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Teoria e konstruksioneve ii

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Teoria e konstruksioneve ii

  1. 1. TEORIA E KONSTRUKSIONEVE II 2009 * [Type the company name] 5/5/2009
  2. 2. TEORIA E KONSTRUKSIONEVE II 2009 VEDAT RAMADANI 2 4.50m 4.50m 4.50m 13.50m 4.50m 4.50m 4.50m 13.50m F=120kn F=120kn F=120kn F=120kn X1=1kn X2=1kn YA=120kn YB=120kn 540knm 540knm M
  3. 3. 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
  4. 4. 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
  5. 5. TEORIA E KONSTRUKSIONEVE II 2009 VEDAT RAMADANI 5 4.50m 4.50m 4.50m 13.50m 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
  6. 6. TEORIA E KONSTRUKSIONEVE II 2009 VEDAT RAMADANI 6 3.00m 3.00m 6.00m X1=90.62kn 132.167knm M 45.31kn 45.31kn Q YA=45.31kn YB=45.31kn
  7. 7. TEORIA E KONSTRUKSIONEVE II 2009 VEDAT RAMADANI 7 155.00mm 23.00mm 377.00mm 23.00mm 400.00mm 14.00mm 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
  8. 8. Teoria e Konstruksioneve II Mars2009 VEDAT RAMADANI 1 1.50m 5.00m 5.00m 5.00m 1.50m 18.00m 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
  9. 9. 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 10*0.5 3*0.510*0.5 10*0.5 K1=0.25*1.30*160000=5200 K2=0.50*1.30*160000=10400
  10. 10. Teoria e Konstruksioneve II Mars2009 VEDAT RAMADANI 3 A A PRERJAA-A PARAQITJA3DETHEMELIT
  11. 11. Teoria e Konstruksioneve II Mars2009 VEDAT RAMADANI 4 DIAGRAMET TE PUNUARA ME PROGRAMIN STAAD 303.08 383.24 MOMENTETEPERKULJESTETHEMELIT Mo DIAGRAMET TRANZVERZALET T(Q) DIAGRAMET 507.69 302.47279.85 407.39 395.75 310.46 516.52 559.89 519.7 569.65 605.77 532.5 335.27
  12. 12. 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= min*Ab=0.25/100*60*110.6=16.59cm2 Posht Kh= = =3.25
  13. 13. 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
  14. 14. 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=min u* = =0.6cm2 st=ø10=0.79cm2 >0.6cm2 dmth e ploteson kushtin
  15. 15. Teoria e Konstruksioneve II Mars2009 VEDAT RAMADANI 8 132RØ14 132RØ14 144RØ25 11 4RØ12 11 4RØ12 132RØ14 143RØ25 132RØ14 11 4RØ12 11 4RØ12 pRØ10 132RØ14 143RØ25 132RØ14 144RØ25 11 4RØ12 11 4RØ12 pRØ10 132RØ14 132RØ14 144RØ25 11 4RØ12 11 4RØ12 132RØ14 143RØ25 132RØ14 11 4RØ12 11 4RØ12 pRØ10 132RØ14 132RØ14 144RØ25 11 4RØ12 11 4RØ12
  16. 16. Teoria e konstruksioneve II 2009 VEDAT RAMADANI 1 д ч 1:д се определ т ре кции и ди гр ми н ст тичките големини Mr,Mt и угиб w з д ден т рмир но бетонск кружн плоч со примен н т блиците н Marcus. одул н ел стичност : E=31*106 KN/m2 , =0.2 a=4m b=2m q=20kn/m P=15kn t=0.16m Q=1 N= E*h3 12*(1- 3 ) = 31*106*0.163 12*(1-0.23 ) =16533.33 Mr=0 Mt= q*a2 8 (1- )= 20*42 8 (1-0.2)=32knm W’=- q*a3 8*N*(1+ ) = 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
  17. 17. Teoria e konstruksioneve II 2009 VEDAT RAMADANI 2 Q= r a = 2.0 4.0 =0.5 N= E*h3 12*(1- 3 ) = 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+ )* φ1= 20*42 16 *(3+0.2)*0.75=48knm Mt= q*a2 16 *[2(1- )+(1+3* )* φ1= 20*42 16 *[2(1-0.2)+(1+3*0.2)*0.75]=56 knm W”= q*a4 64*N*(1+ ) *[2(3+ )* φ1-(1+ )* 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+ )= 20*42 16 *(3+0.2)=64knm W= q*a4 64*N * (5+ ) (1+ ) = 20*44 64*16533.33 * (5+0.2) (1+0.2) =0.0209m Qr=- q*a 2 *Q= 20*4 2 *1=-40kn
  18. 18. Teoria e konstruksioneve II 2009 VEDAT RAMADANI 3 q 2a + 48 48 64 Mt 64 5656 3232 Mr -0.008 0.0148 0.02 0.0148 -0.008W
  19. 19. Teoria e konstruksioneve II 2009 VEDAT RAMADANI 4 β= b a = 2.0 4.0 =0.5 x1=(3+ )*(1- β2 )+2*(1+ )* β2 *ln β =(3+0.2)*(1-0.52 )+2*(1+0.2)*0.52 *ln0.5 =1.98 x2=(1- )*(1- β2 )-2*(1+ )*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+ ) *( 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- )*(1- β2 )= 15*2 4 *(1-0.2)*(1- 0.52 )=4.5knm P P 2b 2a
  20. 20. Teoria e konstruksioneve II 2009 VEDAT RAMADANI 5 + 16.59 16.59 Mt 16.5916.59 4.54.5 Mr 0.00056 7 0.0007 0.00598 0.0007 0.00056 7 W P P 2b 2a W’= P*b*a 8*N*(1+ ) *(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+ ) *x1= 15*2*42 8*16533.33*(1+0.2) *1.98=0.00598m
  21. 21. Teoria e konstruksioneve II 2009 VEDAT RAMADANI 6 Ly д ч 2: се определ т ст тичките големини з д ден т пр во голн рмир но бетонск плоч ,тов рен со р мномерен тов р п,со примен н т блиците н rcus. се определи м ксим лен угиб н плоч т . =31*106 kN/m2 : =0.2 Lx=5m Ly=6m P=12kn/m T=0.16m λ= Ly Lx = 6 5 =1.2 N= E*h3 12*(1- 2 ) = 31*106 *0.163 12*(1-0.22 ) = 11022.22 vx=vy= v з д ден т плоч (вклештен во две стр ни и две стр ни слободни потпрен ) ги чит ме следниве под тоци v=1- 15 32 * λ2 1+λ4 =1- 15 32 ∗ 1.22 1+1.24 =0.78 px=p* λ4 1+λ4 =12* 1.24 1+1.24 =8.09kn/m Lx
  22. 22. Teoria e konstruksioneve II 2009 VEDAT RAMADANI 7 py= p* 1 1+λ4 =12* 1 1+1.24 =3.90kn/m max во долн т зон 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 орн т зон 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 ресметк н 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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