ĆWICZENIE PROJEKTOWE NUMER 1. |
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1.1 Wyznaczyć parametry (wartość średnia, odchylenie standardowe, współczynnik zmienności losowych cech geometrycznych przekroju płaskownika stalowego (szerokość b(ω) i grubość t(ω) na podstawie pomiarów, tablica 1 ((bi) t(i)) |
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Tablica 1 |
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b(ω) |
t(ω) |
bi-bśr |
(bi-bśr)2 |
ti-tśr |
(ti-tśr)2 |
(bi-bśr)(ti-tśr) |
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91,1936 |
15,6348 |
3,41677000000001 |
11,6743172329001 |
-0,358 |
0,1280 |
-1,22252 |
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86,6876 |
15,5702 |
-1,08922999999999 |
1,18642199289997 |
-0,422 |
0,1784 |
0,46009 |
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86,2879 |
15,9569 |
-1,48893 |
2,21691254489999 |
-0,036 |
0,0013 |
0,05315 |
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93,2523 |
16,2801 |
5,47547000000002 |
29,9807717209002 |
0,288 |
0,0827 |
1,57420 |
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82,9681 |
16,7526 |
-4,80872999999998 |
23,1238842128998 |
0,760 |
0,5776 |
-3,65463 |
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88,2194 |
16,2081 |
0,442570000000003 |
0,195868204900003 |
0,216 |
0,0464 |
0,09537 |
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90,2794 |
16,0471 |
2,50257000000001 |
6,26285660490003 |
0,054 |
0,0030 |
0,13639 |
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82,3006 |
15,6673 |
-5,47622999999999 |
29,9890950128999 |
-0,325 |
0,1058 |
1,78142 |
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85,3053 |
15,751 |
-2,47152999999999 |
6,10846054089994 |
-0,242 |
0,0584 |
0,59712 |
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91,2741 |
16,0579 |
3,49727000000001 |
12,2308974529001 |
0,065 |
0,0043 |
0,22837 |
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suma: |
877,7683 |
159,926 |
0,00 |
122,97 |
0,00 |
1,19 |
0,04896 |
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n- ilość pomiarów |
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n= |
10 |
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Obliczenia przeprowadzone dla wyników pomierzonych szerokości: |
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a) wartość średnia pomierzonych szerokości b(ω): |
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b,śr= |
87,77683 |
[mm] |
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b) odchylenie standardowe sx: |
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sb= |
3,70 |
[mm] |
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c) współczynnik zmienności ν: |
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νb= |
0,0421 |
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νb[%]= |
4,21 |
[%] |
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Obliczenia przeprowadzone dla wyników pomierzonych grubości: |
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d) wartość średnia pomierzonych grubości t(ω): |
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t,śr= |
15,99 |
[mm] |
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b) odchylenie standardowe sx: |
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st= |
0,36 |
[mm] |
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c) współczynnik zmienności ν: |
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νt= |
0,0227 |
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νt[%]= |
2,27 |
[%] |
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1.2 Wyznaczyć parametry losowej wytrzymałości stali w(ω) na podstawie pomiarów w próbie rozciągania- tablica 2 |
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Tablica 2 |
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w(i) [Mpa] |
w(i)-wśr |
(wi-wśr)2 |
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437,467 |
3,449 |
11,896 |
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457,208 |
23,190 |
537,776 |
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408,364 |
-25,654 |
658,128 |
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457,969 |
23,951 |
573,650 |
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450,898 |
16,880 |
284,934 |
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456,319 |
22,301 |
497,335 |
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456,708 |
22,690 |
514,836 |
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437,832 |
3,814 |
14,547 |
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426,015 |
-8,003 |
64,048 |
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400,749 |
-33,269 |
1106,826 |
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450,512 |
16,494 |
272,052 |
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460,477 |
26,459 |
700,079 |
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396,801 |
-37,217 |
1385,105 |
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411,492 |
-22,526 |
507,421 |
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401,459 |
-32,559 |
1060,088 |
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suma: |
6510,270 |
0,000 |
8188,721 |
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n- ilość pomiarów |
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n= |
15 |
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Obliczenia przeprowadzone dla wyników pomierzonych wytrzymałości: |
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a) wartość średnia : |
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w,śr= |
434,02 |
[MPa] |
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b) odchylenie standardowe sx: |
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sw= |
24,18 |
[MPa] |
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c) współczynnik zmienności ν: |
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νw= |
0,0557 |
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νw[%]= |
5,57 |
[%] |
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1.2 Wyznaczyć parametry losowej siły rociągającej P(ω) na podstawie pomiarów w próbie rozciągania- tablica 3 |
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Tablica 3 |
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P(i) [kN] |
P(i)-Pśr |
(Pi-Pśr)2 |
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451,639 |
3,81 |
14,53 |
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483,391 |
35,56 |
1264,74 |
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412,785 |
-35,04 |
1228,00 |
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466,441 |
18,61 |
346,45 |
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366,102 |
-81,73 |
6679,11 |
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451,689 |
3,86 |
14,91 |
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487,641 |
39,81 |
1585,09 |
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457,485 |
9,66 |
93,26 |
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445,617 |
-2,21 |
4,89 |
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477,879 |
30,05 |
903,07 |
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442,275 |
-5,55 |
30,83 |
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405,944 |
-41,88 |
1754,25 |
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470,706 |
22,88 |
523,41 |
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492,504 |
44,68 |
1995,96 |
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405,319 |
-42,51 |
1807,00 |
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suma: |
6717,417 |
0,000 |
18245,505 |
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n- ilość pomiarów |
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n= |
15 |
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Obliczenia przeprowadzone dla wyników pomierzonych sił rozciągających: |
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a) wartość średnia : |
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P,śr= |
447,83 |
[kN] |
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b) odchylenie standardowe sx: |
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sP= |
36,10 |
[kN] |
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c) współczynnik zmienności ν: |
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νP= |
0,0806 |
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νP[%]= |
8,06 |
[%] |
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2.1 Wyznaczyć współczynnik korelacji liniowej pomiędzy b(ω) i t(ω) |
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a) współczynnik korelacji liniowej: |
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x |
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ρbl € <-1;1> |
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ρbl= |
0,0041 |
więc zmienne nie są skorelowane |
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2.2 Aproksymacja za pomocą rozkładu normalnego parametrów losowych pola przekroju płaskownika A(ω) |
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z(ω)=b(ω)*t(ω) |
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zśr=bśr*tśr |
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xśr=87,77683[mm] |
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Aśr=87,7768*15,99= |
1403,78 |
[mm2] |
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sy=0,36[mm] |
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sx=3,70[mm] |
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yśr=15,99[mm] |
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poprawic wzor kwadrat |
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sA=Sz= |
67,17 |
[mm2] |
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vx=0.0421 |
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vy=0,0227 |
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νA=Vz= |
0,0478 |
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νA[%]= |
4,78 |
[%] |
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2.3 Wyznaczenie obliczeniowego pola przekroju Ao,tak aby prawdopodobieństwo |
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wystapienia przekroju mniejszego niż Ao nie przekroczyło: |
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a= |
69 |
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tab.3 |
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a) |
0,05 |
ai= |
0,0569 |
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bezpieczeństwo: |
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B=1-0,0569 |
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B= |
0,9431 |
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0,94295 |
0,9431 |
94408 |
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94407,05705 |
0,00015 |
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1,58 |
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1,59 |
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0,01 |
x |
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x= |
1,58886427230383E-11 |
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t= |
1,58000 |
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Aśr=1403,78 |
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Ao=Aśr-t*sA |
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615,7301 |
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Ao= |
1297,65 |
[mm2] |
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b) |
0,01 |
ai= |
0,0169 |
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bezpieczeństwo: |
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B=1-0,0169 |
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B= |
0,9831 |
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0,983 |
0,9831 |
0,98341 |
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0,00041 |
0,00010 |
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2,15 |
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2,16 |
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0,01 |
x |
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x= |
0,002439024390244 |
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t= |
2,15244 |
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Ao=Aśr-t*sA |
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599,7085 |
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Ao= |
1259,20 |
[mm2] |
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c) |
0,0013 |
ai= |
0,001369 |
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bezpieczeństwo: |
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B=1-0,001369 |
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B= |
0,998631 |
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0,998605 |
0,998631 |
0,99865 |
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0,000045 |
0,000026 |
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2,99 |
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3 |
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0,01 |
x |
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x= |
0,005777777777786 |
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t= |
2,99578 |
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Ao=Aśr-t*sA |
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Ao= |
1202,56 |
[mm2] |
575,6761 |
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Pole nominalne: |
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An=bn*tn |
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bn= |
85 |
tn= |
16 |
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An= |
1360 |
[mm2] |
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Porównanie w % o ile zmniejszyło się pole przekroju: |
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Pole [mm^2] |
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% |
Aśr |
An |
Aoa |
Aob |
Aoc |
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Aśr |
1403,78 |
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Aśr |
x |
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An |
1360 |
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An |
-3,12% |
x |
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Aoa |
1297,65 |
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Aoa |
7,56% |
4,58% |
x |
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Aob |
1259,20 |
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Aob |
10,30% |
7,41% |
2,96% |
x |
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Aoc |
1202,56 |
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Aoc |
14,33% |
11,58% |
7,33% |
4,50% |
x |
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2.4 Wyznaczyć parametry losowego wskaźnika wytrzymałości przy zginaniu Wx(ω) przekroju płaskownika |
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bśr= |
87,77683 |
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sb= |
3,70 |
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tśr= |
15,99 |
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st= |
0,36 |
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Wx=1/6*tśr*(bśr)2 |
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Wx= |
20536,556 |
[mm3] |
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14,6294716666667 |
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467,926577152667 |
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sWx= |
1791,34493242564 |
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νWx=sWx/Wx |
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νWx= |
0,0872 |
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νWx[%]= |
8,72 |
[%] |
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Parametry losowego momentu bezwładności przy zginaniu Ix(ω) przekroju przy zginaniu |
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Ix=(tśr*bśr3)/12 |
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Ix= |
901316,88 |
[mm4] |
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sIx= |
115689,685191417 |
[mm^4] |
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νIx= |
0,1284 |
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νIx[%]= |
12,84 |
[%] |
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Porównanie wyników współczynnika zmienności: |
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νb= |
4,21 |
[%] |
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νt= |
2,2697 |
[%] |
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νA= |
4,7848 |
[%] |
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νWx= |
8,72 |
[%] |
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νIx= |
12,84 |
[%] |
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3.1 Wyznaczyć parametry losowej nośności na rozciąganie Nrt(ω) płaskownika stalowego o losowym przekroju(A(ω), zad.2.2) oraz losowej wytrzymałości stali w(ω) |
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Nrt,śr+t*Sn
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t-indeks niezawodności Cornella(ufności) |
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(obiektywna miara bezpieczeństwa elementu) |
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Nrt(ω)=A(ω)*W(ω)-losowa nośność pręta rozciąganego osiowo |
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Nrt,śr=Aśr*Wśr-wartość średnia losowej nośności |
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Aśr= |
0,00140378 |
[m2] |
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Wśr= |
434018,00 |
[kN/m2] (z pkt.1.2) |
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Nrt,śr= |
609,27 |
[kN] |
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sA= |
0,00006717 |
[m2] (pkt.2.2) |
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sw= |
24184,88 |
[kN/m2] (z pkt.1.2) |
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sNrt= |
44,78 |
[kN] -odchylenie standardowe(Sn) losowej nośności |
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P,śr=447,83[kN] |
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Sp=36,1[kN] (pkt.1.2) |
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t= |
2,81 |
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t*Sn= |
125,680504756865 |
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N+t*sN= |
734,95 |
[kN] |
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N-t*sN= |
483,59 |
[kN] |
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3.2 Oszacować bezpieczeństwo płaskownika stalowego (wg zad. 3.1) obciążonego osiowo losową siłą rozciągającą P(ω) o parametrach wg zadania 1.3 |
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Nrt,śr=609,27[kN] |
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sNrt= |
44,78[kN] -parametry losowe nośności |
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P,śr=447,83[kN] |
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Sp=36,1[kN] -parametry osiowego obciążenia |
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Wartość dystrybuanty: |
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F(t)=F(2,81)= |
0,997523 |
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odczt. Tab 3 |
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Niezawodność : |
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R= |
0,997523 |
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Awaryjność: |
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A=1-R |
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A= |
0,002477 |
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3.3 Wyznaczyć parametry losowej siły rociągającej P1(ω), aby bezpieczeństwo stalowego płaskownika z zad. 3.2 wyniosłoR= 0,999a zakładając ten sam (jak w zad. 3.2) wsp. zmienności siły osiowej. |
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Dla R=0,99969 |
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ν=sP/Pśr |
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νP= |
0,0806 |
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Parametr t rozkładu normalnego dla dystrybuanty R=F(t)=0,99969 wynosi: |
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t= |
3,43 |
odczt. Tab 3 |
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Dane: |
Nrt,śr=609,27[kN] |
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=> |
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t=3,43 |
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vp=0,0806 |
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sNrt=44,78 |
[kN] |
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P(-)1śr= |
902,22 |
[kN]-"odpada" |
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przy sile P1=417,18[kN] zapas nośności jest większy |
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sP1,śr=νP1,śr*P1,śr |
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P(+)1śr/Nrt,śr=417,18/609,72=0,68 |
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sP1,śr=0,0806*417,18= |
33,63 |
[kN] |
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więc zapas nosności wynosi 31% |
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Sprawdzenie: |
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t= |
3,43 |
ok. |
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4.1 Dany jest łańcuch złożony z 57 elementó (ogniw). Ogniwa wykonano z płaskownika o losowej nośności (wg zad. 3.3). |
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Oszacować: |
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n= |
57 |
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a) Obliczeniową nosność graniczną pojedyńczego ogniwa dla bezpieczeństwa pi=0,999a |
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a= |
69 |
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Dane: |
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pi= |
0,99969 |
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0,9996869 |
0,99969 |
0,9996982 |
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0,000011 |
0,000003 |
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3,42 |
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3,43 |
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0,01 |
x |
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x= |
0,002743362831805 |
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t= |
3,42 |
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dane:(parametry losowe nośności) |
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Nrt,śr= |
609,27 |
[kN] |
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sNrt= |
44,78 |
[kN] |
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Obliczeniowa nośność graniczna dla jednego ogniwa: |
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Noi=Ni,śr-t*sNi |
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Noi= |
456,00 |
[kN] |
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b) bezpieczeństwo łańcucha rozciąganego siłą P2(ω) o parametrach: |
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Dane:(parametry losowe obciążenia) |
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P2śr=1,1*P1śr |
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P1śr= |
417,18 |
[kN] |
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P2śr= |
458,90 |
[kN] |
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sP1,śr= |
33,63 |
[kN] |
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sP2,śr= |
1,1*sP1,śr |
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sP2,śr= |
36,99 |
[kN] |
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indeks niezawodności pojedyńczego ogniwa |
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t= |
2,59 |
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Bezpieczeństwo jednego ogniwa: |
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pi=F(2,59) |
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pi= |
0,995201 |
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Bezpieczeństwo łańcucha złożonego z 57 ogniw: |
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R= |
0,7601789 |
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c) obliczeniową nośność graniczną łańcucha dla bezpieczeństwa Ro=0,999a |
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a= |
69 |
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Ro= |
0,99969 |
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pi= |
0,99999456 |
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Aby bezp.łancucha o n=57 wyniosło |
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Ro=0,99969, bezp.1 ogniwa musi wynosić pi=0,99999456 |
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0,99999433 |
0,99999456 |
0,99999459 |
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0,000000 |
0,000000 |
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4,39 |
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4,4 |
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0,010000000000001 |
x |
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x= |
0,00896373120811 |
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t= |
4,39896 |
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Obliczeniowa nośność graniczna łańcucha dla Ro=0,99969 wynosi: |
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No=NRt,śr-t*sNrt |
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Nrt,śr= |
609,27 |
[kN] |
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No= |
412,29 |
[kN] |
sNrt= |
44,78 |
[kN] |
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d) efekt statystycznego osłabienia systemu |
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Dane: |
Noi= |
456,00 |
[kN] |
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νo=No/Noi |
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νo= |
0,9041 |
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