Projekt Kursi Metalike

2.DIMENSIONIMI I TRAREVE GJATESORE TE MBULESES DHE I LIDHJEVE TE TYRE

NGARKESAT a) Flete llamarine celiku gn=10 daN /m2 g=g n × γ F =10× 1.1=11 daN /m2 b) Listela druri (5x5)cm/25cm (per 1ml listele) (0,05 ×0,05) ×1 ×500 gnlistela = =5 daN /m2 1 ×0,25 glistela =gnlistela × γ F =5× 1,1=5,5 daN /m2 c) Binare druri (6x13)cm/120cm gnbinare=

(0,06× 0,13)× 1× 500 =3,25 daN /m 2 1× 1,2

gbinare=g nbinare × γ F =3,25× 1,1=3,575 daN /m2 -

Ngarkesa e mbuleses n n n n 2 gmbulese =g flete + glistela + gbinare=18,25 daN /m gmbulese =g flete + glistela + gbinare=20 daN /m2

-

Ngarkesa nga pesha vetjake e profilit Shvele gnprofili=14,2 daN /ml ¿

n

g profili=g profili ×γ F =14,2 ×1,1=15,62 daN /ml g profili=g¿profili ×k =15,62× 1,1=17,2 daN /ml -

Ngarkesa nga debora α ≤ 25 ° → c=1 pdebora =75 daN /m2 n

pdebora =p debora ×c=75 × 0,86=64,5 daN /m

2

n

¿

pdebora =p debora ×n p =64,5 ×1,4=90,3 daN /m

2

LLOGJARITJA E TRAUT TE MBULESES -

Ngarkesa totale mbi traun e mbuleses gn 18,25 q n=gnprofili + mbulese + p nmbulese d=15,62+ + 64,5 ×3,5=307,1 daN /m cosα cos 15 °

(

)

(

)

( gcos α + p ) d=17,2+( cos2015 ° + 90,3)× 3,5=405.3 daN /m

q=g profili +

mbulese

mbulese

q x =q cosα=394,1 daN / m q y =q sinα=94,6 daN /m q nx =q n cosα=198,6 daN /m q nx =q n sinα=71,7 daN /m -

Percaktimi i momenteve perkules q x l 2x 394,1 ×62 M x= = =1773,5 daNm 8 8 2

q l 94,6× 32 M y= y y = =106,5 daNm 8 8

DIMENSIONIMI 1. Kushti ne soliditet σ =σ x + σ y ≤ R ↔ σ= nevojshem

Wx

=

Mx M y + ≤R Wx W y

Mx M y 1773,5 106,5 1+ k = 1+7 =1.2 R Mx 2100 1773,5

(

)

(

)

2. Kushti ne deformacion n 4 5 q x lx f x= ≤[f ] 384 E I x

[ f /l ] =

1 600 ; l=600; → [ f ]= =3 cm 200 200 n 4

I

4 5 q x lx 5 198,6∗600 = = =531,96 384 E [ f ] 384 2,1∗106∗3

nevojshem x

3. Dimensionimi Nga Tabela 22SH me kushtin qe : Zgjedh profilin :

{

W xprofili ≥ W nevojshem x profili nevojshem Ix ≥ Ix

 C14a

4. Kontrolli σ=

M x M y 1773.5 106.5 + = + =30.8 ≤ R=2100 daN /m 2 Wx Wy 77.8 13.3

f x=

n 4 5 q x lx =2.92 cm≤ [ f ] =3 cm 384 E I x

LLOGARITJA E TIRANTEVE -

σ=

Dimensionimi i tirantit te drejte N 3tirantit =R y + R y +0.5 R y =2.5 R y =2.5∗( 1.25 q y l y ) =887 daN N trdr N 887 ≤mR ↔ F trdr = trdr = =0.53 cm2 F trdr mR 0.8∗2100

Ftrdr =

-

√

π d 2trdr 4 F trdr 4∗0.53 ↔ d trdr= = =0.82 cm 4 π 3.14

√

Dimensionimi i tirantit te pjerret

∑ F x =0↔ N trdr + R y =2 N trpj cosβ N trpj=

N trdr + R y = 2 cosβ

Ftrpj =

N trpj 1241.8 = =0.74 cm 2 mR 0.8∗2100

d trpj=

√

Pranoj

3.5 R y B/2 2 cos arctg d / cosα

(

)

=¿

√

4 F trdr 4∗0.74 = =0.97 cm π 3.14 d trpj=10 mm dhe d trdr=9 mm

3.DIMENSIONIMI I TRAREVE GJATESOR TE TAVANIT -

PERCAKTIMI I NGARKESAVE 1. a) Trare terthore te rrjetit te trareve gntr =12 daN /m2 n

gtr =gtr ∗n g=12∗1.1=13.2 daN /m

2

b) Termoizoluesit gntermo =24 daN /m2 gtermo =gntermo∗n g=24∗1.2=28.8 daN /m2 c) Rrjete hekuri te rrumbullaket n 2 gheko =15 daN /m gheko =gnheko∗n g=15∗1.1=16.5 daN / m2 d) Suvatim me rrjete teli gnsuva=145 daN /m2 gsuva=g nsuva∗n g=145∗1.2=174 daN / m2

2. Ngarkesa e tavanit

n

n

n

n

n

gtavanit =g tr + g termo + g heko + gsuva =12+15+ 24+145=196 daN /m

2

gtavani=g tr + g termo + gheko +g suva=13.2+16.5+28.8+174=232.5 daN /m2

3. Pesha vetjake Pranojme profilin paraprak N22 (Tab 22SH) daN gnpvN 22=24 m n

n

g pv =g pvN 22∗n=24∗1.1=26.4 daN /m g pv =gnpv∗n g=26.4∗1.1=29 daN /m 4. Mbingarkesa ne tavan pntavani=75 daN /m2 n ptavani= ptavani ∗n p =75∗1.4=105 daN /m2

Ngarkesa e plote ne traun e tavanit 1 196∗1 q n=gnpv + gntavani + p ntavani d=26.4 + +75 3.5=995 daN /m cosβ cos 15

(

)

(

q=g pv + gtavani

(

)

1 232.5 +p d=29+ +101 3.5=1234 daN /m cosβ tavani cos 15

)

(

**Percaktimi i momenteve perkules 2

M=

2

q l 1234 × 6 = =5553 daNm 8 8

DIMENSIONIMI 1. Kushti ne soliditet M σ ≤R↔σ= x ≤R Wx

)

W nevojshem = x

M x 5553 = =2.645 R 2100

2. Kushti ne deformacion n 4 5 q❑ l ❑ f x= ≤[f ] 384 E I x

[ f /l ] =

1 600 ; l=600; → [ f ]= =3 cm 200 200

I nevojshem = x

n 4 5 q❑ l 5 995∗6004 = =2665.17 384 E [ f ] 384 2,1∗106∗3

3. Dimensionimi

{

profili

≥Wx

profili

≥ Ix

Wx

Nga Tabela 22SH me kushtin qe :

Ix

nevojshem

nevojshem

Zgjedh profilin :  C23 4. Kontrolli M 5553 σ = x= =22.95 ≤ R=2100 daN /m2 W x 242 f x=

n 4 5 q x lx =2.76 cm≤ [ f ] =3 cm 384 E I x

4.PERCAKTIMI I NGARKESAVE NE NYJET E KAPRIATES - Ng. Mbulese n 2 gmbul =18,25 daN /m gmbul =20 daN /m 2 -Ng. Tra Mbulese gnc 14=15,62 daN / m gc 14=17,2 daN /m

-Mbingarkesa ne mbulese n 2 pmbul =64,5 daN /m pmbul =90,3 daN /m2 -Ng. Tavani n 2 gTav =196 daN /m gtav=232,5 daN /m2 -Ng. Tra Tavani

n

gc 23=26,4 daN /m gc 23=29 daN / m

-Mbingarkese Tavani n 2 pt av =75 daN /m ptav=105 daN /m2

-

Ngarkesa e ndermjetme n n n n g g g g q ndermjetme= mbul + c14 + Tav + c 23 + pnmbul + p ntav cos 15 d cos 15 d cos 15 cos 15 ¿

-

18.25 15.62 196 26.4 2 + + + + 64.4+75=372daN /m cos 15 3.5 cos 15 3.5 cos 15 cos 15

Pesha vetjake e vete kapriates

(

gnpv =1.5 1.8+

q nderm B 372∗6 daN L=1.5 1.8+ 28=169.4 1000 1000 m

)

(

)

g pv =gnpv∗1.1=169.4∗1.1=186.5

daN m

A. Ngarkesat e perhershme a) Ng. Ne Brezin e Siperm g d Gsiper =gmbulese B + gC 14 a B+ pv d cos 15 2 Gsiper =20∗6∗3.6+17.2∗6+93.25∗3.5=861.5 daN b) Ng. Ne Brezin e Poshtem g d G poshte=g tavani B + g C 23 B+ pv d =5522daN cos 15 2 B. Ngarkesat e perkoheshme a) Ng. Ne Brezin e Siperm Psiper = pmbulese Bd=90.3∗6∗3.5=1896.3daN b) Ng. Ne Brezin e Poshtem P poshte= p tavani Bd=105∗6∗3.5=2205 daN

5.SKEMAT E NGARKIMIT

Gsiper Gsiper

Gsiper Gsiper

Gsiper

Gsiper

Gsiper

Gsiper/2

3.50

Gsiper/2

15° Gposhte

Gposhte

Gposhte

Gposhte

28.00

Gposhte/2

Gposhte

Gposhte Gposhte Gposhte/2

Skema 1

P siper P siper

P siper P siper

P siper

P siper

P siper

Psiper /2

3.50

15°

28.00

Skema 2

Psiper /2

3.50

15° Ptavani

Ptavani/2

P tavani

Ptavani

Ptavani

28.00

Ptavani

Ptavani Ptavani Ptavani/2

Skema 3

7. PERCAKTIMI I LIDHJEVE NE BREZIN E SIPERM DHE BREZIN E POSHTEM

TE KAPRIATES

-

3.50

15°

3.50

6.00

6.00

6.00

6.00

Lidhjet ne BS

3.50

3.50

3.50

3.50

3.50

3.50

3.50

Planimetria e lidhjeve ne BS

-

Lidhjet ne BP

3.50

15°

3.50

6.00

6.00

6.00

6.00

3.50

3.50

3.50

3.50

3

Planimetria e lidhjeve ne BP

Percaktimi i trashesise δ [mm] te pllakes se nyjes N max =603 kN ↔ δ=12mm

8. DIMENSIONIMI I SHUFRAVE TE KAPRIATES

-

BREZI I SIPERM a) Kushti ne Soliditet N N 60220 nevojshme 2 ( F dobs=0 ) ↔ σ = F ≤ R↔ F neto = R = 2100 =28.7 cm neto

-

b) Kushti ne Qendrueshmeri Pranojme : λ p =λ x =λ y =(80 ÷ 100)

λ x=

l ox l ox 362 l ox 362 ≤ [ λ ] ↔ r nev = =4.02↔ r min = =3.02 x = x = rx λ x 90 [ λ ] 120

λ y=

l oy l oy 362 l oy 362 ≤ [ λ ] ↔ r nev = =4.02 ↔ r min = =3.02 y = y = ry λ y 90 [ λ ] 120 λmax =max ( λx , λ y ) =max ( λ p ) =90

-

φmin =f ( λmax )=0.69 σ=

N φmin F

≤ R ↔ F nevojshme = ❑

N φ min R

=

60220 =41.56 cm2 0.69∗2100

c) Kontrolli 1. Seksion barabrinjes Perzgjedh Profilin : 2L 130x10x13 i. Soliditet N 60220 σ= = =1320.62≤ R=2100 F neto 2∗25.3 ii.

Qendrueshmeri

λ x=

l ox 362 = =90.3 ≤ [ λ ] =120 r x 4.01

λ y=

l oy 362 = =62.5≤ [ λ ] =120 r y 5.79 λmax =max ( λx , λ y ) =max ( λ p ) =90.3 φmin =f ( λmax )=0.69

σ=

N φmin F

=

60220 =1724.8 ≤ R=2100 0.69∗50.6

2. Seksion jobarabrinjes Perzgjedh Profilin : 2L 130x90x10 i. Soliditet N 60220 σ= = =1413.6 ≤ R=2100 F neto 42.6 ii.

Qendrueshmeri

λ x=

l ox 362 = =87.87 ≤ [ λ ] =120 r x 4.12

λ y=

l oy 362 = =93.3 ≤ [ λ ] =120 r y 3.88 λmax =max ( λx , λ y ) =max ( λ p ) =93.3 φmin =f ( λmax )=0.68

σ=

N φmin F

=

60220 =2078.8≤ R=2100 0.68∗42.6

BREZI I POSHTEM

a) Kushti ne Soliditet N

( F dobs=0 ) ↔ σ = F

nevojshme

≤ R↔ F neto

neto

=

N 56580 2 = =26.95 cm R 2100

b) Kushti ne Qendrueshmeri -

Pranojme :

λ p =λ x =λ y =(80 ÷ 100)

λ x=

l ox l ox 362 l ox 362 ≤ [ λ ] ↔ r nev = =4.02↔ r min = =3.02 x = x = rx λ x 90 [ λ ] 120

λ y=

l oy l oy 362 l oy 362 ≤ [ λ ] ↔ r nev = =4.02 ↔ r min = =3.02 y = y = ry λ y 90 [ λ ] 120

-

λmax =max ( λx , λ y ) =max ( λ p ) =90 φmin =f ( λmax )=0.69

σ=

N N 56580 ≤ R ↔ F nevojshme = = =39.05 cm2 ❑ φmin F φ min R 0.69∗2100 c) Kontrolli

1. Seksion barabrinjes Perzgjedh Profilin : 2L 130x10x13 i. Soliditet N 56580 σ= = =1120 ≤ R=2100 F neto 2∗25.3 ii.

Qendrueshmeri

λ x=

l ox 362 = =90.3 ≤ [ λ ] =120 r x 4.01

λ y=

l oy 362 = =62.5≤ [ λ ] =120 r y 5.79 λmax =max ( λx , λ y ) =max ( λ p ) =90.3 φmin =f ( λmax )=0.69

σ=

N φmin F

=

56580 =1620 ≤ R=2100 0.69∗50.6

2. Seksion jobarabrinjes Perzgjedh Profilin : 2L 130x90x10 i. Soliditet N 56580 σ= = =1329 ≤ R=2100 F neto 42.6 ii.

Qendrueshmeri l 362 λ x = ox = =87.87 ≤ [ λ ] =120 r x 4.12 λ y=

l oy 362 = =93.3 ≤ [ λ ] =120 r y 3.88 λmax =max ( λx , λ y ) =max ( λ p ) =93.3 φmin =f ( λmax )=0.68

σ=

N φmin F

=

56580 =1954 ≤ R=2100 0.68∗42.6

SHUFRAT E RRJETIT Profili me i vogel qe mund te perdoret eshte 2L 50x5 me te dhena F1 L=9.6 cm2 r x =1.53 cm r y ( δ=12mm )=2.53 cm a. Element ne Shtypje GRUPI I

1) Kushti ne Soliditet N σ= ≤ R ↔ N max soliditet =F∗R=9.6∗2100=20160 daN F neto 2) Kushti ne Qendrueshmeri λ x=

l ox ≤ [ λ ] ↔ l max ox =[ λ ]∗r x =1.53∗150=229.5 cm rx

λ y=

l oy ≤ [ λ ] ↔ l max oy =[ λ ]∗r y =2.53∗150=379.5 cm ry λmax =max ( λx , λ y ) =[ λ ] =150

-

φmin =f ( λmax )=0.32

σ=

N φmin F

max

≤ R ↔ N qendr =φ min F2 L R=0.32∗9.6∗2100=6451daN

3) Dimensionimi

Marr:

max 2 N max =min ( N max qendr , N soliditet )=6451 daN / cm max

l ox =[ λ ]∗r x =229.5 cm l max oy =[ λ ]∗r y =379.5 cm

Te tille Dimensionohen Elementet 24,30

GRUPI II (Shufrat ne mbeshtetje) Elementet 17 dhe 37

a) Kushti ne Soliditet N N 36520 nevojshme 2 ( F dobs=0 ) ↔ σ = F ≤ R↔ F neto = R = 2100 =17.4 cm neto b) Kushti ne Qendrueshmeri Pranojme : λ p =λ x =λ y =(80 ÷ 100)

λ x=

l ox l ox 280 l ox 280 ≤ [ λ ] ↔ r nev = =3.12 ↔ r min = =2.33 x = x = rx λ x 90 [ λ ] 120

λ y=

l oy l oy 350 l oy 350 ≤ [ λ ] ↔ r nev = =3.89 ↔ r min = =2.92 y = y = ry λ y 90 [ λ ] 120 λmax =max ( λx , λ y ) =max ( λ p ) =90

-

φmin =f ( λmax )=0.69 σ=

N φmin F

≤ R ↔ F nevojshme = ❑

N φ min R

c) Kontrolli Seksion barabrinjes

=

36520 =25.21 cm2 0.69∗2100

Perzgjedh Profilin : 2L 75x10x9

σ=

-

Soliditet

-

Qendrueshmeri

N 36520 = =1320.62≤ R=2100 F neto 2∗14.1

λ x=

l ox 280 = =90.3 ≤ [ λ ] =120 r x 2.26

λ y=

l oy 350 = =62.5≤ [ λ ] =120 r y 5.79 λmax =max ( λx , λ y ) =max ( λ p ) =90.3 φmin =f ( λmax )=0.69

σ=

N φmin F

=

36520 =1877 ≤ R=2100 0.69∗28.2

GRUPI III Elementet : 19, 21, 23 , 32, 33, 35

a) Kushti ne Soliditet N N 27710 nevojshme 2 ( F dobs=0 ) ↔ σ = F ≤ R↔ F neto = R = 2100 =13.2 cm neto

-

b) Kushti ne Qendrueshmeri Pranojme : λ p =λ x =λ y =(80 ÷ 100)

λ x=

l ox l ox 288 l ox 288 ≤ [ λ ] ↔ r nev = =3.2 ↔ r min = =1.92 x = x = rx λ x 90 [ λ ] 150

λ y=

l oy l oy 360 l oy 360 ≤ [ λ ] ↔ r nev = =4 ↔r min = =2.4 y = y = ry λ y 90 [ λ ] 150

-

λmax =max ( λx , λ y ) =max ( λ p ) =90

φmin =f ( λmax )=0.69 σ=

N φmin F

≤ R ↔ F nevojshme = ❑

N φ min R

=

27710 =19.13 cm2 0.69∗2100

c) Kontrolli Seksion barabrinjes Perzgjedh Profilin : 2L 75x8x9 - Soliditet N 27710 σ= = =1205 ≤ R=2100 F neto 2∗11.5 - Qendrueshmeri l ox 288 λ x= = =126 ≤ [ λ ] =150 r x 2.28 λ y=

l oy 360 = =101≤ [ λ ] =150 r y 3.57 λmax =max ( λx , λ y ) =max ( λ p ) =126 φmin =f ( λmax )=0.43

σ=

N 27710 = =1747 ≤ R=2100 φmin F 0.69∗23

a) Element ne terheqje GRUPI I 1) Kushti ne Soliditet N σ= ≤ R ↔ N max soliditet =F∗R=9.6∗2100=20160 daN F neto 2) Kushti ne Qendrueshmeri l [¿ ¿ ox]=r x λ x =1.53∗400=612 cm l λ x = ox ≤ [ λ ] ↔¿ rx

λ y=

l oy ≤ [ λ ] ↔ [ l oy ]=r y λ y =2.53∗400=1012 cm ry 3) Dimensionimi max max Marr: N =N soliditet =20160 daN l max ox = [ λ ]∗r x =612 cm max

l oy =[ λ ]∗r y =1012cm Te tille Dimensionohen Elementet 25,29 GRUPI II Elementet : 18,20,27,34,36

a) Kushti ne Soliditet N N 40280 nevojshme 2 ( F dobs=0 ) ↔ σ = F ≤ R↔ F neto = R = 2100 =19.18 cm neto -

b) Kushti ne Qendrueshmeri Pranojme : λ❑=400

λ x=

l ox l ox 430 ≤ [ λ ] ↔ r min = =1.1 x = rx [ λ ] 400

λ y=

l oy l oy 538 ≤ [ λ ] ↔ r min = =1.35 y = ry [ λ ] 400

c) Kontrolli Seksion barabrinjes Perzgjedh Profilin : 2L 75x8x9 - Soliditet N 40280 σ= = =1751 ≤ R=2100 F neto 2∗11.5 -

Qendrueshmeri

λ x=

l ox 430 = =189 ≤ [ λ ]=400 r x 2.28

λ y=

l oy 538 = =151≤ [ λ ] =400 r y 3.57

GRUPI III

Elementet : 22,26,28,31

a) Kushti ne Soliditet N N 5260 nevojshme 2 ( F dobs=0 ) ↔ σ = F ≤ R↔ F neto = R = 2100 =2.51 cm neto -

b) Kushti ne Qendrueshmeri Pranojme : λ❑=400

λ x=

l ox l ox 430 ≤ [ λ ] ↔ r min = =1.1 x = rx [ λ ] 400

λ y=

l oy 1+ l oy2 l oy1 +l oy 2 882 ≤ [ λ ] ↔ r min = =2.21 y = ry 400 [λ]

c) Kontrolli Seksion barabrinjes Perzgjedh Profilin : 2L 50x5 - Soliditet N 5260 σ= = =548 ≤ R=2100 F neto 2∗4.8 -

Qendrueshmeri

λ x=

l ox 430 = =281≤ [ λ ] =400 r x 1.53

λ y=

l oy 882 = =349≤ [ λ ] =400 r y 2.53

9. DIMENSIONIMI I NYJES SE KAPRIATES Nyja 2 Elementi 1 Elementi 2

N 1=−35820 daN N 2=−50490 daN

Elementi 19

N 19=−27710 daN

Elementi 20

N 20=29820 daN

Elementi 19 '

T =0.72 N 19=19952 daN T ' ' =0.28 N 19 =7758 daN δ 2=12 mm; δ 1=10 mm ↔ δ min =10 mm h't =1.2∗δ min =12 mm

h't ' =δ 1−s=10−2=8 mm

'

T 19952 l= +1 cm= +1 cm=8.92 cm 9 cm ' t 2520 2(0.7 ht ) Rt ' t

''

''

lt =

T 7758 ' 'min +1 cm= + 1cm=5.62cm ↔l t =6 cm ' t 1680 2(0.7 h t ') Rt

Elementi 20 '

T =0.72 N 20=21470 daN T ' ' =0.28 N 20 =8350 daN δ 2=12 mm; δ 1=10 mm ↔ δ min =10 mm h't =1.2∗δ min =12 mm l 't= ''

h't ' =δ 1−s=10−2=8 mm

'

T 21470 +1 cm= + 1cm=9.52 cm ' t 2520 2(0.7 ht ) Rt

lt =

''

T 8350 ' ' min +1 cm= +1 cm=5.97 ↔l t =6 cm ' t 1680 2(0.7 h t ' ) Rt

Saldimet ne Brezin e Siperm

N H =N 2−N 1 +Qsin 15=50490−35820+1799 sin 15=15090 daN Q=G mbul +0.9 P mbul =( 629+0.9∗1300 )=1799 daN N v =Qcos 15=1799cos 16=1750 daN '

T =0.72 N H =10865 daN V ' =V ' ' =

T ' '=0.28 N H =4225 daN

;

N V 1750 = =875 daN 2 2

2

2

2

2

R' =√ T ' 2 +V ' 2= √108652 +8752=10900 daN R' ' =√ T ' ' 2+ V ' ' 2=√ 42252 +8752 =4315 daN h't =1.2∗δ min =12 mm ; h't ' =δ 1 −s=10−2=8 mm l 't=

'

R 10900 +1 cm= + 1cm=5.4 cm↔ l 'min =6 cm t ' t 2520 2(0.7 ht ) Rt

l 't' =

R'' 4315 +1 cm= +1 cm=3.6 cm ↔l 't ' min =6 cm ' t 1680 2(0.7 h t ' ) Rt