e ramit karakteristik përfshirë ndikimet nga era dhe ato sizmike 2. Dimensionimi i elementeve karakteristike të ramitrigeli!sht"lla dhe themeli!përfshirë detajet e armimit#. $. %e dimensionimi i rigelit të merret paras"shë gjendja kufitare e aftësisë mbajtëse&'(# dhe ajo e shf"tëzueshmërisë ('(#. %ë dhënat: 

)apësira në drejtimin************ drejtimin*************... *...



Destinimi45ategoria e objektit**********. 5ategoria e truallit************** truallit**************... ... 6ona gjeografike45ategoria e terenit*******. 6ona sizmike*********** sizmike*****************. ******. 5lasa e betonit *********** **************** ***** 'loji 8 ;elikut për armim********** armim************* *** 'artësia e katit të parë *********** ************... *... 'astësia e katit të d"të ************... ************...
       

+,-,+ ',/.0m +"-"+ '"3.0m Afarizëm D 7 89 7-$4$ (-0 h1$./m h2$.2m =Pa σ

Planimetria e bjektit

m @  0 3  " '

A

A

m @  0 3  " '

P6878<8=8 8 B'B=B<%BCB Pozi@ionimi i 5atit

P( 13 P( (1

0  1 ( A P

P( (1

1 1  ( P A

/  1 ( A P

( P A

P( (1> m @  0 3  " '

0  1 ( A P

1 1 

P( 1$ P( (13

P( (10

P( (1/

A

A 0  1 ( A P

1 1  ( P A

P( 12

/  1 ( A P

( P A

1 1 

0  1 ( A P

m @  0 3  " '

Prerja A-A P( 1$ P( 10

P( 1/

P( 10

P( (13

P( (10

P( (1/

P( 0

P( /

P( 0

P( (3

P( (0

P( (/

P( $

P( %$

P( %2

P( %1

',/0@m

',/0@m

m @  2 $  2 h

m @  / $  1 h

PE?7A5%8=8 8 D8=B<(8
/ 1

hr = 

 0

Përvetësojë hr 0 @m

 1 1  1 1 16.66 ) cm ÷  hr =  ÷  50 = ( 25 ÷16 2 3 2 3    

b=

Përvetësojë b$ @m Dimensionet e sht"llës h sh ∢0.6hr h sh ∢0.6*50 h sh ∢30cm

$

A

S = µ i * Ce * Ct * Sk

Për Prishtinë lartësia mbidetare H mbi = ( 600 ÷ 700 700 ) m

C e 1.

Përvetësojë H mbi = 650m '

C t 1. µ i 1. S k = 0.75 + S k = 0.75 +

H − 500 400 650 650 − 500 400

S k = 1.125 kN m2

= 1.0*1.0*1.0*1.125 S =

5ategoria e terenit - 888 0

= 0.3m

zmin = 5m max

= 200m

5oefi@ienti i vrazhdësisë  C r ( z ) #

5oefi@ienti i turbulen@ës



  për zmin ≤ z ≤ z max  0

Cr ( z ) = kT * l n 

Cr ( z ) = C r ( zmin )

për

≤ z min

0.07

 0.3  = 0.19    0.05 



   z 0 

min

≤ z = 6.8 ≤ z max

C0( z ) * l n 

k I = 1.0

<ë rastin tonë z = 6.80m > z min  z  k T = 0.19  0   z   0, II 

k I

I v =

C 0( z ) = 1.0 -koefi@ienti orografisë 0.07

= 0.215

I v =

1.0

 6.8    0.3 

1.0* l n 

5oefi@ientët e jashtëm të sht"pjes së erës a# =uret Certikale

d Për A > 10m 2 ⇒ C pe = C pe,10 6.8 13 = 0.52 0.523 3 Për h d = = 6.8

I

D

B

0.25 < h d = 0.523 ≤ 1.0 nga %ab..1 le,ojmë:

b

6ona: D: C pe ,10 = + 0.8 B:

C pe ,10 = − 0.5

b# Për kulmet e rrafshta me strehë të mpreftë 3 4 e

H

θ = 0° -drejtimi i erës e = min(b : 2h) F

)

8

b

e = min(b : 13) ⇒ e = 13.6m

IF-$.23 k<4m2

I)-2.1/ k<4m2

I8-1. k<4m2

2 m 4 < k 1 $ : . 1  D I

2 m 4 1 $ : . 1  B I

',/0@m

',/0@m

m @  2 $  2 h

m @  / $  1 h

GA6A B 5&'=8% 'artësia e pllakës nga kushti 8 uljeve sipas B7-2

∅8mm -armatura për armim c0 = 15mm -shtresa mbrojtëse = λ = d

l ! l "

=

l e## 32

450 650

=

= 0.692 -pllaka punon në d" drejtime

450 32

h = d + c0 + ∅

= 14.06cm 2

= 14.06 + 1.5 + 0.8 = 15.96cm 2

Përvetësojmë lartësin e pllakës h1/.@m ?ërë lumi fraksion 8C 0 @m# )idroizolim (htresa për ramje 0@m# %ermoizolim 1 @m# Pllaka beton-arme 1/ @m# (uvatimi 2 @m#

GA6A B PE?D)B(E(

Pllaka keramike . @m#
1

Pllaka e kulmit

*1.00 = 6.765 kN m ' Gk = 6.765 * b = 6.765 *1

Prerja tërthore e rigelit / 1

Pesha vetiake e ramit

 0

g r = ( hr − h pll ) * br * ! b = ( 0.50 − 0. 0 .16 ) * 0. 0.30 *2 * 25 r

$

= 2.55 kN m '

(hpërndarja e ngarkesave l min

4.50 * λ = 6.765 * * 2 = 30.44 kN m 2 2 4.50 G" = 6.765 * = 15.22 kN m 2 4.50 l min 3 G trapez G trapez (1 2 2 )

G trapez = Gk *

0.346

G = 10.935 kN m

l30@m

105 = ' A %&S ,G

ql 10.935*4.5 2

=

2

= 24.60kN

105 2 * ' A %&S = 49.20kN -
P( 1/

P( 1$

" G E$ = 9.51 9.51* * 2 = 19.0 19.02 2 kN m T

= 1.425 kN m



Pllaka e përdhesës

0*1.00 = 5.90 kN m G $ = 5.90* b = 5.90*1 r

= 2.55 kN m

(hpërndarja e ngarkesave

l min

4.50 * λ = 5.90 * * 2 = 26.55 kN m 2 2 4.50 G" = 5.90 * = 13.275 kN m 2 4.50 l trapez = 0.346 G E$ = G trapez (1 − 2# 2 + # 3 ) # = min = 2l max 2 * 6.50

G trapez = Gk *

trapez G E$ = 26.55 (1 − 2* 0.346 2 + 0.3 0.3463 )

trapez = 21.30 kN m G E$

" = G E$

5 8

* G" =

G = G trapez + g

5 8

*13.275 = 8.29 kN m

G = 28.91 kN m

l30@m

005 ' A %&S = ,G

ql 2

=

28.91*4.5 2

= 65.05kN

005 2 * ' A %&S = 130.11kN -ngakresa në n"je të ramit. ,G

P( /

P( $

.2k<

>2.k<

3>.2k<

F2>.>/k<4m

1$.11k<

1$.11k<

1.k< F2$.0k<4m

G $ SH = 2.25 kN m

'

/0

G $ SH = 2.25 kN m

'

/0

G $ SH

m @  2 $  2 h

m @  / $  1 h

b#

Sk = 1.125 kN m 5*1.0 = 1.125 kN m S = Sk * b = 1.125*1

(hpërndarja e ngarkesave S trapez = S *

l min

4.50 * λ = 1.125 * *2 2 2

S trapez = 5.06 kN m S" = 1.125 *

# =

l min 2l max

=

4.50

= 2.531kN m

2 4.50

2 * 6.50

= 0.346

trapez S E$ = S trapez (1 − 2# 2 + # 3 ) trapez S E$ = 5.0 5.06 (1 − 2*0.3 2*0.346 2 + 0.346 3 )

trapez

l30 @m " S E$ = 1.58 1.581*2 1*2 = 3.16 3.16kN m

"

S E$ = 3.16 kN m

l30@m 106 = ' A %&S , S

ql 2

=

3.16*4.5 2

= 7.11kN

106 2 * ' A %&S = 14.22kN -ngakresa në n"je të ramit. , S

@#
(hpërndarja e ngarkesave l min

4.50 * λ = 5.0 * * 2 = 22.50 kN m' 2 2 4.50 = 11.25 kN m' )" = 5.0 * 2 4.50 l = 0.346 # = min = 2l max 2 * 6.50

)trapez = )k *

trapez = ) trapez (1 − 2# 2 + # 3 ) ) E$ trapez = 22.50 (1 − 2* 0.346 2 + 0.3 ) E$ 0.3463 )

trapez ) E$ = 18.04 kN m

" ) E$ =

5 8

* )" =

5 8

*11.25 = 7.03 kN m

l30 @m "

) E$ = 7.03* λ = 7.03*2 = 14. 14.06 kN m " ) E$ = 14.06 kN m

l30@m 006 = ' A %&S ,)

ql 2

=

14.06*4.5 2

= 31.63kN

006 2 * ' A %&S = 63.27 kN -ngakresa në n"je të ramit. ,)

A
H2

m2

H1

m1

 2 . $  2 h

$./m  / . $  1 h

',/.0

  . /  h

',/.0 '1$.

PB()A B GMB5%8%



=bulesa

23.31J1$.$1.$$ k<

5ombinimi i ngarkesave
= 1.0 $ =

% 2i = 0.3

*2 = 1.0 * Gk2 + $ *% 2i * S k 2 = 1.0 *5 * 567.28 + 1.0 *0 * 0.3 * 65 65.81

*2 = 587.0kN

= 0.5 $ =

% 2i = 0.3

*1 = 1.0 *Gk1 + $ *% 2i * )k 1 = 1.0 *6 * 689.82 + 0.5 * 0. 0.3 * 36 361.05

*1 = 743.97kN

=asa e ramit m2 = *2 g = 587.0 9.81 = 59.83 kN se c2 m1 = *1 g = 743 743.97 9.81 = 75. 75.83 kN se c2

6ona sizmike 89-të a gr = 0.30

! I = 1.20

EI sh = 21262.5kNm3 k = 3 (12*2126 12*21262. 2.5 5 3.60 3.603 ) = 1640 16406. 6.25 25 kN m & 1 =

T 1 =

k m1

2'

&

=

16406.25

=

75.83

2 *3 *3.14 14.70

= 14.70 rad sec

0.427 7 sec sec < 2.0sec 2.0sec = 0.42

0.427sec 7sec < 2.0sec 2.0sec T 1 = 0.42

T +( S ) = 0.20 T C ( S ) = 0.80

q = 3.0*

# ,

për

T + ≤ T1 ≤ TC ⇒ Sd (T ) # ,

# 1

# 1

= 1.30

3.0*1.30 30 = 3.90 3.90 q = 3.0*1.

(

( q) = a * S ( 2.5 q)

T + ≤ T2 ≤ TC ⇒ Sd (T ) = ag * S 2.5

S d (T ) = 0.36*1.35 2.5

3.90

)

g

H223.11k<  2 . $  2 h

H11//.3k<

 / . $  1 h

',/.0

',/.0

  . /  h

N3ark0 ( Pesha Vetiake 8 9 . 3 2 1 -

9 5 . 1 6 -

61.59 4 4 . 6 6 -

3 6 . 6 6

8 7 . 0 9 -

9 5 . 1 6 -

-61.59 4 4 . 6 6 -

3 6 . 6 6

-45.49

45.49

20.95

-20.95

5 3 . 7 4

5 3 . 7 4

-10.50

10.50

Ndikimet në tra ma4 %:5 ++0+: * min %:5 8(&:06; kNm

N3ark0 ( Pesha Vetiake 7 9 . 6 0 1

6 2 .

7 7 . 7 8

7 7

N3ark0 & N3arkesa n3a /#ra<#mplet= 6 2 . 7 1 -

5 3 . 7 -

5 3 . 7 -

7.35

-7.35

2 5 . 1 -

5 3 . 9

5 3 . 9

-2.46

2 5 . 1 -

2.46

-0.94

0.94 6 9 . 0

0.40

-0.40

Ndikimet në tra ma4 %:5 60:) * min %:5 8(70&+ kNm N3ark0 & N3arkesa n3a /#ra<#mplet= 9 6 . 4 1

3.07

4 6 . 1 1

-3.07

N3ark0 : N3arkesa n3a /#ra
2 6 . 7 -

4.85

6 2 . 0

-7.62

7 1 . 1 -

5 6 . 0 -

-1.39

-0.23

-2.71

-0.44

5 0 . 2 1

1.07 0.72

1 6 . 1

0.11

0.22

-0.28

Ndikimet në tra ma4 %:5 (&0') * min %:5 8((0'+ kNm N3ark0 : N3arkesa n3a /#ra
0 0 . 1

0.35

2.36

-2.71

N3ark0 9 N3arkesa n3a era
7 4 . 5 -

9 8 . 2 -

-7.04

-2.70 2 0 . 9 -

9 3 7 9 6 . 4 . 5 5 . 8 -

4 0 . 7

2.89

3.27

0.88

-7.31

-7.29

-8.14

3 9 . 4

0 2 . 0 1

15.89

-2.30

7.44

16.24

Ndikimet në tra ma4 m a4 %:5 (+0&9 * min %:5 860'& kNm N3ark0 9 N3arkesa n3a era
6 9 . 4

2.20

-1.87 4 4

7

5 1

6 0 . 2 -

N3ark0 ) N3arkesa n3a era
9 8 . 2 -

2.70 2.30

2 0 . 9 -

7.06

9 3 7 9 6 . 4 . 5 . 5 8 -

-0.88

6 0 . 7

-3.27

8.14

7.29 3 9 . 4

-16.24

-2.89 7.31

0 2 . 0 1

-7.44

-15.89

Ndikimet në tra ma4 %:5 ('0&' * min %:5 8(+0&9 kNm N3ark0 ) N3arkesa n3a era
-2.20

6 9 . 4

1.86 3

N3ark0 + N3arkesa sh>r.te?uese<#mplet=

2 2 . 6 -

2 2 . 6 -

0 9 . 2 1 7 9 .

6.22 2 7 . 2 4 -

-6.22 2 7 . 2 4 -

3

-21.27

21.27

21.45

-21.45

6 4 . 7 3

6 4 . 7 3

-10.32

10.32

Ndikimet në tra ma4 %:5 :709+ * min %:5 87&06' kNm N3ark0 + N3arkesa sh>r.te?uese<#mplet=

6 5 . 1

6 5 . 1 -

7 2 . 3 6

9 9 . 3 5

N3ark0 7 N3arkesa sh>r.te?uese
7 3 . 2 -

2.37

6 8 . 3 -

8 3 -7.95 . 1 3 5 9 2 . 5 5 . 1 2 -

8 8 . 4 4 -

-22.93

-3.86

15.15

21.95

-14.70 6 1 . 2

4 1 . 7 4

-5.86

9.14

4.47

Ndikimet në tra ma4 %:5 970(9 * min %:5 8)(0:; kNm N3ark0 7 N3arkesa sh>r.te?uese
1 5 . 1

7.91

5 0 . 0 -

3 6 . 9 5

-7.22

-0.69

N3ark0 ; N3arkesa n3a si?mika
9 . 0 5 1 -

8 6 . 4 8 -

-150.91

79.35

1 9 . 0 5 1

8 6 . 2 8 7 6 1 . 4 8

348.38

-169.36

164.07

188.97

-181.28

8 6 . 2 7 1

-212.74

348.38

Ndikimet në tra ma4 %:5 :9;0:; * min %:5 8&6&0'6 kNm N3ark0 ; N3arkesa n3a si?mika
-71.96

-104.20

-150.91

79.35

-212.74

9 0 . 2 9 2

9 0 . 2 9 2 -

0 5 . -71.96 1 7

N3ark0 6 N3arkesa n3a si?mika
150.91

8 6 . 4 8 -

9 0 . 2 9 2 -

169.36

8 6 . 2 8 7 6 1 . 4 8

-79.35 212.74

150.91 1 9 . 0 5 1

-164.07 181.28

-348.38

-79.35

212.74

8 6 . 2 7 1

9 0 . 2 9 2

-188.97

-348.38

Ndikimet në tra ma4 %:5 &6&0'6 * min %:5 8:9;0:; kNm N3ark0 6 N3arkesa n3a si?mika
71.96

104.20

5 2 . 6 3 -

71.96

?BA5(8
N3ark0 ( Pesha Vetiake

10.50

10.50

8.74

8.74

5 1 . 6 5 3

Reaksi#net e m/ështetës$e

N3ark0 & N3arkesa n3a /#ra<#mplet=

5 7 . 4 6 5

5 1 . 6 5 3

N3ark0 : N3arkesa n3a /#ra
0.11 0.09

0.22 0.18

7 5 . 0

Reaksi#net e m/ështetës$e N3ark0 9 N3arkesa n3a era
0.28 0.28

2 4 . 1 2

0 7 . 9 1

N3ark0 + N3arkesa sh>r.te?uese<"#mplet=

10.32

10.32

8.83

8.83

8 1 . 7 8

Reaksi#net e m/ështetës$e N3ark0 7 N3arkesa sh>r.te?uese
0 7 . 6 8 1

8 1 . 7 8

N3ark0 6 N3arkesa n3a si?mika
348.38

188.97

155.87

102.85

5 7 . 7

Reaksi#net e m/ështetës$e

348.38 155.87

5 7 . 7

KOMBINIMI I NGARKESAVE SIPAS GJENDJES KUFITARE MBAJTËSE (ULS) 1.SITUATA E PËRHERHSHME DHE PËRKOHSHME LLOGARITËSE Ssid 



Gj

* GKj K j   p * Pk   Qi Qi * QKi Ki 

j 1



Qi

* 0,i * Q i

 Gj

i 1

NYJA 1

Mo me me nt nte t (k Nm Nm ) M G= G= -1 -1 0. 0. 50 50 MS=0 M W= W= -1 -1 6. 6. 24 24 M Q= Q= -1 -1 0. 0. 32 32

Transversalet (kN) Aks ia ia le le t ( kN kN ) T G= G= -8 -8 .7 .7 4 TS=0 T W= W= -9 -9 .8 .8 9 T Q= Q= -8 -8 .8 .8 3

gGj

 1.35  Qi  1.50

Gkj

Ssd,T= -33.95 Ssd,N= -634.94

gQi Y0,i Qw

Y0,i Qs

1 .3 .3 5 - 1 0. 0. 50 50 1. 1. 50 50 - 10 10 .3 .3 2 1 .5 .5 0 0 .6 .6 0 - 16 16 .2 .2 4 0 .7 .7 0 1 .3 5 - 8. 8. 74 74 1 .5 0 - 8. 8. 83 83 1 .5 0 0 .6 0 1.35 1.35 -356 -356.15 .15 1.50 1.50 -89. -89.31 31 1.50 1.50 0.60 0.60

MG=MG=-45 45.4 .49 9 MS=0 MW=0 MQ=-21.27

Gkj

1 .3 .3 5

S sd sd ,T ,T S sd sd ,N ,N

1 .3 .3 5 - 3 3. 3. 46 46 1. 1. 50 50 1 .3 .3 5 - 14 14 4. 4. 17 17 1 .5 .5 0

Momentet (kNm) MG=47.35 MS=0 MW=2.30 MQ=47.14

0 .0 .0 0 - 4 4. 4. 27 27 S sd sd ,M ,M

Ssd,T= -64.64

Ssd,N= -216.66

Y0,i Qs

0 .0 .0 0 0 .7 .7 0

- 8. 8. 83 83 1 .5 .5 0 0 .6 .6 0 - 3 .3 .3 3 0 .7 .7 0 - 1. 1. 56 56 1. 1. 50 50 0. 0. 60 60 0 .0 .0 0 0 .7 .7 0

gQi Qki

- 93 93 .3 .3 2 S sd sd ,M ,M

Momentet (kNm)

Ssd,N= 37.27

gQi Y0,i Qw

MG=0 MS=0 MW=3.27 MQ=15.15

Y0,i Qs

S sd,N= -1087.43

gQi Y0,i Qw

Y0,i Qs

7 .4 .4 4 0 .7 .7 0

0 .2 .2 2

2 0. 0. 64 64 S sd sd ,M ,M

gQi Qki

1 .3 5

4 7. 7. 35 35 1. 1. 50 50 4 7. 7. 14 14 1 .5 0 0 .6 0

2 .3 0 0 .7 0

0 .0 0 13 13 6. 6. 70 70 S sd sd ,M

1. 35 35

1.35 1 .3 .3 5

0.00 1.50 2 4. 4. 73 73 1. 1. 50 50

2.15 0.70 0 .0 .0 0 0 .7 .7 0

0.43 3 .4 .4 4

1. 35 35 0.00 1.50 1 .3 .3 5 - 31 31 3. 3. 14 14 1. 1. 50 50

1 .3 .3 5

S sd,M= 68.53

MG=-66.44 MS=-1.52 MW=-9.02 MQ=-44.88

S sd,T= -28.34 S sd,N= -634.94

gQi Y0,i Qw

Y0,i Qs 8 .1 .1 4 0. 0. 70 70

gQi Y0,i Qw

Y0,i Qs 0 .0 .0 0

4 4. 4. 27 27

NG=24 NG=24.73 .73 NS=3.44 NW=0 NQ=0. NQ=0.18 18

gQi Qki

Ssd,M= -166.73 Ssd,T= 188.42 Ssd,N= 37.27

gQi Y0,i Qw

Y0,i Qs

6 8. 8. 53 53 S s d, d, M

1 .3 .3 5

- 66 66 .4 .4 4 1. 1. 50 50 - 44 44 .8 .8 8 1. 1. 50 50 0 .6 .6 0

- 9. 9. 02 02 0 .7 .7 0

- 1. 1. 52 52 - 16 16 6. 6. 73 73

- 3. 3. 66 66 0. 0. 70 70 0 .0 .0 0 - 2 8. 8. 34 34 S sd sd ,T ,T - 0. 0. 09 09 0 .7 .7 0 - 19 19 .1 .1 3 - 63 63 4. 4. 94 94 S s d, d, N

1 .3 .3 5 1 .3 .3 5

7 3. 3. 77 77 1. 1. 50 50 5 7. 7. 63 63 1. 1. 50 50 0. 0. 60 60 2 4. 4. 73 73 1 .5 .5 0 0 .1 .1 8 1. 1. 50 50 0 .6 .6 0

2 .1 .1 5 0 .7 .7 0 0 .0 .0 0 0. 0. 70 70

0 .4 .4 3 1 88 88 .4 .4 2 3 .4 .4 4 3 7. 7. 27 27

S sd,M= 25.67 S sd,T= 12.51

S sd,N= -468.51

gQi Y0,i Qw

7.22 1.50 0.60 0 .0 .0 0 1 .5 .5 0 0 .6 .6 0

Ssd,N= -632.34

1 0. 0. 50 50 1. 1. 50 50 1 0. 0. 32 32 1. 1. 50 50 0. 0. 60 60 1 6. 6. 24 24 0. 0. 70 70

Gkj

Transversalet (kN) A ks ks ia ia le le t (k N) N)

Mo me me nt nte t (kN m) m)

0 .0 0 1. 1. 50 50 1 5. 5. 15 15 1 .5 0 0 .6 0

Ssd,M= 44.27 Ssd,T= 33.95

Transversalet (kN) Aksialet (kN)

NYJA 6P

gQi Qki

TG=73 TG=73.77 .77 TS=0.43 TW=2.15 TQ=57 TQ=57.63 .63

gGj 0 .0 .0 0

NG=-356.15 NS=-19.70 NW=-0.09 NQ=-87.18

1 .3 5 8. 74 74 1 .5 0 8. 83 83 1 .5 0 0 .6 0 9 .8 9 0 .7 0 0 .0 0 3 3. 3. 95 95 1.35 1.35 -356 -356.1 .15 5 1.50 1.50 -87.1 -87.18 8 1.50 1.50 0.60 0.60 -0.0 -0.09 9 0.70 0.70 -19. -19.70 70 -63 -632. 2.34 34

Momentet (kNm)


Gkj

Gkj

NYJA 4HD

N G= G= -3 -3 56 56 .1 .1 5 NS=-19.13 NW=-0.09 NQ=-89.31

NG=-313.14 NS=-43.59 NW NW=0 NQ=0

T G= G= 8. 8. 74 74 TS=0 T W= W= 9. 9. 89 89 T Q= Q= 8. 8. 83 83

gGj

4. 09 09 0 .7 0 0 .1 .18 1 4. 4. 70 70 Ss Ss d, d, T 0.00 0.00 0.70 0.70 -42. -42.83 83 -1087.43 Ssd, Ssd,N N

2 0. 0. 95 95 1 .5 .5 0 2 1. 1. 95 95 1 .5 .5 0 0. 0. 60 60

Ssd,T S sd sd ,N ,N

2. 39 39 Ss Ssd,T 3 7. 7. 27 27 S sd sd ,N ,N

9 .1 .1 4 1 .5 .5 0 0 .6 .6 0

gQi Qki

TG=0 TS=0 TW=1.87 TQ=7.22

gGj

M G= G= 10 10 .5 .5 0 MS=0 M W= W= 16 16 .2 .2 4 M Q= Q= 10 10 .3 .3 2

S sd,M= 20.64 S sd,T= 14.70


S sd sd ,M

0. 00 00 1.50 0.60 0 .1 .1 8 1 .5 .5 0 0 .6 .6 0

Momentet (kNm)

1 .3 .3 5 - 8. 8. 74 74 1. 1. 50 50 - 8. 8. 83 83 1. 1. 50 50 0. 0. 60 60 1 .3 .3 5 - 35 35 6. 6. 15 15 1 .5 .5 0 - 89 89 .3 .3 1 1. 1. 50 50 0 .6 .6 0

NYJA 6L

Ssd,M= 136.70

0 .0 .0 0 1 .5 .5 0

Gkj 1 .3 .3 5

- 3. 3. 07 07 - 64 64 .6 .6 4 Ss Ss d, d, T - 18 18 .7 .7 5 - 21 21 6. 6. 66 66 S s d, d, N

Ssd,T= 2.39

gQi Qki

T G= G= -8 -8 .7 .7 4 TS=0 TW ==- 3. 3. 66 66 TQ=-8.83

gGj 0 .0 .0 0

NG=-564.75 NS=-42.83 N W= W= 0 NQ=- 18 18 6. 6. 70 70


NYJA 3

1. 35 35 0 .0 0 1 .5 0 7 .2 2 1 .5 0 0 .6 0 1.35 1.35 -564 -564.7 .75 5 1.50 1.50 -186.70 1.50 5 0 0.60

M G= G= 20 20 .9 .9 5 MS=0 MW =8 =8 .1 4 MQ=21.95

Ssd,M= -93.32


Gkj

1 .3 .3 5

0 .0 0 - 33 33 .9 5 S sd sd ,T -19. -19.13 13 -634 -634.9 .94 4 Ssd,N Ssd,N

Momentet (kNm)

gQi Y0,i Qw

NG=24.73 NS=3.44 NW=0 NQ=0.18

Gkj

NYJA 4P

- 45 45 .4 .4 9 1. 1. 50 50 - 21 21 .2 .2 7 1 .5 .5 0 0 .6 .6 0

TG=0 TS=0.43 TW=2.15 TQ=0

gGj

NG=NG=-14 144.1 4.17 7 NS=-18.75 NW=0 NQ=-1.56

gQi Qki

S sd sd ,M ,M

NYJA 5

- 9. 9. 89 89 0 .7 0 -0.0 -0.09 9 0.70 0.70

Transversalet (kN) Aks ia ia le le t ( kN kN ) TG=0 TS=0. 18 18 T W= W= 4. 4. 09 09 TQ=7 .2 .22

gGj

Transversalet (kN) Aksialet (kN) TG=TG=-33 33.4 .46 6 TS=-3.07 TW=-3.33 TQ=-8.83

gGj

MG=0 MS =0 =0. 22 22 MW=7.44 MQ =9 =9. 14 14

Ssd,M= -44.27

S sd sd ,T Ssd, Ssd,N N

Momentet (kNm)

0.6 -për erë 0.7 -për borë

M om om en ent et et ( kN kN m) m)

S sd sd ,M ,M

NYJA 4L





NYJA 2

N G= G= -3 -3 56 56 .1 .1 6 NS=-19.13 N W= W= -0 -0 .0 .0 9 N Q= Q= -8 -8 9. 9. 31 31

gQi Qki

 0,i  0, i

MG=0 MS ==- 0. 0. 44 44 MW=-7.29 MQ=-14.70

Y0,i Qs 3. 27 27 0 .7 0

gGj 0 .0 .00

2 5. 5. 67 67 S sd sd, M

1. 87 87 0.70 0. 00 00 12.51 Ss Ssd, T 0 .0 .0 0 0 .7 .7 0 - 43 43 .5 .5 9 - 46 46 8. 8. 51 51 S sd sd ,N ,N

TG=0 TS =0 =0 .1 8 T W= W= 4. 4. 09 09 TQ=7.22 Gkj 1 .3 5

NG=-564.75 NS ==- 42 42 .8 .83 N W= W= 0 NQ=-186.70

gQi Qki

Ssd,M= -29.07 Ssd,T= 14.70 Ssd,N= -1087.43

gQi Y0,i Qw

0. 00 00 1. 1. 50 50 - 14 14 .7 .70 1 .5 0 0. 0. 60 60

Y0,i Qs

- 7. 7. 29 29 0 .7 0

- 0. 0. 44 44

- 29 29 .0 7

1.35 0. 00 00 1.50 7. 22 22 1.50 0.60 4.09 0.70 0.18 14.70 1 .3 .3 5 - 56 56 4. 4. 75 75 1. 1. 50 50 -186.70 1.50 1.50 0.60 0.00 0.00 0.70 0.70 -42. -42.83 83 -1087.43

KOMBINIMI I NGARKESAVE SIPAS GJENDJES KUFITARE MBAJTËSE (ULS) 1.SITUATA E PËRHERHSHME DHE PËRKOHSHME LLOGARITËSE Ssid    Gj * GKj K j   p * Pk   Qi Qi * QKi K i    Qi * 0,i * Q i j 1

NYJA 6HM

Momentet (kNm) MG=MG=-90 90.7 .78 8 MS=0 MW ==- 5. 5. 63 63 MQ=-72.90

S sd sd ,M ,M S sd sd ,T S sd sd ,N



1.35

 Qi  1.50

Transversalet (kN) Aksialet (kN) TG=TG=-81 81.2 .26 6 TS=0 TW ==-2 .4 4 TQ=-63.27

gGj

Gkj

NG=2 NG=24. 4.73 73 NS=3.44 NW =0 =0 NQ=0

gQi Qki

Momentet (kNm) M G= G= 45 45 .4 .4 9 MS=2.46 MW=0.88 MQ=21.27 gGj S sd sd ,M ,M S sd sd ,T S sd sd ,N ,N

Gkj

gQi Qki

Ssd,N= 37.00

Y0,i Qs

Momentet (kNm) MG=6 MG=61. 1.59 59 MS =7 =7 .3 5 MW=1.00 MQ=6.22 gGj

Gkj

gQi Qki

1 .3 .3 5 1. 35 35 1. 35 35

M G= G= -2 -2 0. 0. 95 95 MS=0 M W= W= -8 -8 .1 .1 4 MQ=-21.95

Ssd,M= 96.69 Ssd,T= 64.64

Ssd,N= -217.70

gQi Y0,i Qw

NG=NG=-14 144.1 4.17 7 NS ==- 18 18. 75 75 NW=0 NQ=-1.56

Gkj

Momentet (kNm)

Y0,i Qs

0 .8 .8 8 0 .7 .7 0 3 .3 3 0 .7 0 0 .0 .0 0 0 .7 .7 0

M G= G= -6 -6 1. 1. 59 59 MS ==-7 .3 5 MW ==- 1. 1. 00 00 MQ=-6.22

Ssd,N= -216.66

gQi Y0,i Qw

1 .3 .3 5 6 1. 1. 59 59 1. 1. 50 50 6 .2 .2 2 1 .5 .5 0 0 .6 .6 0 1 .3 5 - 33 33 .4 6 1 .5 0 - 8. 8. 83 83 1 .5 0 0 .6 0 1 .3 .3 5 - 14 14 4. 4. 17 17 1. 1. 50 50 - 1. 1. 56 56 1 .5 .5 0 0 .6 .6 0

Y0,i Qs

1 .0 .0 0 0 .7 .7 0 0 .0 0 0 .7 0 0 .0 .0 0 0 .7 .7 0

7 .3 .3 5 10 10 1. 1. 09 09 S sd sd ,M ,M - 3. 3. 07 07 -6 1. 1. 64 64 S sd sd ,T - 18 18 .7 .7 5 - 21 21 6. 6. 66 66 S sd sd ,N ,N

S sd,N= 37.00

gQi Y0,i Qw

N G= G= -3 -3 56 56 .1 .1 5 NS=-19.13 NW=-0.09 NQ=-89.31

gGj

- 5. 5. 63 63 0 .7 .7 0 2. 44 44 0. 0. 70 70 0 .0 .0 0 0 .7 0

S sd,M= -68.53 S sd,T= 28.34 S sd,N= -634.94

gQi Y0,i Qw

Gkj

N G= G= -3 -3 3. 3. 46 46 NS=-3.07 NW=0 NQ=-8.59

gQi Qki

1 .3 .3 5 - 6 1. 1. 59 59 1. 1. 50 50 1. 35 35 8 7. 7. 77 77 1. 1. 50 50 1 .3 .3 5 - 33 33 .4 .4 6 1. 1. 50 50

MG=4 7. 7. 35 35 MS=0 MW =2 =2 .3 0 M Q= Q= 37 37 .4 .4 6

TG=0 TS=0.05 TW =2 =2 .4 4 T Q= Q= 3. 3. 64 64

Y0,i Qs

gGj - 0. 0. 65 65 - 23 23 7. 7. 65 65 S s d, d, M 0 .0 .05 2 06 06 .8 5 Ss Ss d, d, T 3 .4 .4 4 37.00 S sd sd, N

Gkj

S sd,N= -61.28

gQi Y0,i Qw

MG=MG=-66 66.44 .44 MS ==- 1. 1. 52 52 M W= W= -9 -9 .0 .0 2 MQ=MQ=-42 42.72 .72

TG=TG=-73 73.77 .77 TS ==- 0. 0. 38 38 T W= W= -2 -2 .1 .1 5 TQ=TQ=-53 53.99 .99

gGj

Gkj

- 6. 6. 22 22 1. 1. 50 50 0. 0. 60 60 1 .5 6 1 .5 0 0 .6 0 - 8. 8. 59 59 1 .5 0 0 .6 .6 0

NG=24.73 NS NS =3 =3 .4 4 NW=0 NQ=0

gQi Qki

MG=66.63 MS=9.35 MW=0 MQ=0

TG=0 TS=0 TW=1.45 TQ=1.56

gGj

Gkj 1 .3 .3 5 1 .3 5 1 .3 .3 5

NG=-33.46 NS=-3.07 NW=0 NQ=-8.59

gQi Qki

Y0,i Qs

2 .3 .3 0 0. 0. 70 70 0 .0 .0 0 1 2 2. 2. 18 18 2 .4 4 0 .7 0 0 .0 5 7 .7 1 0 .0 0 0 .7 0 3 .4 4 3 7. 7. 00 00

Ssd,M= -163.49 Ssd,T= -182.91

Ssd,N= 37.00

1 .3 .3 5 - 66 66 .4 .4 4 1. 1. 50 50 - 42 42 .7 .7 2 1. 1. 50 50 0 .6 .6 0 1 .3 5 - 7 3. 3. 77 77 1. 1. 50 50 - 53 53 .9 .99 1. 1. 50 50 0. 0. 60 60 1 .3 .3 5 2 4. 4. 73 73 1 .5 .5 0 0 .0 .0 0 1. 1. 50 50 0 .6 .6 0


- 1. 1. 00 00 0. 0. 70 70 - 7. 7. 35 35 - 10 10 1. 1. 09 09 S sd sd ,M ,M 0. 00 00 0. 0. 70 70 1 1. 1. 64 64 1 33 33 .0 5 Ss Ss d, d, T 0 .0 .0 0 0 .7 .7 0 - 3. 3. 07 07 - 61 61 .2 .2 8 Ss Ss d, d, N

Ssd,N= 37.00

gQi Y0,i Qw

Momentet (kNm)

Y0,i Qs

gQi Y0,i Qw


NYJA 10

S sd,M= -101.09

gQi Qki

Ssd,M= 122.18 Ssd,T= 7.71

1 .3 .3 5 4 7. 7. 35 35 1 .5 .5 0 3 7. 7. 46 46 1 .5 .5 0 0. 0. 60 60 1 .3 5 0. 00 00 1. 1. 50 50 3. 64 64 1. 1. 50 50 0. 0. 60 60 1 .3 5 2 4. 4. 73 73 1. 1. 50 50 0. 00 00 1. 1. 50 50 0. 0. 60 60

- 8. 8. 14 14 0 .7 .7 0 0 .0 .0 0 - 68 68 .5 .5 3 S sd sd ,M ,M 3. 66 66 0. 0. 70 70 0 .0 .00 2 8. 8. 34 34 S sd sd, T - 0. 0. 09 09 0 .7 .7 0 - 19 19 .1 .1 3 - 63 63 4. 4. 94 94 S s d, d, N

S sd,T= 133.05

NG=24.73 NS NS=3.44 NW=0 NQ=0

Momentet (kNm)

Y0,i Qs

Transversalet (kN) Aksialet (kN) T G= G= 87 87 .7 .7 7 TS =1 =1 1. 1. 64 64 TW =0 =0 TQ=1.56


NYJA 8HM

1 .3 .3 5 - 20 20 .9 .9 5 1. 1. 50 50 - 21 21 .9 .9 5 1. 1. 50 50 0 .6 0 1. 35 35 8 .7 4 1 .5 0 8 .8 3 1 .5 0 0 .6 0 1 .3 .3 5 - 35 35 6. 6. 15 15 1 .5 .5 0 - 89 89 .3 .3 1 1. 1. 50 50 0 .6 .6 0

Momentet (kNm) Ssd,M= 101.09

S sd,M= -237.65 S sd,T= 206.85


gQi Qki

Momentet (kNm)

NYJA 7

- 90 90 .7 .7 8 1. 1. 50 50 - 72 72 .9 .9 0 1. 1. 50 50 0 .6 .6 0 8 1. 1. 26 26 1. 1. 50 50 6 3. 3. 27 27 1. 1. 50 50 0. 0. 60 60 2 4. 4. 73 73 1 .5 0 0.00 1. 1. 50 50 0 . 60 60

Gkj

NYJA 9HD

Ssd,T= -61.64

N G= G= 24 24 .7 .7 3 NS=3.44 NW=0 NQ=0

gQi Qki

T G= G= 8. 8. 74 74 TS=0 T W= W= 3. 3. 66 66 TQ=8.83

gGj

2 .4 .4 6 9 6. 6. 69 69 S s d, d, M 3 .0 7 64 .6 .64 S sd sd ,T - 19 19 .7 .7 4 - 21 21 7. 7. 70 70 S s d, d, N

Transversalet (kN) Aksialet (kN) TG=TG=-33 33.4 .46 6 TS ==- 3. 3. 07 07 TW=0 TQ=-8.83

gGj 0 .0 .0 0 - 23 23 6. 6. 97 97 S s d, d, M 0 .0 0 - 20 20 6. 6. 80 80 S sd sd ,T 3 .4 4 37 .0 .00 Ss Ss d, d, N

Transversalet (kN) Aksialet (kN) T G= G= 81 81 .2 .2 6 TS =0 =0 .0 .05 T W= W= 2. 2. 44 44 TQ=63.27

NYJA 8P

1 .3 .3 5 4 5. 5. 49 49 1 .5 0 2 1. 1. 27 27 1 .5 .5 0 0 .6 .6 0 1 .3 5 3 3. 3. 46 46 1. 1. 50 50 8. 83 83 1 .5 0 0 .6 0 1 .3 .3 5 - 14 14 4. 4. 17 17 1 .5 .5 0 - 1. 1. 56 56 1. 1. 50 50 0. 0. 60 60

NYJA 9P

0.6 -për erë 0.7 -për borë

M G= G= -9 -9 0. 0. 78 78 MS ==-0 .6 5 M W= W= -5 -5 .6 .6 3 MQ=-72.90

Ssd,M= -236.97 Ssd,T= -206.80

gQi Y0,i Qw

N G= G= -1 -1 44 44 .1 .1 7 NS=-19. 74 74 NW=0 NQ=-1.56





Momentet (kNm)

Transversalet (kN) Aksialet (kN) T G= G= 33 33 .4 .4 6 TS=3.07 TW=3.33 T Q=8.83

 0, i  0, i

NYJA 6HD

1 .3 .3 5 - 90 90 .7 .7 8 1. 1. 50 50 - 72 72 .9 .9 0 1 .5 .5 0 0 .6 .6 0 - 5. 5. 63 63 0. 0. 70 70 1 .3 5 - 8 1. 1. 26 26 1. 1. 50 50 - 63 63. 27 27 1 .5 0 0 .6 0 - 2. 2. 44 44 0 .7 0 1 .3 5 2 4. 4. 73 73 1 .5 0 0. 00 00 1 .5 0 0 .6 0 0 .0 0 0 .7 0

NYJA 8L

S sd sd ,M ,M S sd sd ,T S sd sd ,N ,N

 Gj

i 1

Y0,i Qs

- 9. 9. 02 02 0 .7 .7 0 - 2. 2. 15 15 0. 0. 70 70 0 .0 .0 0 0. 0. 70 70

- 1. 1. 52 52 - 16 16 3. 3. 49 49 - 0. 0. 38 38 - 18 18 2. 2. 91 91 3 .4 .4 4 3 7. 7. 00 00

Ssd,M= 99.77 Ssd,T= 3.65 Ssd,N= -61.28

gQi Y0,i Qw

Y0,i Qs

6 6. 6. 63 63 1. 1. 50 50 0 .0 .0 0 1 .5 .5 0 0 .6 .6 0 0 .0 .0 0 0 .7 .7 0 9 .3 .3 5 9 9. 9. 77 77 0. 00 00 1. 1. 50 50 1. 56 56 1. 1. 50 50 0. 0. 60 60 1 .4 5 0 .7 0 0 .0 0 3 .6 5 - 33 33 .4 .4 6 1 .5 .5 0 -8.59 1.5 1.50 0 0.60 0.00 0.00 0.70 0.70 -3.0 -3.07 7 -61.28

KOMBINIMI I NGARKESAVE SIPAS GJENDJES KUFITARE MBAJTËSE (ULS)

1.SITUATA E PËRHERHSHME DHE PËRKOHSHME LLOGARITËSE i S sid    Gj * GK j   p * Pk  Q i *QK i    Qi * 0,i * Q j 1

i 1

NYJA 11P

Momentet (kNm) MG=0 MS=0 MW=-2. 70 70 MQ=-7.95

S sd sd ,M ,M S sd sd ,T ,T S sd sd ,N ,N

Gkj

Momentet (kNm) MG=66.63 MS=12.05 MW=0 MQ=0

1 .3 .3 5 1 .3 5 1 .3 .3 5

 1.35

 0, i

 0.6 -për erë

 Qi

 1.50

 0, i

 0.7

Momentet (kNm) Ssd,M= -14.36

Ssd,T= 12.21

Ssd,N= -468.51

gQi Y0,i Qw

- 7. 7. 95 95 1. 1. 50 50 0. 0. 60 60 7 .2 .2 0 1 .5 .5 0 0 .6 .6 0 0 .0 .0 0 1 .5 .5 0 0 .6 .6 0

Gkj

NG=-33.46 NS=-3.07 NW=0 NQ=-8.59

gQi Qki

-për borë

NYJA 11HM

MG=-12 MG=-123.9 3.98 8 MS=-17.26 MW=0 MQ=-2.02

Y0,i Qs

- 2. 2. 70 70 0. 0. 70 70 1 .5 .5 7 0 .7 .7 0 0 .0 .0 0 0 .7 .7 0

0 .0 .0 0 - 14 14 .3 .3 6 S sd sd ,M ,M 0 .0 .0 0 1 2. 2. 21 21 S sd sd ,T ,T - 43 43 .5 .5 9 - 46 46 8. 8. 51 51 S sd sd ,N ,N

Gkj


Ssd,N= -61.28

gQi Y0,i Qw

6 6. 6. 63 63 1 .5 .5 0 0 .0 .0 0 1 .5 .5 0 0 .6 .6 0 0 .0 .0 0 0 .7 .7 0 0 .0 0 1 .5 0 - 1. 1. 56 56 1 .5 0 0 .6 0 - 1. 1. 45 45 0 .7 0 - 33 33 .4 .4 6 1. 1. 50 50 - 8. 8. 59 59 1 .5 .5 0 0 .6 .6 0 0 .0 .0 0 0 .7 .7 0

S sid , M  1.35 * M G 1.50M Q  1.50  0.6MW  0.7M S S sid ,T  1.35 *TG 1.50TQ  1.50  0.6TW  0.7TS

MG=MG=-61 61.5 .59 9 MS ==-7 .6 2 MW=-1.00 MQ=-6.22

Y0,i Qs





S sid , N  1.35 * N G  1.50 N Q  1.50  0.6 NW  0.7 N S 

gGj 1 2. 2. 05 05 1 0 2. 2. 60 60 S s d, d, M 0 .0 0 -3 .6 5 Ss Ss d, d, T - 3. 3. 07 07 - 61 61 .2 .2 8 S sd sd ,N ,N

NG=-3 NG=-33.4 3.46 6 NS=-3. NS=-3.07 07 NW=0 NQ=-8.59

gQi Qki

NYJA 11HD

Momentet (kNm) S sd,M= -188.53 S sd,T= -159.83

S sd,N= -61.28

gQi Y0,i Qw

1 .3 .3 5 - 12 12 3. 3. 98 98 1 .5 .5 0 - 2. 2. 02 02 1 .5 .5 0 0. 0. 60 60 1 .3 .3 5 - 10 10 6. 6. 97 97 1. 1. 50 50 0 .0 .0 0 1 .5 .5 0 0 .6 .6 0 1 .3 .3 5 - 3 3. 3. 46 46 1. 1. 50 50 -8.59 1.5 1.50 0 0.60

NYJA 13P

Ssd,T= -3.65

Transversalet (kN) Aksialet (kN) TG=-1 TG=-106. 06.97 97 TS=-14 TS=-14.69 .69 TW=0 TQ=0

gGj

Transversalet (kN) Aksialet (kN) TG=0 TS=0 TW=-1.45 TQ=-1.56

gGj

NG=-313.14 NS=-43.59 NW=0 NQ=0

gQi Qki

1 .3 .3 5 0 .0 .0 0 1. 1. 50 50 1 .3 .3 5 0 .0 .0 0 1 .5 .5 0 1 .3 .3 5 - 31 31 3. 3. 14 14 1. 1. 50 50

NYJA 12

 Gj

Transversalet (kN) Aksialet (kN) TG=0 TS=0 TW=1.57 TQ=7.22

gGj

S sd sd ,M ,M S sd sd ,T S sd sd ,N ,N

Y0,i Qs

Gkj

NG=NG=-14 144. 4.17 17 NS ==- 19 19 .7 4 NW=0 NQ=-1.56

gQi Qki

1 .3 .3 5 - 61 61 .5 .5 9 1. 1. 50 50 1. 35 35 3 3. 3. 46 46 1. 1. 50 50 1 .3 .3 5 - 14 14 4. 4. 17 17 1 .5 .5 0

TG=10 TG=106.9 6.97 7 T S= S= 14 14 .6 .6 9 TW=0 TQ=0

gGj

0 .0 .0 0 0. 0. 70 70 - 17 17 .2 .2 6 - 18 18 8. 8. 53 53 S s d, d, M 0 .0 .0 0 0 .7 .7 0 - 14 14 .6 .6 9 - 15 15 9. 9. 83 83 S sd sd ,T ,T 0.00 0.00 0.70 0.70 -3.0 -3.07 7 -61.28 S sd sd ,N ,N

Transversalet (kN) Aksialet (kN) TG=3 TG=33.4 3.46 6 TS =3 =3 .0 .07 TW=0 TQ=8.83

MG=-1 MG=-123. 23.98 98 M S= S= -1 -1 7. 7. 26 26 MW=0 MQ=-5.93

Gkj

Momentet (kNm)

S sd,M= -101.38

S sd,N= -217.70

gQi Y0,i Qw

- 6. 6. 22 22 1 .5 .5 0 0. 0. 60 60 8 .8 3 1 .5 0 0 .6 0 - 1. 1. 56 56 1 .5 .5 0 0. 0. 60 60

MG=-61.59 MS=-7.62 MW=-1.00 MQ=-6.22

Y0,i Qs

- 1. 1. 00 00 0 .7 .7 0 - 7. 7. 62 62 - 10 10 1. 1. 38 38 S s d, d, M 0. 00 00 0. 0. 70 70 3 .0 .07 6 1. 1. 64 64 S sd sd, T 0 .0 .0 0 0. 0. 70 70 - 19 19 .7 .7 4 - 21 21 7. 7. 70 70 S s d, d, N

NG=-3 NG=-33.4 3.46 6 N S= S= -3 -3 .0 .0 7 NW=0 NQ=-8.59

gQi Qki

Ssd,M= -194.39 Ssd,T= 159.83 Ssd,N= -61.28

gQi Y0,i Qw

Y0,i Qs

1 .3 .3 5 - 12 12 3. 3. 98 98 1 .5 .5 0 - 5. 5. 93 93 1 .5 .5 0 0. 0. 60 60 0 .0 .0 0 0. 0. 70 70 - 17 17 .2 .2 6 - 19 19 4. 4. 39 39 1 .3 .3 5 10 10 6. 6. 97 97 1. 1. 50 50 0 .0 .0 0 1 .5 .5 0 0 .6 .6 0 0 .0 .0 0 0 .7 .7 0 14 14 .6 .6 9 1 59 59 .8 .8 3 1 .3 .3 5 - 33 33 .4 .4 6 1 .5 .5 0 - 8. 8. 59 59 1 .5 0 0. 0. 60 60 0 .0 .0 0 0. 0. 70 70 - 3. 3. 07 07 - 61 61 .2 .2 8

NYJA 13HM

S sd,T= 61.64


gGj

Transversalet (kN) Aksialet (kN) TG=-8 TG=-87.7 7.77 7 TS=-12 TS=-12.63 .63 TW=0 TQ=-1.56 Gkj

NG=-3 NG=-33.4 3.46 6 NS=-3. NS=-3.07 07 NW=0 NQ=-8.59

gQi Qki

1 .3 .3 5 - 61 61 .5 .5 9 1. 1. 50 50 1 .3 5 - 8 7. 7. 77 77 1. 1. 50 50 1 .3 .3 5 - 33 33 .4 .4 6 1. 1. 50 50

Ssd,M= -101.38 Ssd,T= -134.09 Ssd,N= -61.28

gQi Y0,i Qw

- 6. 6. 22 22 1 .5 .5 0 0. 0. 60 60 -1 .5 .56 1. 1. 50 50 0. 0. 60 60 - 8. 8. 59 59 1 .5 .5 0 0. 0. 60 60

Y0,i Qs

- 1. 1. 00 00 0 .7 .7 0 - 7. 7. 62 62 -1 -1 01 01 .3 .3 8 0 .0 0 0 .7 0 - 12 12 .6 3 - 13 13 4. 4. 09 09 0 .0 .0 0 0. 0. 70 70 - 3. 3. 07 07 - 61 61 .2 .2 8

KOMBINIMI I NGARKESAVE SIPAS GJENDJES KUFITARE MBAJTËSE (ULS) 2.SITUATA SIZMIKE LLOGARITËSE

S sd   GKj   ( 2,i * Qi )   I * Ad  Pk j 1

Vlera e  nuk merret parasyshe gjatë kombinimit pasi I që është marrë gjatë llogaritjes së sizmikës.

 2, i

 0.6

për destinim të objektit: Afarizëm nga Tab.A1.1 fq.51 EN1990:2002

i 1

Vërejtje: Për situatën sizmike llogaritëse do të bëjmë kombinimin e ngarkesave vetëm për elementet vertikale,pasi që ndikimi në elementet horizontale nga ngarkesa e sizmikës është i vogël.

NYJA 1

Momentet (kNm) MG=-10.50 MQ=-10.32 MA=-348.38 Gkj Ssd,M Ssd,T Ssd,N

Momentet (kNm) MG=-45.49 MQ=- 22 22 .9 3 MA=-79.35 Gkj Ssd,M Ssd,T Ssd,N

NG=-356.15 NQ=-89.31 NA=-107.75

Y2,i Qki

Ad

Momentet (kNm) MG=0 MQ=-14.70 MA=-181.28 Gkj

-1 -10.32 -8.83 - 89 89.31

S sd,T= -169.91 Ssd,N= -517.49

MG=0 MQ=9.14 MA=188.97

Ssd,M Ssd,T Ssd,N

Y2,i Qki

Ad

-2 -22.93 -8.83 -1.56

S sd,T= -110.72 Ssd,N= -181.36

MG=20.95 MQ=21.95 MA=212.74 Gkj

-7 -79.35 -7 -71.96 -3 -36.25

Transversalet (kN)

Momentet (kNm) Ssd,M= -138.60

-138.60 -110.72 -181.36

Ssd,M Ssd,T Ssd,N

TG=0 TQ=7.22 TA=104.20

NG=-564.75 NQ NQ=-186.70 NA=0

Y2,i Qki

Ad -181.28 104.20 0.00

Momentet (kNm) Ssd,M= -190.10 S sd,T= 108.53 Ssd,N= -676.77

M G= G=45 .4 .49 MQ=21.27 MA=79.35 Gkj

-190.10 108.53 -676.77

Ssd,M Ssd,T Ssd,N

NYJA 3

Aksialet (kN)

TG=0 TQ=7.22 TA=104.20

NG=-564.75 NQ=-186.70 NA=0

Y2,i Qki

Ad

Ssd,N= -676.77

MG=1 0. 0. 50 50 MQ=10.32 MA=348.38

Ssd,M Ssd,T Ssd,N

Ad

Ssd,T= -169.91 Ssd,N= -517.49

MG=0 MQ=15.15 MA=164.07 Gkj

212.74 -155.87 -107.75

Transversalet (kN)


246.86 -169.91 -517.49

Ssd,M Ssd,T Ssd,N

TG=3 3. 3. 46 46 TQ=8.83 TA=71.96

NG=-144.17 NQ=-1.56 NA=-36.25

Y2,i Qki

Ad

Momentet (kNm) Ssd,M= 137.60 Ssd,T= 110.72 Ssd,N= -181.36

MG=-20.95 MQ=-21.45 MA=-212.74 Gkj

79.35 71.96 -36.25

13 137.60 110.72 -181.36

Ssd,M Ssd,T Ssd,N

Aksialet (kN)

TG=8 .7 4 TQ=8.83 TA=155 TA=155.87 .87

NG ==-3 56 56 .1 5 NQ=-87.18 NA=-10 NA=-107.7 7.75 5

Y2,i Qki

Ad

10.32 8 .83 -87.18

348.38 155.87 -107. 75 75

Transversalet (kN)

Ssd,T= 169.91 Ssd,N= -516.21

Aksialet (kN) NG=-313.14 NQ=0 NA=0 NA=0

Y2,i Qki

Ad

15.15 7 .2 .22 0.00

164.07 104.20 0.00

Transversalet (kN)

Ssd,M= 173.16 Ssd,T= 108.53 Ssd,N= -313.14

173.16 108.53 -313.14

Aksialet (kN)

TG=8.74 TQ=8.83 TA=155.87

NG=-356.15 NQ=-87.18 NA=-107.75

Y2,i Qki

Ad

-20.95 0.60 -21.45 8.74 0.60 8. 8 .83 -356.15 0.60 -87.18

Ssd,M= 365.07

365.07 169.91 -516.21

TG=0 T Q=7.22 TA=104 TA=104.20 .20

0.00 0.60 0.00 0.60 -313.14 0.60

NYJA 8P

Aksialet (kN)

Transversalet (kN)

10.50 0.60 8.74 0.60 -356.15 0.60

NYJA 6L

Y2,i Qki

21.27 8. 8 .83 -1 -1.56

Ssd,T= 108.53

Aksialet (kN) NG=-356.15 NQ=-89.31 NA=-107.75

45.49 0.60 33.46 0.60 -144.17 0.60

Ssd,M= 194.45

194.45 108.53 -676.77

TG=-8.74 TQ=-8.83 TA=-155.87

21.95 -8.83 -89.31

Momentet (kNm)

Gkj 188.97 104.20 0.00

Transversalet (kN)

20.95 0.60 -8.74 0.60 -356.15 0.60

NYJA 8L

Aksialet (kN)

Transversalet (kN)

0.00 0.60 9.14 0.00 0.60 7. 7 .22 -564.75 0.60 -186. 70 70

NYJA 4P

Aksialet (kN) NG=-144.17 NQ=-1.56 NA=-36.25

- 14 14.70 7.22 -186.70

Ssd,M= -365.07

-365.07 -169.91 -517. 49 49

TG=-33.46 TQ=- 8. 8. 83 83 TA=-71.96

0.00 0.60 0.00 0.60 -564.75 0.60

Momentet (kNm)

Gkj

-348.38 -155.87 -107.75

Transversalet (kN)

-45.49 0.60 -33.46 0.60 -144.17 0.60

NYJA 6P

NYJA 2

Aksialet (kN)

TG=-8.74 TQ=-8.83 TA=-155.87

-10.50 0.60 -8.74 0.60 -356.15 0.60

NYJA 4L

Ssd,M Ssd,T Ssd,N

Transversalet (kN)

-212.74 155.87 -107.75

Ssd,M= -246.56 Ssd,T= 169.91 Ssd,N= -516.21

-246.56 169.91 -516.21

KOMBINIMI I NGARKESAVE SIPAS GJENDJES KUFITARE MBAJTËSE (ULS) 2.SITUATA SIZMIKE LLOGARITËSE

NYJA 9P

Momentet (kNm) MG=61.59 MQ=6.22 MA=150.91 Gkj Ssd,M Ssd,T Ssd,N

Transversalet (kN)

NYJA 11P

Aksialet (kN)

TG=-33.46 TQ=-8.83 TA=-71.96

NG=-144.17 NQ=-1.56 NA=-36.25

Y2,i Qki

Ad

61.59 0.60 -33.46 0.60 -144.17 0.60

6.22 -8.83 -1.56

S sd ,M  M G  0.6M Q  M A S sd ,T  TG  0.6TQ  T A S sd ,N  NG  0.6 N Q  N A

150.91 -71.96 -3 - 36.25

Momentet (kNm) Ssd,M= 216.23 S sd,T= -110.72 Ssd,N= -181.36

MG=0 MQ=-7.95 MA=-169.36 Gkj

216.23 -110.72 -181.36

Ssd,M Ssd,T Ssd,N

Transversalet (kN)

NYJA 13P

Aksialet (kN)

TG=0 TQ=7.22 TA=104.20

NG=-313.14 NQ=0 NA=0

Y2,i Qki

Ad

0.00 0.60 0.00 0.60 -313.14 0.60

-7 -7.95 7. 7 .22 0.00

-169.36 104.20 0.00

Momentet (kNm) Ssd,M= -174.13 Ssd,T= 108.53 Ssd,N= -313.14

MG=-61.59 MQ=-6.22 MA=-150.91 Gkj

-174.13 108.53 -313.14

Ssd,M Ssd,T Ssd,N

Transversalet (kN) (kN)

Aksi Aksial alet et (kN) (kN)

TG=33.46 TQ=8.83 TA TA=71.96

NG=-144.17 NQ=-1.56 NA=-36.25

Y2,i Qki

Ad

-61.59 0.60 33.46 0.60 -144.17 0.60

-6 -6.22 8 .83 -1 -1.56

-150.91 71.96 -3 - 36.25

Ssd,M= -216.23 Ssd,T= 110.72 Ssd,N= -181.36

-216.23 110.72 -181.36

1.Diagrami llogaritës nga situata e përhershme dhe e për kohshme llogartitëse

2.Diagrami llogaritës nga situata sizmike llogaritëse

Dimensionimi i rigelit P( $ 0 / . : $ 2 -

$ : . / / 1 -

> 3 . $ / 1 -

5lasa e betonit 7-$4$ 5lasa e ;elikut (-0 Prerja tërthore e rigelit 1/. $3.

D 1 . 2 2 1

 : . / $ 1

0.

$.

a. Dimensionimi në fushë =sd1$/. k
# ck ! c #

=

30 1.5

= 20 N mm2

500

=

= 0.7* l = 0.7*6. 7*6.50 = 4.55m l0 = 0. 1

l + 2b =

1

* 4 55 + 2 *0 *0 3 = 1 51m

Armatura e nevojshme A sl = A sl =

sds

k * d * # !d

+

N sd # !d

128.93 *106 0.97 0.975* 5* 460* 460* 434. 434.78 782 2

+

37.0 *1 *103 434. 434.78 782 2

prov A sl = 746.28mm 2 Përvetësojë 4∅16 me A sl = 804 mm 2

b= b=

bw − 2c0 − 2∅ s + c

d1 = c0 + ∅ s + ∅ 2

c+∅ 300 − 2 * 20 20 − 2 *8 * 8 + 30

d 1 = 20 + 8 + 16 2

30 + 16

b = 5.0 sh,#ra rend

d1 = 36mm ( d1 s = 40mm

b. Dimensionimi mbi mbështetës =sd2$./0 k
=

=

Armatura e nevojshme A sl = A sl =

sds

k * d * # !d

+

N sd # !d

229.88 *1 *106 0.87 0.874* 4* 460* 460* 434. 434.78 782 2

+

37.0 *1 *103 434. 434.78 782 2

prov A sl = 1400.2 mm 2 Përvetësojë 7∅16 me A sl = 1407 mm 2

0-shufra rendi i parë 2-shufra rendi i d"të d 1 = d 1 = d1 =

5 ( c0 + ∅ s + ∅ 2 ) + 2 ( c0 + ∅ s + ∅ + c + ∅ 2) 7 5 ( 20 + 8 + 16 2) + 2 ( 20 + 8 + 16 + 30 + 16 2) 7 344 7

= 49.14mm ( d1 s = 40mm

(8F&?8=8
sw,max =

A s1 ρ * bw

=

1.0 0.0011* 30 30

= 30.30cm

0.5- #cd * z * bw - 'd 2 = 0. - = 0.7 −

# ck

= 0.7 −

30

= 0.55 > 0.50 200 200 0.874* 4* 46 = 40.2 40.204 04cm = 0.87 30 # cd = = 20 N mm2 = 2.0 kN cm2 1.5

bw = 30cm - 'd 2 = 0.5*0.55*2.0*40.204*30 - 'd 2 = 663.36kN 0.2- 'd 2 < -sd* < 0.67- 'd 2

132. 132.67 67 < 169. 169.69 69 < 444. 444.45 45

sw,max = 0. 0.6d = 0.6 * 46 46 = 27.6cm sw,max ≤ 30 cm

Përvetësojë ∅ s 8 20 [ cm]

D8=B<(8<8=8 8 ()%K''E( 0.

2.

1/.

5lasa e betonit C − − 30 37 − 500 5lasa e ;elikut S − Dimensionet e sht"llës b h = 30 30 [cm ]

0.

$1.

-Për shq"rtim do të marrim tri raste: 1.(ituata e përhershme dhe përkohshme llogaritëse

2.(ituata sizmike llogaritëse P( (0

P( (0

?asti 1. P( (0

= -2>.

< -1.3$k<

7aktimi i shtresës mbrojtëse sipas B<1>>2-1-1:23 B# cnom = cmin + "cdev

.cmin,b / cmin = max 0cmin,d,r -1.3$k< 2./3 /10mm 1 cmin -përvetësohet në bazë të diametrit të shufrës së përvetësuar për armim nëse agregati i përdorur është më i madh se $2mm! atëherë duhet të rritet për 0mm#.%abela 3.2 Përvetësojë armaturën gjatësore ∅25mm cmin,b = 25mm cmin,d,r min, d,r -@aktohet kga agresiviteti 8 jashtëm.%abela 3.3<

5lasa e ekspozimit 971 ⇒ cmin, d,r = 10 mm

()LK?%8=8 8 5<(%?&5(8<8%
2 N ≤ 0.2 + 0.1n për n < 4 2 EI

2

-shuma e ngarkesave vertikale

2 EI -shuma e B8 për të gjitha elementet vertikale # = htot

2 N ≤ 0.6 për n  4 2 EI

n - numri i etazheve

htot -lartësia totale e objektit

2 N = 1330.97kN

2 EI = 6 *

3.15*10 3.15*107 * 0.3* 0.3* 0.3 0.33 12

= 127575kNm3

htot = 6.8m O n = 2 # = 6.8 *

1330.97 127575

= 0.694 ≤3 0.4 -konstruksioni analizohet si konstruksion me zhvendosje.

''FA?8%MA B ==B<%BCB D)B FMA%E(8(E (E BPMB( a#

l0 =  * l

B7-2 0..$.2 fq./#

rigeli

 3.15*1 5*107 *0. *0.3*0. 3*0.33   3.15*1 5*107 *0. *0.3*0. 3*0.33   4  +  5 2 . 7 0 * 1 2 3.10 *12 4 a  b 5 O k 1 =  2* 3.15*1 3.15*10 07 *0.3*0.53  42 2 * 5 6.50*12   trari

 EI   EI   4 l  +  l  5   a  b  k 1 =  2 EI   2 2 * 4 l 5 trari k 1 =

7875 7875 + 6858. 6858.87 87 60576.92

k 1 = 0.2432 k 2 = 0.1

./

 = max 0 1 + 10 *

1/

0.2432 * 0.1  0.2432   0 .1  * / ; 1+ * 1 +     0.2432 + 0.1  1+ 0.2432   1+ 0.1 , /

1.307;1.3037  = max 61.307;1.303

 = 1.307 b * h3

l0 =  * l l 0 = 1.307*3.10 l0 = 4.05m

= λ =

l 0 i

i=

I A

=

12 = b*h

h2 12

=

h 12

O i =

0.30 12

= 0.0866

= λ =

4.05 0.0866

= 46.76

( 01 = ( + ei * N ED

( 02 = ( + ei * N ED

( = − 29.07 kNm

( = 20.64kNm

( 01 = − 29.0 29.07 7 + 6.42 6.429 9

( 02 = 20.6 20.64 4 + 6.42 6.429 9

( 01 = − 22.64kNm

( 02 = 27.062kNm

λ lim =

λ lim =

A = 0.7

20 A+C 20 A+C n

r m =

( 02

=

−22.64 27.062

+ = 1.1

C = 1.7 − ( −0.836)

C = 1.7 − r m

C = 2.536

20*0.7*1.1*2.536 0.604

λ lim = 50.25 λlim = 50.25 > λ = 46.76 -nuk paraqitet epja.

Hor@at meritore për dimensionim kNm

( 01

= −0.836

n=

N ED Ac * # cd

# cd =

# ck ! c

=

=

1087.43*103 300 *3 *300 *2 * 20

30 1.5

= 20 N / mm2

= 0.604

2

lim 'd

(

lim 'd

 249.5  −6 =   *300*0.85*20*10  1.634  = 118.90kNm -=omenti përballues i betonit në zonën e sht"pur.

lim " ( = ( sds − ( 'd

" ( = 135.26 135.26 − 118.90 118.90 " ( = 16.36 Nm A s1 = A s1 =

lim ( 'd *

k * d * # !d

+

d 2 = 50.5mm

"( − ED ( d − d2 ) # !d # !d

118.83 *106 0.74 0.740* 0* 249. 249.5* 5* 434. 434.78 782 2

+

16.36 *106 249.5 5 − 50.5 50.5) 434. 434.78 782 2 ( 249.

−

1087.43 *103 434. 434.78 782 2

A s1 = 1480. 1480.3 3 + 189. 189.08 08 − 2501. 2501.09 09 A s1 = − 831.71mm2 A s 2 = 189.08mm2 A s min =

0.1 N ED # !d

=

0.1*1087.43*1 3*103 434.782

= 250.10mm2

A s min = 2 0 0 Ac = 0.02 *3 * 300 *3 * 300 = 1800mm2

-armatura minimale.

?asti 2. a# P( (0 a#

P( (0

=

− 30 37 5lasa e betonit C − 5lasa e ;elikut S − − 500 Dimensionet e sht"llës b h = 30 30 [cm ]

-1>.1

< -//.k<

(htresa mbrojtëse cnom = 30 + 8 = 38mm

= λ =

4.05 0.0866

= 46.76

l0 = 4.05m

1>3.30

''FA?8%MA B ==B<%BCB ( 01 = ( + ei * N ED ( = −190.10kNm ei = 0.005913 ( 01 = −190.10 190.10 + 0.00591 0.005913* 3* 676.77 676.77 ( 01 = −186.09kNm

( 02 = ( + ei * N ED ( = 194.45kNm ( 02 = 194.45 194.45 + 0.005913*676.77 0.005913*676.77 ( 02 = 198.45kNm

-//.k<

7aktimi i armaturës së nevojshme cnom = 38mm O ∅25mm

d1 = cnom + ∅sw + ∅ sl 2

d = h − d 1

d 1 = 30 + 8 + 25 2

d = 300 300 − 50.5 50.5

d 1 = 50.5

d = 249.5mm

k d =

d

249.5

=

( sds

265.78*10

b *# * # cd

h  − d 1  2   0.3  ( sds = 198.45 + 676.77  − 0.0505  2  ( sds = 265.78kNm ( sds = ( ED + N ED 

6

= 1.092

8 = * s

# !d E

=

434.782*106 9

200*10

= 2.17 0 00

300*0.85*20

le,ojmë deformimet: 8 c 8 s = −3.5 2.1 0 00 ⇒ k d * = 1.634 ! k "* = 0.625 ! k z * = 0.740 2

lim 'd

 d  =  *  * b *# * # cd  k d  2

lim 'd

(

lim

 249.5  −6 =   *300*0.85*20*10  1.634  118.90kNm 118.90kNm -=omenti përballues i betonit në zonën e sht"pur.

?asti 2. b# P( (3 a#

P( (3

= 5lasa e betonit C − − 30 37 5lasa e ;elikut S − − 500 Dimensionet e sht"llës b h = 30 30 [ cm ]

< -01.3>

23/./

a# rigeli 1 P( (3

(htresa mbrojtëse cnom = 30 + 8 = 38mm

2

-$/0.

-01.3>

''FA?8%MA B ==B<%BCB D)B FMA%E(8(E (E BPMB(

l0 =  * l

/.

 = max 0 1 + 10 *

/1

* k1 * k2  k   k  / ;  1 + 1  *  1 + 2  -për konstruksion me zhvendosje. k + k  1 + k   1 + k / ,

k1 sht"lla e shq"rtuar b# k2

''FA?8%MA B ==B<%BCB ( 01 = ( + ei * N ED

( 02 = ( + ei * N ED

( = 246.86kNm

( = −365.07kNm

ei = 0.005913 ( 01 = 246.86 246.86 + 0.0059 0.005913* 13* 517.49 517.49 ( 01 = 249.91kNm

λ lim =

λ lim =

20 A+C 20 A+C n

A = 0.7

λ lim = 68.703

( 02 = − 362.01kNm

r m =

( 01 ( 02

=

249.91

−362.01

+ = 1.1

C = 1.7 − ( −0.690)

C = 1.7 − r m

C = 2.390

20*0.7*1.1*2.390 0.287

( 02 = − 365.07 365.07 + 0.0059 0.005913* 13* 517.49 517.49

= −0.690

n=

N ED Ac * # cd

# cd =

# ck ! c

=

=

517.49*103 300 *3 * 300 *2 * 20

30 1.5

= 20 N / mm2

= 0.287

2

lim 'd

 d  =  *  * b *# * # cd  k d  2

lim 'd

(

lim 'd

 249.5  *300*0.85*20*10−6 =    1.634  = 118.90kNm -=omenti përballues i betonit në zonën e sht"pur.

lim " ( = ( sds − ( 'd

" ( = 413.50 413.50 − 118.90 118.90 " ( = 294.6kNm A s1 = A s1 =

lim ( 'd

k * * d * # !d

+

d 2 = 50.5mm

"( − ED ( d − d2 ) # !d # !d

118.83 *106 0.74 0.740* 0* 249. 249.5* 5* 434. 434.78 782 2

+

294.6 *106 249.5 5 − 50.5 50.5) 434. 434.78 782 2 ( 249.

A s1 = 1480.3 1480.3 + 3404.9 3404.92 2 − 1190.2 1190.22 2 A s1 = 3695mm2 A s 2 = 3404.92mm2

−

517.49 *103 434. 434.78 782 2

Përvetësojë dimensionet b h = 30 40 [cm ] 5lasa e betonit C − − 30 37 5lasa e ;elikut S − − 500 Dimensionet e sht"llës (htresa mbrojtëse cnom = 30 + 8 = 38mm d = h − d 1

d 1 = 30 + 8 + 25 2

d = 400 400 − 50.5 50.5

d 1 = 50.5

d = 349.5mm

k d =

349.5

=

( sds b *# * # cd

9'0''

h  − d 1  2   0.4  − 0.0505 ( sds = 362.01 + 517.49   2  ( sds = 439.37 kNm ( sds = ( ED + N ED 

d1 = cnom + ∅sw + ∅ sl 2

d

:'0''

439.37*106 300*0.85*20

= 1.190

8 = * s

# !d E

=

434.782*106 9

200*10

= 2.17 0 00


lim 'd

 d  =  *  * b *# * # cd  k d 

lim 'd

 349.5  *300*0.85*20*10−6 =    1.634 

2

(

Përvetësojë dimensionet b h = 40 40 [cm ] 5lasa e betonit C − − 30 37 5lasa e ;elikut S − − 500 Dimensionet e sht"llës (htresa mbrojtëse cnom = 30 + 8 = 38mm d = h − d 1

d 1 = 30 + 8 + 25 2

d = 400 400 − 50.5 50.5

d 1 = 50.5

d = 349.5mm

k d =

349.5

=

( sds b *# * # cd

9'0''

h  − d 1  2   0.4  − 0.0505 ( sds = 362.01 + 517.49   2  ( sds = 439.37 kNm ( sds = ( ED + N ED 

d1 = cnom + ∅sw + ∅ sl 2

d

9'0''

439.37*106 400*0.85*20

= 1.374

8 = * s

# !d E

=

434.782*106 9

200*10

= 2.17 0 00


lim 'd

 d  =  *  * b *# * # cd  k d 

lim 'd

 349.5  *400*0.85*20*10−6 =    1.634 

2

(

=ën"ra tjetër e llogaritjes së armaturës.

d1 h = 50.5 50.5 400 400 = 0.12 0.126 6 me interpoli

mes vlerave d1 h = 0.10 dhe d1 h = 0.15 @akt jme vlerat.

- sd = N ED bh# cd = 517. 517.49*1 49*10 03 400* 400* 400* 20 = 0.16 .161 µ sd = ( ED bh 2 # cd = 439. 439.37*10 37*106 400 400 4002 *20 = 0.343

=e anë të interpolimit @aktojmë & = = 0.70 & tot =

A s ,tot b*h

*

# !d

A s ,tot = 0.702 *

Për C − − 30 37 ⇒ ) rd = 0.34 N mm2 k = 1.6 − d k = 1.6 1.6 − 0.34 0.3495 95 k = 1.2505

= ρ =

A s bw * d

σ cp =

N A

=

=

6*491 300 * 34 349.5

517.49*103 400*400

= 0.0280

= 3.234

mm2

400*400 434.782 20

tot *

= 5166.75mm

Përvetësojë 12∅ 25mm me A s prov = 5892 m 2

(8F&?8=8 8 ()%K''E(
# cd

⇒ A s ,tot = As1 + As 2 =

b*h # !d # cd

D8=B<(8<8=8 8 %)B=B'8% 

(ituata sizmike llogaritëse

( G = 10.50kNm

G

( ) = 10.32kNm

)

= 356.15kN H G = 8.74kN = 89.31kN H ) = 8.83kN

E

= 107.75kN

E

= ±348.38kNm

H E = ± 155.87kN

?asti 1

?asti 2 <

< =

)

)

=

'f1.2m hth.m

G

'f1.2m hth.m

G

( 0* = ( 0 = 507.95kNm N 0* = N $ + N TH TH

= . * + * . # * ! mes = 3.30 * 3.30 *1. 0*20 = 261.36kN

N 0* = 553. 553.21 21+ 261. 261.36 36 = 814. 814.57 57kN

e0 =

( 0* N 0*

=

507.95 814.57

= 0.623m

e0 = 0.623m >

. 6

=

3.30 6

= 0.55m Mashtëqendërsi e m dhe.

Vërejtje: Pasi ngarkesa nga sizmika është ngarke së alternative atëherë projektojmë themel simetrik .

(for@imet nën taban të themelit. c=

+

− e0 =

2 c = 1.027m σ =

2 N 0*

=

3.3 2

0.623 = 1.02 1.027 7m − 0.623

2*814.57

= 160.23 <

a

= 170 kN m2

3c. 3 *1 *1.027 * 3. 3.30 2 N sd σ sd = O sd = ! G N G + % 2, i N ) + ! r d N E Oku ! G = 1.0 ! % 2,i = 0.6 ! ! rd = 1.40 3c.

'logaritja e sipërfaqes së armaturës. drejtimi ,-, d " = hth − d 1" O d1 " = c0 +

∅ " 2

= 50 +

12 2

= 56mm

d " = 800 − 56 = 744mm d

k d =

req sl

A

" − " l

s

I − I ( sd

744

=

= 3.232

540.43*106

b# # cd

600*0.85*20

=

( Isd − I

540.43*106

=

As1

k z d# !d

A sl req

=

0.941*744*434.782

*. =

113 1775.43

⇒ kd = 3.225; k" = 0.158; k z = 0.941

= 1775.43mm 2

* 3300 = 210.03mm Përvetësojë ∅12 20 [ cm]

drejtimi "-" d ! = hth − d 1 ! O d1 ! = c0 + ∅ " + d ! = 800 − 68 = 732mm

∅12mm me A s1 = 113mm2

∅ ! 2

= 50 + 12 +

12 2

= 68mm

?asti 2 < )

Dimensionet e themelit G '$.$4$.$ m

= 'f1.2m hth.m

G

7aktimi i ndikimeve në themelë. $

= N G + N ) + N E = 356.1 356.15 5 + 89.31 89.31 107. 107.75 75 = 553. 553.21 21kN

H $ = H G + H ) − H E = 8.74 + 8.83 − 155.87 = − 138.3kN

= ( G + ( ) − ( E = 10.5 10.50 0 + 10.3 10.32 2 348. 348.38 38 = − 327. 327.56 56kNm

$

9

0

*0.80 = 438. 438.20 20kNm = ( $ + H $ * hth = 327.56 + 138.3*0.80

9 N 0 = N $ + N TH = 553. 553.21 21+ 3.3 3.32 *1.2* *1.2* 20 = 814. 814.57 57kN e0 =

( 0 N 0

=

438.20 814.57

= 0.537 m

(for@imet nën taban të themelit.

e0 = 0.537m <

. 6

=

3.30 6

= 0.55m Mashtëqendërsi e vo ël.

σ IG− I = 56.91 kN m 2 E$ , I − I G

σ

=

σ G1 + σ G I − I 2

, II − II σ E$ = G

σ G1 + σ G 2 2

σ I)− I = 8.40 kN m 2

= =

56.91 56.91 + 58.75 58.75 2 58.75 58.75 + 54.95 54.95 2

= 57.68kN m

E$ , I − I )

2

σ

=

σ )1 + σ ) I − I

, II − II σ E$ = )

= 56.70 kN m2

2 σ )1 + σ ) 2 2

= =

9.92 9.92 + 8.40 8.40 2 9.92 9.92 + 6.48 6.48 2

= 9.16kN m2 = 8.20 kN m 2

I − I σ E = 17.18 kN m 2 E$ , I − I E

σ

=

σ A1 + σ IA− I

σ E E$ , II − II =

2 1 2

σ A1 =

1 2

=

68.0 68.05 5 + 17.18 17.18 2

= 42.16kN m2

*68.05 *68.05 = 34.0 34.025 25kN m2

'logaritja e ndikimeve statike. ( IG− I = σ GE$ , I − I * . * "I − I * "c I − I G

= 57.68*3.30*1 57.68*3.30*1.45*0.91 .45*0.914 4 = 252.2 252.26 6 kNm

− II ( II = σ GE$ , II − II * + * !I −I * !c G II − II G

= 56.70* 56.70* 3.30*1.4 3.30*1.45* 5* 0.91 0.914 4 = 247.9 247.97 7 kNm

I − I )

9.16* 3.30*1.4 3.30*1.45* 5* 0.914 0.914 = 40.06 40.06kNm = 9.16*

II − II )

8.20* 3.30*1.45*0.91 3.30*1.45*0.914 4 = 35.86 35.86kNm = 8.20*

" − " l

s

=

As1 A sl req

*. =

113 1762.90


drejtimi "-" d ! = hth − d 1! O d1 ! = c0 + ∅ " +

∅ ! 2

= 50 + 12 +

12 2

= 68mm

d ! = 800 − 68 = 732mm

d

k d =

req sl

A

! − ! l

s

II − II sd

=

=

732

=

6

(

477.806*10

b# # cd

600*0.85*20

− II ( II sd

477.806*106

k z d# !d

As1 A sl req

=

= 3.382 ⇒ kd = 3.372; k" = 0.150; k z = 0.945

0.945*732*434.782

*+ =

113 1588.68

= 1588.68mm2

∅12mm me A s1 = 113mm2




(ituata e përhershme dhe e përkohshme llogaritëse

a0 b0 = 40 40 [ cm ] qa = 170 kN m + = 3.30 m . 3.30 [ ] 2

G )

= 10.50 kNm = 26.56 kNm

N G = 356.15kN

H G = 8.74 kN

N ) = 108.53kN

H ) = 18.72kN

=

'f1.2m hth.m

G

7aktimi i ndikimeve në themelë. N $ = NG + N) = 356. 356.15 15 + 108. 108.53 53 = 464. 464.69 69kN H $ = H G + H ) = 8.74 .74 + 18.7 18.72 2 = 27.4 27.46 6kN ( $ = ( G + ( ) = 10.5 10.50 0 + 26. 26.56 = 37.0 37.06 6kNm

9

0

= ( $ + H $ * hth = 37.06 + 27. 27.46*0. 6*0.80 = 59.028kNm

9 N 0 = N $ + NTH = 464. 464.69 69 + 3.3 3.32 *1. *1.2* 2* 20 = 726 726.05 .05kN (

59.028

.

3.30

)

σ IG− I = 56.91 kN m 2 E$ , I − I G

σ

=

σ I)− I = 10.49 kN m 2

σ G1 + σ G I − I

=

2

, II − II σ E$ = G

56.91 56.91 + 58.75 58.75

σ G1 + σ G 2 2

2

=

58.75 58.75 + 54.95 54.95 2

= 57.68kN m

E$ , I − I )

2

σ

=

σ )1 + σ I)− I

, II − II σ E$ = )

= 56.70 kN m2

2 σ )1 + σ ) 2 2

= =

14.3 14.39 9 + 10.4 10.49 9 2 14.3 14.39 9 + 5.53 5.53 2

= 9.96 kN m2

'logaritja e ndikimeve statike. ( IG− I = σ GE$ , I − I * . * "I − I * "c I − I G

= 57.68*3.30*1 57.68*3.30*1.45*0.91 .45*0.914 4 = 252.2 252.26 6 kNm

− II ( II = σ GE$ , II − II * + * !I −I * !c G II − II G I − I

c

= !c = 0.914m

5ombinimi i ngarkesave * (

* (

12.44 4 *3.30*1.45* *3.30*1.45* 0.914 0.914 = 54.40 54.40kNm = 12.4

II − II )

= 56.70* 56.70* 3.30*1.4 3.30*1.45* 5* 0.91 0.914 4 = 247.9 247.97 7 kNm

= !I − I = 1.45m O

I − I )

1.35

1.50

= 12.44kN m2

9.96* 3.30*1.4 3.30*1.45* 5* 0.91 0.914 4 = 43.55 43.55kNm = 9.96*

drejtimi "-" d ! = hth − d 1! O d1 ! = c0 + ∅ " +

∅ ! 2

= 50 + 12 +

12 2

= 68mm

d ! = 800 − 68 = 732mm

d

k d =

II − II sd

732

=

400.08*10

(

A

! − ! l

s

=

=

− II ( II sd

k z d# !d

As1 req sl

A

= 3.696 ⇒ kd = 3.691; k" = 0.134; k z = 0.951

600*0.85*20

b# # cd req sl

6

=

400.08*106 0.951*732*434.782

*+ =

113 1321.85

= 1321.85mm2

∅12mm me A s1 = 113mm2

* 3300 = 282.10mm Përvetësojë ∅12 27 [ cm ]

KOMBINIMI I NGARKESAVE SIPAS GJENDJES KUFITARE TË SHFRYTËZIMIT (SLS) KOMBINIMI KUAZI6VAZHDUESHËM KUAZI6VAZHDUESHËM Ed = ∑ GKj ⊕ ∑ (ψ 2,i * QKi ) ⊕ P k j ≥1

ψ 2, i

= 0.6

i >1

për de!"#"$ !ë %&'e!"! A*+r",ë$ #-+ T+&.A/./ *0.1/ EN/2234334

NYJA 4HD

Për %$&"#"$"# =+,"6>+,?d=e?ë$ d% !e &ë'$ë %$&"#"$"# e #-+re+>e >e!ë$ për e@e$e#!"#(RIGELI) 0ë d% !ë %#!r%@@%?e! #-+ p@++r"!'e! d?e =@'e!.

NYJA 5

M%$e#!e! (N$)

Tr+#>er+@e! (N)

A"+@e! (N)

NYJA 6HM

M%$e#!e! (N$)

Tr+#>er+@e! (N)

A"+@e! (N)

M%$e#!e! (N$)

Tr+#>er+@e! (N)

A"+@e! (N)

MG5677.88

TG59:.99

NG548.9:

MG589.:1

TG53.3

NG548.9:

MG5623.9<

TG56
NG548.9:

M;5688.<<

T;519.7:

N;53./<

M;589./8

T;53.3

N;53./<

M;5694.23

T;567:.49

N;563.4:

Ed,M= -93.37 Y4"

G'

;"

Ed,M= 75.63 Y4"

G'

Ed,T= 108.35

;"

Ed,M= -134.52

Ed,N= 24.84

Y4"

G'

Ed,T= 0.00

;"

Ed,T= -119.22

Ed,N= 24.84

Ed,N= 24.59

EdM

677.88

3.73

688.<<

62:.:9

EdM

89.:1

3.73

89./8

91.7:

EdM

623.9<

3.73

694.23

6/:8.14

EdT

9:.99

3.73

19.7:

/3<.:1

EdT

3.33

3.73

3.33

3.33

EdT

6
3.73

67:.49

6//2.44

EdN

48.9:

3.73

3./<

48.<8

EdN

48.9:

3.73

3./<

48.<8

EdN

48.9:

3.73

63.4:

48.12

NYJA 6HD

NYJA 7

M%$e#!e! (N$)

Tr+#>er+@e! (N)

A"+@e! (N)

NYJA 8HM

M%$e#!e! (N$)

Tr+#>er+@e! (N)

A"+@e! (N)

M%$e#!e! (N$)

Tr+#>er+@e! (N)

A"+@e! (N)

MG5623.9<

TG5
NG548.9:

MG589.:1

TG53.3

NG548.9:

MG5677.88

TG569:.99

NG548.9:

M;5694.23

T;57:.49

N;563.4:

M;5:9.87

T;53.3

N;563.4:

M;5684.94

T;561:.22

N;563.4:

Ed,M= -134.52 Y4"

G'

;"

Ed,M= 69.83 Y4"

G'

Ed,T= 119.22

;"

Ed,M= -92.07

Ed,N= 24.59

Y4"

G'

Ed,T= 0.00

;"

Ed,T= -106.16

Ed,N= 24.59

Ed,N= 24.59

EdM

623.9<

3.73

694.23

6/:8.14

EdM

89.:1

3.73

:9.87

72.<:

EdM

677.88

3.73

684.94

624.39

EdT

3.73

7:.49

//2.44

EdT

3.33

3.73

3.33

3.33

EdT

69:.99

3.73

61:.22

6/37./7

EdN

48.9:

3.73

63.4:

48.12

EdN

48.9:

3.73

63.4:

48.12

EdN

48.9:

3.73

63.4:

48.12

KOMBINIMI I NGARKESAVE SIPAS GJENDJES KUFITARE TË SHFRYTËZIMIT (SLS) KOMBINIMI I RRALË Ed = ∑ GKj ⊕ QK 1 ⊕ PK ⊕ ∑ (ψ 0, i * QKi ) j ≥1

i >1

ψ 0, i

=

NYJA 4HD

NYJA 5

M%$e#!e! (N$) MG5677.88 M;/5688.<< 56 . G ' SdM SdT SdN

677.88 9:.99 48.9:

Tr+#>er+@e! (N) TG59:.99 T;/519.7: 5 . ; /

NG548.9: N;/53./< 56 .

Y0"

688.<< 19.7: 3./<

A"+@e! (N)

3.93 3.93 3.93

NYJA 6HM

M %$ %$ e# e# !e !e ! ( N N $) $) MG589.:1 M;/589./8 5 .

; 4 62.34 6//9.7: 4./1 /:4.2/ 63.:4 48.72

SdM SdT SdN

T r+ r+ # # >e >e r r +@ +@ e! e! ( ( N) N) TG53.3 T;/53.3 56 .

G '

;/ 89.:1 3.33 48.9:

A "" +@ +@ e! e! ( ( N) N) NG548.9: N;/53./< 56 .

Y0"

89./8 3.33 3./<

MG5623.9< M;/5694.23 56 .

; 4 3.93 3.93 3.93

4.:3 64.88 64.11

27./3 6/.9/ 4:./:

MG5623.9< M;/5694.23 M;4561.7: G ' 623.9<
Tr+#>er+@e! (N) TG5
NG548.9: N;/563.4: 56 .

Y0"

694.23 67:.49 63.4:

;4 3.93 3.93 3.93

61.7: 6/79.74 64.88 6/87.48 64.11 44.94

Ssd,M= 96.10 Ssd,T= -1.71

Ssd,T= -146.24

Ssd,N= 24.69

Ssd,N= 23.13

Ssd,N= 22.72

A"+@e! (N)

MG589.:1 M;/5:9.87 M;454.:3


Tr+#>er+@e! ( (N) TG53.3 T;/53.3 T;454.88

G ' 61.7: 6/79.74 64.88 /84.<4 64.11 44.94

Ssd,M= -167.62

NYJA 8HM

M%$e#!e! ( (N$)

; 4 3.93 3.93 3.93

623.9< 6
;/

A"+@e! (N)

Ssd,T= 132.91

NG548.9: N;/563.4: N;4564.11

Y0"


Tr+#>er+@e! (N) TG56
G '

NYJA 7

M%$e#!e! (N$)

M%$e#!e! (N$)

Ssd,M= -117.63

NYJA 6HD


0.7

;/ 89.:1 3.33 48.9:

:9.87 3.33 63.4:

A"+@e! ( (N) NG548.9: N;/563.4: N;4564.11

Y0"

M%$e#!e! ( (N$) MG5677.88 M;/5684.94 M;4562.34

; 4 3.93 3.93 3.93

TG569:.99 T;/561:.22 T;4564./1

G ' 4.:3 4.88 64.11

<7.84 /.9/ 44.94


Tr+#>er+@e! (N)

677.88 69:.99 48.9:

;/ 684.94 61:.22 63.4:

A"+@e! (N) NG548.9: N;/563.4: N;4563.:4

Y0"

;4 3.93 3.93 3.93

62.34 6//1.89 64./1 6/42.49 63.:4 48.4<

Ssd,M= -167.62

Ssd,M= 86.42

Ssd,T= 142.82

Ssd,T= 1.71

Ssd,M= -115.47 Ssd,T= -129.27

Ssd,N= 22.72

Ssd,N= 22.72

Ssd,N= 24.28

''FA?8%MA B P'A(A?8%MBCB 

''FA?8%MA B P'A(A?8%MBCB
beff101@m 21/

/ 1 m @  0  3 h $

31/ bI$@m



•

A
d = 464mm A s1 = 804mm 2 A s 2 = 402mm2

•

5ontrollimi i nderjeve në beton

σ cd , perm perm =

( ED , perm be## * !0, II

σ cd , perm perm =

( 3d − ! ) + # A 0 , II

6

E

( d − d 2 )

s2

!0, II

O * ( !0, II − d2 )

75.63*106 1510*63.78 6

464 − 63.78) + 6.261* 40 402 ( 3 * 46

( 646 − 36) 6 3.7 8

σ cd , perm ≤ 0.45 # ck për kuazi kombinimin

për kombinimin e rrallë σ cd , perm ≤ 0.60 # ck σ cd , perm 3.47 ≤ 0.45*3 0.45*30 0 = 13. 13.50 50 N mm 2 perm = 3.4

•

5ontrollimi i nderjeve në armaturë

-5ontrollimi i nderjeve në armaturë bëhet nga kombinimi i rrallë σ s ≤ 0.8 # !k ED

= 75.63kNm

ED , perm

= 96.10kNm

= 3.47 N mm2 *( 63.78− 36)

16 16.8 16.8 ; ( # cm ) = = = 2.725 # cm 38 ; ( t 0 ) =

1

( 0.1 + t 00.2 )

=

1

( 0.1 + 280.2 )

= 0.488

$ ( t: , t0 ) = $ 'H * ; ( # cm ) * ; ( t0 ) = 1.796 * 2.725 * 0.488 = 2.389 $ ( t: , t 0 ) = 2.389

 ( t − t0 ) / t 1  ; c ( t: , t 0 ) = 4 5 4 ; H + ( t − t0 ) / t 1 5 p0r t = :

Ec ,e## =

# E ,: =

t − dita dita k,r k,r sh!r sh!rtoh tohet et t0 = 28 dit0 , (dita k,r ngarkohet ) t1 = 1dit0

⇒ ; c ( t: , t 0 ) = 1.0 E cm

1 + $ (t: , t 0 )

E s E c ,e##

=

=

31938.7 1 + 2. 2.389

200000 9424.22

804 



402

21 22 * 0 0011 1

= 9424.22(%a

= 21.22

 

+ II = 21.22 * 0. 0.0011 1+ A

0.3

402 

 = 0.0350 *

36 



0 0242



''FA?8%MA B P'A(A?8%MBCB

•

=B%DA 8
'logarojmë sipërfaqen minimale të armaturës

A s min =

kc  k  #c ,e##  Act σ s



k c = 0.4 41 +

4

k1 = 1.5 k1 =

h h*

2

 5 ≤ 1. 0 k1 + # c ,e## 5 σ c

aksiale n0 n0 sh s ht!p1e − k,r kemi #orc0 ak

0aksialen0t0 en0t0rheq1e − k,r kemi #orc0aks

3

h* = h p0r h < 1.0m h* = 1.0m p0r h  1.0m

7aktojmë pozitën e qendrës rëndëse për prerjen e betonit pa armaturë beff101mm

!0 s =

(b

w

* h2 ) 2 + ( be## − bw ) hp2ll  2





bw * h + hpll ( be## − bw ) * 500 ) (300 *5 2

2 + (1510 − 30 3 00 )160 2  2

s  "

 / 1 m m   0

 # ct ,e##  kc  hcr ; hcr = !0 p = 34.57cm ∅ s = ∅*s  *  2.9  2(h − d ) 4*34.57  3.0  0.4*34 * ∅ s = ∅*s    2.9  2(50 − 46.4)

∅ s = 1.986∅*s ⇒ ∅*s =

16 1.986

= 8.05mm ⇒ ∅ s = 8mm

⇒ σ s = 360 %a A min =

kc  k  #c ,e##  Act

σ sd , perm

*

As

23

2( h − d ) b  #ct ,e##

= 22 *

211.13

*

804

2(5 2(500 − 464 464) 300*3 300*3

1/

211.1$

2

22.$

20

= 57.63  22mm  ∅ p0rv. = 16mm
sl = 238mm

+ t = = : Q σ sd , perm ,: = z = d −

( ED , perm

! II ,: 3

σ sd , perm ,: =

prov

z * A s

= 464 −

87.09 3

75.63*106 434.97*804

= 434.97 mm

= 216.26 %a

sr , max = 3.4c + 0.425k1k2 * ∅ ρ e## < 5( c + ∅ 2) c = h − d − ∅ 2 = 500 − 464 − 16 2 = 28mm

ρ e## =

A s Ac ,e##

=

804 27000

= 0.029

3.4* 28 + 0.42 0.425* 5* 0.8* 0.8* 0.5*1 0.5*16 6 0.02 0.029 9 sr ,max = 3.4* 188.99 sr ,max =

σ sd , perm − k c

# ct ,e##

(1 + # E ,:  ρ e## )

c -shtresa mbrojtëse për armaturë gjatësore

∅ -diametri i armaturës kr"esore k 1 = 0.8 *

 Geton-Arme

k 2 = 0.5,

k 1 = 0.5 *

 Geton i paranderur

k 2 = 1.0 ,

5<%?''8=8 5<%?''8 =8 8 &'MBCB beff101mm

5lasa e betonit 7-$4$ 5lasa e ;elikut (-0 d1 = d 2 = 36mm

As2 21/   0  h

As1 31/

d = 464mm 2 A s1 = 804mm

A s 2 = 402mm2 d1

bI$

ρ = = ρ ' =

A s1 bw * d A s 2 bw * d

ρ 0 = 10

−3

-është raporti i armaturës së kërkuar të tërheqjes në mesin e hapësirës për ti rezistuar momentit. -është raporti i armaturës së kërkuar sht"pëse në mesin e hapësirës për ti rezistuar momentit. # ck -është raporti i armaturës referente.

# ck -rezisten@a e betonit në sht"pje =Pa#

%abela .3< B7 2:23!fq.12# (istemi strukturës 1# %rari i thjeshtë! pllaka e mbështetur lirisht me armim në një drejtim apo d" drejtime

5

'aporti i le1,ar le## d g1eg10sisht l1 d (l2 d )

 1.5% ρ 

≤ 0.5% ρ ≤

(kema statike

leff

1.

13

2

1 l

l2

2# Husha e parë e trarit kontinual apo pllakës kontinuale të armuar në d" drejtime me kontinuitet nga ana më e gjatë.

leff

1.$

1

2/

1 l

l2

$# Husha e brendëshme e trarit apo pllakës e armuar në një drejtim apo d" drejtime.

leff

1.0

2

$

1 l

l2

3# Pllakat pa trarë!për anën më të gjatë.

1.2

1

1 l

23 l2

0# 5onzola

leff

.3

/



1 l

''FA?8%MA B (A5%E B &'MBCB I# -7aktojmë aksin neutral për gjendjen e nderur nderur 8#!ndërsa për gjendjen 88# e kemi nga llogaritja e plasaritjeve. beff101@m As2

=

, / 1

 f h

=

m @ d  0  h

5lasa e betonit 7-$4$ 5lasa e ;elikut (-0 d1 = d 2 = 36mm d = 464mm A s1 = 804mm 2 A s 2 = 402mm2

As1 bI$@m

= : Për t = 2

 be##  h   A  b  h − 1   #  + A I O + I = # E: ρ I 1 + s 2  O D I =  e## − 1  # + +I A I = # E: ρ I 1 +   O C I = 0.5  h A s1 d   bw   h   A s1   bw  h 0.5 + C d 

As 2 d 2 

=omentet e iner@isë I 0 -=omenti i iner@isë së prerjes se betonit pa armaturë I 0 =

bw 3

(!

3 0 , p

+!

3 0, s

)

(b +

e##

− bw )  h3# 12

h   + ( be## − bw ) h# *  !0,s − #  2  

2

(151 − 30 )  163

2

16   + (151− 30) 16 * 15.42 −  = 598055.12cm 4 I 0 = 34.58 + 15.42 ) + ( 3 12 2  I I -=omenti i iner@isë së prerjes për gjendjen e nderur 8 30

I I =

bw 3

3

(!

3 I , p ,:

+!

3 I , s ,:

)+

(b

e##

− bw ) h3# 12

2

h   + ( be## − bw ) h# *  !I ,s ,: − #  + (# E , : − 1) *  As1 ( d − !I ,s , : )2 + As 2 ( !I , s , : − d )2  2  

151 − 30 ) 16 163 (151

2

16   + (151− 30) 16 10.95−  + I I = 39.05 + 10.95 ) + ( 3 12 2  21.22 2 − 1)   8.04 8.04(4 (46. 6.4 4 − 10.9 10.95 5)2 + 4.02 4.02((10.9 10.95 5− 46.4 46.4))2  ( 21.2 30

3

3

I I = 875444.95cm4

I II -=omenti i iner@isë së prerjes për gjendjen 88

I II =

be## * !I3I , s ,:

+ # E A 1 ( d − !II

2

) + (#

E

(

II , s , :

< h # )

− 1) A 2 ( !II

,

− d 2 )

2

= ( ED, perm,: = 75.63kNm

sd

Ec.e## = 9424.22 (pa = 94242.2 kN cm 2

I I = 875444.96cm4 1

75.63 *100

=

r I

94242.2*875444.96

= 9.16*10−8

1 cm

σ sd , perm,: = 211.13 %a σ sd , perm,:

8 s =

E s

II , s ,:

1 r II

=

=

211.13 200000

= 1.05*10−3

= 8.709cm 1.05 *10−3 46.4 46.4 − 8.70 8.709 9

( cr = # ctm *

I 0

= 2.78*10−5

1 cm

për prerje drejtkëndeshe ( cr = # ctm *

0, p

bh 2 6

-=omenti i plasaritjes.

598055.12 = 5015.49kNcm 34.58

cr

= 0.29 

sd

= 75.63kNm <3 ( cr = 50.15kNm duhet të llogariten uljet nëse

sd

< ( cr ateherë prerja mbetet në gjendjen

 1  'akueshmëria nga tkurrja e betonit    r csm  1

= (1 − = )

rcs ,m 1 rcs

=

1

+ =

rcs , I

1 r cs, II

8 cs:  # E  S I

8 cs: -deformimet nga tkurrja! S -momenti statik i sipërfaqes së armaturës ndaj aksit neutral! I -momenti i iner@isë.

1 rcs,1 1 rcs ,2 1 r cs, m 1 rcs m

=

8 cs:  # E  S I

=

8 cs:  # E  S II

I I I II

=

0.00059 * 21.22 * 255.47

=

875444.96

= 3.65*10−6

0.00059 * 21.22 * 282.60 277743.43

1 cm

cm

= 1.27*10−5

0.769 9) *3.65* *3.65*10 10−6 + 0.76 0.769*1 9*1.2 .27*10 7*10−5 = (1 − 0.76

= 1.06*10−5

1 1 cm

$ = ; = $ = =

5 48

(1 − 0.1; ) O

93.27 93.27 + 134.5 134.52 2 75.63 5 48

l 250

=

1 r tot

=

650 250

A

+ ( +

( C

= 3.011

0.1*3.01 011 1) = (1 − 0.1*3.

# tot = $  l 2  # le1 =

; =

3.4945 48

3.4945 48 *650 *6502 *3. *3.19*10 9*10−5 = 0.981cm

= 3.25cm

# tot = 0.981cm < #le1 = 3.25cm - Plotësohet kushti.

Detali i armimit te rigelit POS 003 P 1:25 Prerja 1-1 P 1:10 1 7 2ø 16

2 18ø8 /20,stafë

3 3ø 16

5 2ø 16

2 2 28ø8 /12,stafë

6 2ø 16

4 3ø 16

3 2 18ø8 /20,stafë

7 2ø 16

3 ø 16 3 2 13ø8 /12,stafë

13ø 8/12,stafë 2

6 1

2 ø 16 5 1 2 ø 16

8 2ø 12

9 2ø 16

1 2ø 16 1

40

61 0

8 2ø 12

2 ø 16

9 2ø 16

2

40

3

6 0 5

2 ø 12 8

610

40

4 3

28ø 8/12,stafë 2

1340 4

4

4 3ø 16L=499cm

1 4ø 16L=735cm

0 1

45° 4

5

720

45°

5

720

6 4

6 4

4

5

440

30

1 0 26

3

45°

2 90ø 8L=191cm

10

4 4

7 4ø 16L=219cm

3 3ø 16L=1059cm

45°

5

45° 4

1000

4

5

160 5 4

5 4

45°

10

4

5

5 2ø 16L=370cm

45°

5

6 2ø 16L=311cm

45°

360 4

4

5

45°

Prerja 2-2 P 1:10

4

4

9 4ø 16L=510cm

5

10 5

300

4 4

4 4

2 ø 16 1 5 4

0 1

3

8 4ø 12L=735cm

0 1

5

500

45°

3 ø 16 3

45°

N k 8 7 . 4 7 1 = 6

d s F N k 8 7 . 4 7 1 = 4

d N s k F 5 9 . 6 3 4 = N 0 k d s 7 F 1 . 2 6 2 =

7

N k 8 7 . 4 7 1 = 5

d s F

N k 8 7 . 4 7 1 N = 1 k 6 5 . 9 4 3 =

1

d s F

N 0 k d 8 s 7 F . 4 7 1

0 5

2 ø 12 8 4 3

2 ø 16 9

2 ø 16 1

6

30

6

Prerja 3-3 P 1:10

3 ø 16 4

5

5

4 1

2 ø 16 7

7

0 5

2 ø 12 8

4 3

2 ø 16 1

30

3

d s F3

6 1

1 8 ø8 / 2 0 , s at f ë 2

4

1

-

Betonstahllisten Nr.

-|

Zu Plan Nr.

Rami 4-H

Datu Datum: m: 31.0 31.01. 1.20 2016 16 gez.:

KRR

BSt 500 S(B)

gepr.: rev.:

Zusammenfassung der Betonstahlliste Metergewicht

Fix- und Lagerlängen Ø mm

Länge [m]

Bearbeitet Ø mm

Gewicht [kg]

Länge [m]

Gewicht [kg]

6

0.222

6

8

0.395

8

524.29

207.09

10

0.617

10

345.60

213.24

12

0.888

12

58.80

52.21

14

1.210

14

Plan Nr.

STAHL


BSt 500 S(B) Posi Posittion ion

Anza Anzahl hl

Ø mm

abgew. Länge

Totallänge Form (Außenmaße in cm) ohne besondere Angaben werden m die Betonstähle nach SIA-NORM 262 abgebogen

1

51

10

3.60

183.60

2

45

10

3.60

162.00

3

36

25

2.89

104.04

4

36

25

4.35

156.60

5

105

8

2.01

211.05

Plan Nr.

STAHL


BSt 500 S(B) Posi Posittion ion

Anza Anzahl hl

Ø mm

abgew. Länge

Totallänge Form (Außenmaße in cm) ohne besondere Angaben werden m die Betonstähle nach SIA-NORM 262 abgebogen

13

8

16

2.20

17.60

14

6

16

4.99

29.94

15

6

16

10.60

63.60

Elaborati Ne Konstruksionet e Betonit III(Ultimate Limit State & Serviceability Limit State)

Recommend Documents