Teoria e konstruksioneve ii
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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
TEORIA E KONSTRUKSIONEVE II 2009
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
TEORIA E KONSTRUKSIONEVE II 2009
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
TEORIA E KONSTRUKSIONEVE II 2009
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
TEORIA E KONSTRUKSIONEVE II 2009
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
TEORIA E KONSTRUKSIONEVE II 2009
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
Teoria e Konstruksioneve II Mars2009
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
Teoria e Konstruksioneve II Mars2009
VEDAT RAMADANI 3
A
A
PR
ERJA
A-A
PA
RA
QIT
JA 3
D E
TH
EMEL
IT
Teoria e Konstruksioneve II Mars2009
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
Teoria e Konstruksioneve II Mars2009
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
Teoria e Konstruksioneve II Mars2009
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
Teoria e Konstruksioneve II Mars2009
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
Teoria e Konstruksioneve II Mars2009
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
Teoria e konstruksioneve II 2009
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
Teoria e konstruksioneve II 2009
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
Teoria e konstruksioneve II 2009
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
Teoria e konstruksioneve II 2009
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
Teoria e konstruksioneve II 2009
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
Teoria e konstruksioneve II 2009
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