Teoria e konstruksioneve ii

TEORIA E KONSTRUKSIONEVE II

2009

* [Type the company name]

5/5/2009

TEORIA E KONSTRUKSIONEVE II 2009

VEDAT RAMADANI 2

4.50 m 4.50 m 4.50 m

13.50 m

4.50 m 4.50 m 4.50 m

13.50 m

F=120kn F=120kn

F=120kn F=120kn

X1=1kn X2=1kn

YA=120kn YB=120kn

540knm 540knm

M


VEDAT RAMADANI 3

X2=1kn

9.00 m 4.50 m

13.50 m

X1=1kn

4.5knm

M1

YA=1.0kn YB=1.0kn

4.5knm

3.00 m 3.00 m

6.00 m

X1=1kn

4.5knm

M2


VEDAT RAMADANI 4

EKUACIONI KANONIK X1δ11+δ10=0

δ10=(- *4.52*540)*2-4.52*540=-14580

δ11= 4.53)*2+4.53+2*( 4.52*3+ 4.52*3)=160.875

X2= =90.62kn X2= X1


VEDAT RAMADANI 5

4.50 m 4.50 m 4.50 m

13.50 m

F=120kn F=120kn

X1=90.62kn X2=90.62kn

132.167knm 132.167knm

M

29.37kn

29.37kn

Q

YA=29.37kn YB=29.37kn


VEDAT RAMADANI 6

3.00 m 3.00 m

6.00 m

X1=90.62kn

132.167knm

M

45.31kn

45.31kn

Q

YA=45.31kn YB=45.31kn


VEDAT RAMADANI 7

155.00 mm

23

.00

mm

37

7.0

0 m

m

23

.00

mm

40

0.0

0 m

m

14.00 mm

DIMENSIONIMI I TRAUT 1

Mmax=1.7*132.167=225.539 knm

lej

Wpot= = =0.0014096m3=1409.6cm3

Hopt= = =37.54cm

Pervetesohet h=400mm

Karakteristikat e profilit

H=400mm

B=155mm

D=14mm

T=23mm

Ix =29210cm4

Wx=1460cm3

G=1949.4kg/m

=154478.76≤ lej=160000 kn/m2

fmax= = =0.000973m

flej= = =0.054m

Teoria e Konstruksioneve II Mars2009

VEDAT RAMADANI 1

1.50 m 5.00 m 5.00 m 5.00 m 1.50 m

18.00 m

Nr=P1+P2+P3+P4

Nr=950+1100+1300+1000=4450kn

Nt=0.15*4450+4450=5117.5kn

Mr=P1*7.5+P2*2.5-P3*2.5-P4*7.5=

Mr=950*7.5+1200*2.5-1300*2.5-1000*2.5=625knm

A=B*L=B*18

NR

P1 P2 P3 P4

Mr


VEDAT RAMADANI 2

At=0.30*0.40+0.60*0.8=1.04m2

t=1.04*25*18=450kn/m2

3 *0 .5 1 0* 0.5 3 *0.51 0 *0 .5 1 0* 0.5

K1=0.25*1.30*160000=5200

K2=0.50*1.30*160000=10400


VEDAT RAMADANI 3

A

A

PR

ERJA

A-A

PA

RA

QIT

JA 3

D E

TH

EMEL

IT


VEDAT RAMADANI 4

DIAGRAMET TE PUNUARA ME PROGRAMIN STAAD

30

3.0

83

83

.24

MO

ME

NT

ET E

PER

KU

LJES

TE

TH

EM

ELI

TM

o

DIA

GR

AM

ET

TR

AN

ZVE

RZA

LET

T(Q

) D

IAG

RA

MET

50

7.6

93

02

.47

27

9.8

5

40

7.3

9

39

5.7

5

31

0.4

6

51

6.5

2

55

9.8

9

51

9.7

56

9.6

5

60

5.7

7

53

2.5

33

5.2

7


VEDAT RAMADANI 5

Mumax=1.7*507.69=863.073knm

Mumax=1.7*407.39=692.563knm

a=ao+øu+ =2.0+1.0+ =9.4cm

hstat=h-a=120-9.4=110.6cm

lart

Kh= = =2.90

Per MB 3O Kz=0.956 Ea/Eb=10/1.4‰

Aa= = =20.51cm2

I 2ø14 me Aa=3.08cm2

II 4ø25 me Aa=19.64cm2

Aa=22.72>20.51cm2

Aamin=μmin*Ab=0.25/100*60*110.6=16.59cm2

Posht

Kh= = =3.25


VEDAT RAMADANI 6

Per MB 3O Kz=0.962 Ea/Eb=10/1.2‰

Aa= = =16.27cm2

I 2ø14 me Aa=3.08cm2

II 3ø25 me Aa=14.73cm2

Aa=17.81>16.27cm2

Aa=17.81> Aamin==16.59cm2

Amon=4ø12/30cm

Stafat

eu=2/3 h≤30cm

eu=2/3*120=80>30cm

pervetesohen

st=ø10/20/10cm

Tu=1.7*569.65=968.405kn

τn= = =1.49MPa

Per MB 30 τr=1.1MPa< τn=1.49MPa

Behet caktimi se per cilat nderje behet fjale

τr=1.1MPa< τn=1.49MPa<3 τr=3.3MPa


VEDAT RAMADANI 7

Ka nevoje te behet sigurimi nga nderjet kryesore te pjerreta me forca te

reduktuara

n τau

τru

λ

500cm

λ= *5.0=1.30m=130cm

forcat tranzverzale qe pranon betoni

τru= Tu-Tbu

Tbu=1/2(3 τr- τn)*b0*z=1/2(3*1.1-1.49)*60*0.962*110.6=586.7kn

Nderjet ne rreshqitje te reduktuara jane

τru= = =0.905MPa

min Au=minμu* = =0.6cm2

st=ø10=0.79cm2>0.6cm2

dmth e ploteson kushtin


VEDAT RAMADANI 8

1

32

RØ

14

13

2R

Ø1

4

14

4R

Ø2

5

11 4RØ12

11 4RØ12

13

2R

Ø14

14

3R

Ø2

5

13

2R

Ø1

4

11 4RØ12

11 4RØ12

pR

Ø10

13

2R

Ø1

4

14

3R

Ø2

5

13

2R

Ø1

4

14

4R

Ø2

5

11 4RØ12

11 4RØ12

pR

Ø1

0

13

2R

Ø1

4

13

2R

Ø1

4

14

4R

Ø2

5

11 4RØ12

11 4RØ12

13

2R

Ø1

4

14

3R

Ø2

5

13

2R

Ø1

4

11 4RØ12

11 4RØ12

pR

Ø10

13

2R

Ø1

4

13

2R

Ø1

4

14

4R

Ø25

11 4RØ12

11 4RØ12

Teoria e konstruksioneve II 2009

VEDAT RAMADANI 1

Задача1:да се определат реакции и диаграми на статичките големини Mr,Mt и угиб w за дадената армирано бетонска кружна плоча со примена на таблиците на Marcus.Модул на еластичност : E=31*106 KN/m2, μ=0.2

a=4m

b=2m

q=20kn/m

P=15kn

t=0.16m

За Q=1 N= E*h3

12*(1-μ3 )=

31*106*0.163

12*(1-0.23 )=16533.33

Mr=0

Mt=q*a2

8 (1-μ)=

20*42

8 (1-0.2)=32knm

W’=-q*a3

8*N*(1+μ) =

20*43

8*16533.33*(1+0.2)=-0.00806m

Qr=-q*a

2*Q=

20*4

2*1=-40k

P P

q

2b

2a

q

2a


VEDAT RAMADANI 2

Q=r

a =

2.0

4.0 =0.5 N=

E*h3

12*(1-μ3 )=

31*106*0.163

12*(1-0.23 )=16533.33

φ1=1-Q2=1-0.52=0.75 φ0=1-Q4=1-0.54=0.937

Mr=q*a4

16*(3+μ)* φ1=

20*42

16*(3+0.2)*0.75=48knm

Mt=q*a2

16*[2(1-μ)+(1+3*μ)* φ1=

20*42

16*[2(1-0.2)+(1+3*0.2)*0.75]=56 knm

W”=q*a4

64*N*(1+μ)*[2(3+μ)* φ1-(1+μ)* φ0=

20*44

64*16533.33*(1+0.2)*[2(3+0.2)*0.75-

(1+0.2)*0.937=0.01481m

Q=0

Mr= Mt=q*a2

16*(3+μ)=

20*42

16*(3+0.2)=64knm

W=q*a4

64*N*

(5+μ)

(1+μ) =

20*44

64*16533.33*

(5+0.2)

(1+0.2) =0.0209m

Qr=-q*a

2*Q=

20*4

2*1=-40kn


VEDAT RAMADANI 3

q

2a

+48 48

64

Mt

64 56 56 32 32

Mr

-0.008

0.0148 0.02

0.0148

-0.008 W


VEDAT RAMADANI 4

β=b

a =

2.0

4.0 =0.5

x1=(3+μ)*(1- β2)+2*(1+μ)* β2*ln β =(3+0.2)*(1-0.52)+2*(1+0.2)*0.52*ln0.5 =1.98

x2=(1-μ)*(1- β2)-2*(1+μ)*ln β =(1+0.2)*(1-0.52)-2*(1+0.2)*ln0.5 =2.26

Mr=Mt=P*b

4 *x2=

15*2

4.0 *2.26=16.59knm

Qr=0

W=P*b*a2

8*N*(1+μ)*( x1- x2)+ x2* φ1=

20*2*42

64*16533.33*(1+0.2)*( 1.98- 2.26)+

2.26*0.75=0.000713

Q=1

Mr=0

Mt=P*b

4 *(1-μ)*(1- β2)=

15*2

4 *(1-0.2)*(1- 0.52)=4.5knm

P P

2b

2a


VEDAT RAMADANI 5

+16.59 16.59

Mt

16.59 16.59 4.5 4.5

Mr

0.00056

7

0.00070.00598

0.0007

0.00056

7

W

P P

2b

2a

W’= P*b*a

8*N*(1+μ)*(1- β2)=

15*2*4

8*16533.33*(1+0.2)*(1- 0.52)=0.000567

Q=0

W= P*b*a2

8*N*(1+μ)*x1=

15*2*42

8*16533.33*(1+0.2)*1.98=0.00598m


VEDAT RAMADANI 6

Ly

Задача 2: Да се определат статичките големини за дадената правоаголна армирано бетонска плоча,товарена со рамномерен товар п,со примена на таблиците на Маrcus.Да се определи максимален угиб на плочата.

Е=31*106 kN/m2 : μ=0.2

Lx=5m

Ly=6m

P=12kn/m

T=0.16m

λ=Ly

Lx

= 6

5 =1.2

N= E*h3

12*(1-μ 2 ) =

31*106*0.163

12*(1-0.22) = 11022.22

vx=vy= v

за дадената плоча (вклештена во две страни и две страни слободни потпрена) ги читаме следниве податоци

v=1- 15

32*

λ2

1+λ 4 =1-

15

32 ∗

1.22

1+1.2 4 =0.78

px=p*λ4

1+λ 4 =12*

1.24

1+1.2 4 =8.09kn/m

Lx


VEDAT RAMADANI 7

py= p*1

1+λ 4 =12*

1

1+1.2 4 =3.90kn/m

Мmax во долната зона

Mxmax=9

128 *px*Lx

2*v=9

128 *8.09*5.02*0.78=11.09knm

Mymax=9

128 *py*Ly

2*v=9

128 *3.90*6.02*0.78=7.70knm

Muxmax=11.09*1.8=19.96knm

Muymax=7.70*1.8=13.68knm

Горната зона

Mxmin=- px*Lx2

8 =

8.09*5.02

8 =-25.28knm

Mymin=- px*Lx2

8 =

3,90*6,02

8 =-17.07knm

Muxmin=1.8*(-25.28)=-45.50knm

Muymin=1.8*(-17.07)=--30.72knm

Пресметка на угиб

W=1

N*

px*Lx4

720*(1.064+2.815*v)=

= 1

11022.22*

8.09*5.04

720*(1.064+2.815*0.78)=0.00207m

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Teoria e konstruksioneve ii