pouzdanost u inzenjerstvu - projektni zadatak
DESCRIPTION
projektniTRANSCRIPT
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Visko Tehnička Mašinska Škola Stukovnih Studija Trstenik
Projektni rad
Pouzdanost u inženjerstvu
Student: Mirko Zdravić 27/2010
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Tabela T1 – ređanje po veličini
20,5 24 28 28 31 32,4 34,9 38 41 43,845 46,1 47,2 48,3 49,8 52 53 55 56,1 57,457,9 58,3 59,2 60,8 62,1 63 63,9 65 67 68,269,4 70,3 72,2 74,1 76,7 79,6 80,6 82 83,9 86,788,7 91,2 93,1 95,4 97,2 100,4 101 106,3 109,2 114
Određivanje broja intervala
n=50
z=√50=7,07≈7
z=5logn=5log50=8,49
z=1+3,3logn=1+3,3log50=6,60
usvaja se 7
Određivanje veličine intervala
Δt=t max−t min
z=
114−20,57
=93,5
7=13,35714 * 103 km
Tabela T2
Vreme 103
20,5– 33,85714
33,85714 – 47,21429
47,21429 – 60,57143
60,57143 – 73,92857
73,92857 – 87,28571
87,28571 – 100,6429
100,6429 –114
Broj otkaza
6 7 10 10 7 6 4
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Tabela T3
Vreme 103 km
ti N(Δt) N n-N
R(t) F(t) f(t) Λ(t) nsr R(t)sr F(t)sr
20,5– 33,85714
27,17857 6 6 44 0,88 0,12 0,00898
0,010 50
4744
0,94 0,06
33,85714 – 47,21429
40,535715 7 13 37 0,74 0,26 0,010 0,013 44
40,537
0,81 0,19
47,21429 – 60,57143
53,89286 10 23 27 0,54 0,46 0,014 0,025 37
3227
0,64 0,36
60,57143 – 73,92857
67,25 10 33 17 0,34 0,66 0,014 0,041 27
2217
0,44 0,56
73,92857 – 87,28571
80,60714 7 40 10 0,2 0,8 0,010 0,05 17
13,510
0,27 0,73
87,28571 – 100,6429
933,964305 6 46 4 0,08 0,92 0,00898
0,11 10
74
0,14 0,86
100,6429 –114
107,32145 4 50 0 0 1 0,00598
4
20
0,04 0,96
Σn
R(t)=n−NΣn
R(t)1=4450
=0,88
R(t)2=3750
=0,74
R(t)3=2750
=0,54
R(t)4=1750
=0,34
R(t)5=1050
=0,2
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R(t6)=4
50=0,08
R(t)7=0
50=0
F(t)=1-R(t) f(t)=N (Δt)Σn∗Δt
λ(t)=f (t )R (t)
F(t)1=1-0,88=0,12 f(t)1=6
50∗13,35714=0,00898 λ(t)1=
0,008980,88
=0,010
F(t)2=1-0,74=0,26 f(t)2=7
50∗13,35714=0,010 λ(t)2=
0,0100,74
=0,013
F(t)3=1-0,54=0,46 f(t)3=10
50∗13,35714=0,014 λ(t)3=
0,0140,54
=0,025
F(t)4=1-0,34=0,66 f(t4)=10
50∗13,35714=0,014 λ(t)4=
0,0140,34
=0,041
F(t)5=1-0,2=0,8 f(t)5=7
50∗13,35714=0,010 λ(t)5=
0,0100,2
=0,05
F(t)6=1-0,08=0,92 f(t)6=6
50∗13,35714=0,00898 λ(t)6=
0,008980,08
=0,11
F(t)7=1-0=1 f(t)7=4
50∗13,35714=0,00598 λ(t)7=
0,005980
=
R(t)sr=nsrΣn
F(t)sr=1-R(t)sr
R(t)1sr=4750
=0,94 F(t)1sr=1-0,94=0,06
R(t)2sr=40,550
=0,81 F(t)2sr=1-0,81=0,19
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R(t)3sr=3250
=0,64 F(t)3sr=1-0,64=0,36
R(t)4sr=2250
=0,44 F(t)4sr=1-0,44=0,56
R(t)5sr=13,550
=0,27 F(t)5sr=1-0,27=0,73
R(t)6sr=7
50=0,14 F(t)6sr=1-0,14=0,86
R(t)7sr=2
50=0,04 F(t)7sr=1-0,04=0,96
Određivanje srednjeg vremena tsr
tsr=Σ ¿¿
tsr=(27,17857∗6 )+(40,535715∗7 )+ (53,89286∗10 )+ (67,25∗10 )+ (80,60714∗7 )+(93,964305∗6 )+(107,32145∗4)
50
tsr=163,07142+283,750005+538,9286+572,5+564,24998+563,78583+429,2858
50
tsr=3215,571635
50
tsr=64,3114327
Određivanje disperzije σ
σ=√ ΣN (Δt )∗ti2−tsr2
Σn
σ=√(6∗27,178572−64,31143272)+(7∗40,5357152−64,31143272 )+¿¿¿
√¿¿¿
√¿¿¿=
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σ=
√ 296,0876275+7366,048958+24908,44321+41089,66462+41346,61676+48839,78331+41935,6141450
=
σ=√ 205782,258650
=
σ=√4115,645173=
σ=64,15329432
Određivanje pouzdanosti nakon pređenih 50 000 km Rt(t)
Rt(50000)
Rt(50000)=0,5+φ(z1)
Rt(50000)=0,5+φ(0,2230)
Rt(50000)=0,5+1,16
Rt(50000)=1,66
z1=tsr−50000
σ
z1=64311−50000
64153
z1=1431164153
z1=0,2230