Dane |
Obliczenia |
Wyniki |
N= |
13 |
[kW] |
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n= |
1000 |
[obr/min] |
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i= |
3,5 |
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Lh= |
12000 |
[godz.] |
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1.Obliczenia wstępne |
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Dobór przełożenia na poszczególnych |
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stopniach |
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Uc= |
3,5 |
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Uc= |
3,50 |
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U1= |
2,1 |
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U1= |
2,10 |
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U2= |
1,7 |
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U2= |
1,67 |
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Zmiana obciążenia w czasie |
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Produkcja jednostkowa |
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Dane |
Obliczenia |
Wyniki |
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Obciążenie wałów schematu układu |
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napędowego |
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N= |
13 |
[kW] |
Moc P=N |
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n= |
1000 |
[obr/min] |
Obroty n1=n |
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n1= |
1000 |
[obr/min] |
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n1= |
1000,00 |
[obr/min] |
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n2=n/U1 |
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n2= |
476,190476190476 |
[obr/min] |
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n2= |
476,19 |
[obr/min] |
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n3=n/U2 |
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n3= |
600 |
[obr/min] |
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n3= |
600,00 |
[obr/min] |
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n1>n2>n3 |
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Tj - moment [Nm] |
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T1= |
124,15 |
[Nm] |
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T1= |
124,15 |
[Nm] |
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T2=T1*U1 |
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T2= |
260,715 |
[Nm] |
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T2= |
260,72 |
[Nm] |
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T3=T2*U2 |
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T3= |
434,525 |
[Nm] |
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T3= |
434,53 |
[Nm] |
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T1<T2<T3 |
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Pwyj= |
27,3 |
[kW] |
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Pwyj= |
27,3 |
[kW] |
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Dane |
Obliczenia |
Wyniki |
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1.1.Dobór materiałów na zębnik i koło |
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zębate (obróbka cieplna, naprężenia |
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dop.), stopień pierwszy. |
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Dla jednostkowej produkcji kół zębatych |
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dobieramy: |
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materiał na zębnik-C55 PN-93/H-84019 |
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Twardość = 270 HB |
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Rm=680 [MPa] |
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Re=390 [MPa] |
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materiał koła zębatego-C40 PN-93/H-84019 |
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Twardość = 255 HB |
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Rm=580 [MPa] |
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Re=335 [MPa] |
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2.Dopuszczalne naprężenia stykowe |
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2.1.Podstawa próby zmęczeniowej |
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(bazowa liczba cykli): |
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dla zębnika |
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20*10^6 |
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NHlim1= |
20*10^6 |
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dla koła zębatego |
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17*10^6 |
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NHlim2= |
17*10^6 |
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2.2.Równoważna (ekwiwalentna) |
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liczba cykli obciążenia |
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n1= |
1000 |
[obr/min] |
NHeq1= |
327456000 |
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NHeq1= |
327456000 |
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n2= |
476,190476190476 |
[obr/min] |
NHeq2= |
155931428,571429 |
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NHeq2= |
155931428,571429 |
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gdzie: |
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Lh= |
12000 |
[godz.] |
Lh - liczba godzin pracy przekładni |
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c= |
1 |
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c - liczba zazębień zęba w czasie jednego |
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obrotu(ze schematu napędu) |
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kHeq - współczynnik uwzględniający zmianę |
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obciążenia napędu w czasie |
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kHeq= |
0,4548 |
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kHeq= |
0,45 |
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mH= |
6 |
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mH - współczynnik kierunkowy nachylonego |
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odcinka na wykresie Wohlera |
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Dane |
Obliczenia |
Wyniki |
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2.3.Współczynnik trwałóści pracy |
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NHeq1= |
327456000 |
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NHeq2= |
155931428,571429 |
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ZN1= |
0,010179484673768 |
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NHlim1= |
20*10^6 |
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NHlim2= |
17*10^6 |
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ZN2= |
0,018170380142676 |
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mH= |
6 |
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dla |
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ZN1(2) = 1,0 |
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Przyjmujemy |
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ZN1= |
1 |
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ZN1= |
1,00 |
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ZN2= |
1 |
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ZN2= |
1,00 |
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2.4.Naprężenia krytyczne przy bazowej |
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liczbie cykli |
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sHlim1=2HB1+70 |
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sHlim1= |
610 |
[MPa] |
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sHlim1= |
610,00 |
[MPa] |
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sHlim2=2HB2+70 |
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sHlim2= |
580 |
[MPa] |
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sHlim2= |
580,00 |
[MPa] |
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2.5.Dopuszczalne naprężenia stykowe |
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sH1= |
499,090909090909 |
[MPa] |
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sH1= |
499,09 |
[MPa] |
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sH2= |
474,545454545455 |
[MPa] |
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sH2= |
474,55 |
[MPa] |
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gdzie: |
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SH - współczynnik bezpieczeństwa: |
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`w przypadku normalizacji, ulepszania lub |
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hartowania zebów na wskroś SH=1,1 |
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`dla hartowania powierzchniowego, |
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nawęglania, azotowania SH=1,2 |
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SH= |
1,1 |
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Przyjmujemy SH=1,1 |
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Dane |
Obliczenia |
Wyniki |
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2.6.Obliczeniowe dopuszczalne |
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naprężenia stykowe: |
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`dla przekładni walcowych z kołami o |
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zębach prostych oraz z kołami o zębach |
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skośnych przy niewielkiej różnicy twardości |
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HB1 i HB2 |
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sHP=sH2=474,545 [MPa] |
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sHP= |
474,55 |
[MPa] |
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3.Dopuszczalne naprężenia na |
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zginanie. |
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3.1.Podstawa próby zmęczeniowej |
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NFlim= |
4000000 |
[cykli] |
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3.2.Równoważna (ekwiwalentna) |
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liczba cykli obciążenia |
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c= |
1 |
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Lh= |
12000 |
[godz.] |
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n1= |
1000 |
[obr/min] |
NFeq1= |
223997472 |
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NFeq1= |
223997472 |
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n2= |
476,190476190476 |
[obr/min] |
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mF= |
6 |
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NFeq2= |
106665462,857143 |
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NFeq2= |
106665462,857143 |
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gdzie: |
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KFeq= |
0,3111076 |
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KFeq= |
0,31 |
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3.3.Współczynnik trwałości |
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YN1= |
0,511253325248433 |
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YN2= |
0,578547953260626 |
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dla |
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YN1(2)= |
1,00 |
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Dane |
Obliczenia |
Wyniki |
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3.4.Naprężenia krytyczne |
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sFlim1=1,75HB1 |
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sFlim1= |
472,5 |
[MPa] |
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sFim1= |
472,50 |
[MPa] |
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sFlim2=1,75HB2 |
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sFlim2= |
446,25 |
[MPa] |
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sFlim2= |
446,25 |
[MPa] |
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3.5.Dopuszczalne naprężenia na |
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zginanie |
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sFP1= |
141,75 |
[MPa] |
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sFP1= |
141,75 |
[MPa] |
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sFP2= |
133,875 |
[MPa] |
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sFP2= |
133,88 |
[MPa] |
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gdzie: |
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YA= |
0,75 |
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YA -współczynnik uwzględniający wpływ |
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dwustronnego przekładania obciążenia na |
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ząb |
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`dla przekładni bez zmiany kierunku |
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obracania = 1,0 |
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`dla przekładni za zmianą kierunku |
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obracania = (0,7…0,8) |
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4.Graniczne naprężenia dopuszczalne |
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przy przeciążeniach |
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Dla naprężeń stykowych: |
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sHPmax1=2,8Re1 |
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sHPmax1= |
1092 |
[MPa] |
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sHPmax1= |
1092,00 |
[MPa] |
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sHPmax2=2,8Re2 |
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sHPmax2= |
938 |
[MPa] |
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sHPmax2= |
938,00 |
[MPa] |
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Dla naprężeń gnących |
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sFPmax1=0,8Re1 |
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sFPmax1= |
312 |
[MPa] |
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sFPmax1= |
312,00 |
[MPa] |
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sFPmax2=0,8Re2 |
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sFPmax2= |
268 |
[MPa] |
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sFPmax2= |
268,00 |
[MPa] |
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Dane |
Obliczenia |
Wyniki |
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5.Obliczenie średnicy zębnika i dobór |
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innych parametrów przekładni. |
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5.1.Obliczeniowa średnica zębnika. |
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T1= |
124,15 |
[Nm] |
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U1= |
2,1 |
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sHP= |
474,545454545455 |
[MPa] |
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d`1= |
89,7120927219693 |
[mm] |
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d`1= |
89,71 |
[mm] |
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gdzie: |
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kd= |
77 |
[MPa] |
kd=77 MPa -dla kół o zębach prostych, |
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ybd= |
0,6 |
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ybd- współczynik szerokości wieńca |
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(w stosunku do średnicy zębnika) |
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ybd=b/d1 =f(HB rozmieszczenie kół |
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względem łożysk) = 0,6 |
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kHB= |
1,06 |
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kHB - współczynnik nierównomierności |
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rozkładu obciążenia względem lini styku, |
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kHB= f(HB rozmieszczenie kół względem |
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łożysk ,ybd) = 1,06 |
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kA= |
1,1 |
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kA - współczynnik uwzględniający zew. |
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obciążenie dynamiczne = 1,1 |
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Zazębiene zewnętrzne (+) |
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5.2.Szerokość wieńca koła zębatego. |
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b2 = b = ybd*d`1 |
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b2= |
53,8272556331816 |
[mm] |
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b2= |
53,83 |
[mm] |
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Szerokość wieńca zębnika b1 = b2 + (3…5) |
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b1= |
58,8272556331816 |
[mm] |
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b1= |
57,83 |
[mm] |
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Dla przekładni o zębach prostych (b=0°) |
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z`1= |
19 |
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5.3.Przyjmując wstępnie z`1=19, |
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obliczamy moduł zazębienia |
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m`= |
4,72168909062996 |
[mm] |
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m`= |
4,72 |
[mm] |
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Moduł m` zaokrąglam do wartości zbliżonej |
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do mn ,mm zgodnej z PN-ISO 54:2001 |
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Przyjmuję mn= 4,5 |
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mn= |
4,50 |
[mm] |
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5.4.Liczba zębów zębnika |
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z1= |
19,9360206048821 |
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z1= |
19,94 |
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gdzie: |
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z1 - liczba całkowita (z1 ≥ 17) |
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Dane |
Obliczenia |
Wyniki |
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5.5.Liczba zębów koła |
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U1= |
2,1 |
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188,40 |
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z1= |
19,9360206048821 |
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z2= |
41,8656432702523 |
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z2= |
41,87 |
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gdzie: |
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z2 - liczba całkowita |
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5.6.Odległość zerowa osi |
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aWO= |
139,053743719052 |
[mm] |
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aWO= |
139,05 |
[mm] |
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Zaokrąglam aWO do wielkości aW zgodnej |
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z PN-93/M-88525 |
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Przyjmuję aW= 100 [mm] |
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aW= |
100,00 |
[mm] |
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5.7.Średnice okręgów kół zębatych |
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mn= |
4,5 |
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` tocznych |
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z1= |
19,9360206048821 |
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z2= |
41,8656432702523 |
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dW1= |
89,7120927219693 |
[mm] |
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dW1= |
89,71 |
[mm] |
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dW2= |
188,395394716136 |
[mm] |
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dW2= |
188,40 |
[mm] |
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` wierzchołków zębów |
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da1= |
98,7120927219693 |
[mm] |
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da1= |
98,71 |
[mm] |
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da2= |
197,395394716136 |
[mm] |
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da2= |
197,40 |
[mm] |
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` stóp zębów |
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df1= |
78,4620927219693 |
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df1= |
78,46 |
[mm] |
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df2= |
177,145394716136 |
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df2= |
177,15 |
[mm] |
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5.8.Przełożenie rzeczywiste przekładni |
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urz= |
2,1 |
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urz= |
2,10 |
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Dane |
Obliczenia |
Wyniki |
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6.Sprawdzanie obliczeniowych |
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naprężeń stykowych. |
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6.1.Siła obwodowa w zazębieniu |
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T1= |
124,15 |
[Nm] |
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dW1= |
89,7120927219693 |
[mm] |
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Ft= |
2767,7428144444 |
[N] |
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Ft= |
2767,7428144444 |
[N] |
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6.2.Prędkość obwodowa kół |
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n1= |
1000 |
[obr/min] |
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n= |
4,69731419055842 |
[m/s] |
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n= |
4,70 |
[m/s] |
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6.3.Klasa dokładności = f(n , b) → 8 |
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6.4.Współczynnik międzyzębnego |
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obciążenia dynamicznego |
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kHn=f(n, klasa dokładności, twardość |
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zębów) |
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kHn= |
1,06 |
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kHn= |
1,06 |
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6.5.Współczynnik uwzględniający |
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nierównomierność rozkładu obciążenia |
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między parami zębów w zazębieniu |
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kHa=f(n, klasa dokładności) |
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` dla zębów prostych |
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kHa= |
1 |
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kHa= |
1,0 |
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6.6.Jednostkowa obliczeniowa siła |
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obwodowa |
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kHB= |
1,06 |
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kA= |
1,1 |
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b2= |
53,83 |
[mm] |
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WHt= |
63,5518078843304 |
[N/mm] |
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WHt= |
63,5518078843304 |
[N/mm] |
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6.7.Obliczeniowe naprężenia stykowe |
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U1= |
2,1 |
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sH= |
497,754987951736 |
[MPa] |
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sH= |
497,754987951736 |
[MPa] |
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sH ≤ sHP warunek spełniony |
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gdzie: |
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ZH - współczynnik uwzględniający kształt |
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ZH= |
1,77 |
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stykających się powierzchni zębów |
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ZM - współczynnik uwzględniający własności |
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ZM= |
275 |
[MPa1/2] |
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mechaniczne kół zębatych |
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Zε - współczynnik przyporu |
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Zε= |
1,0 |
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Dane |
Obliczenia |
Wyniki |
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7.Sprawdzanie obliczeniowych |
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naprężeń gnących |
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7.1.Współczynnik międzyzębnego |
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obciążenia dynamicznego przy zginaniu |
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zęba kFn=f(n, klasa dokładności, |
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twardość zębów) |
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kFn= |
1,12 |
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kFn= |
1,12 |
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7.2.Współczynnik nierównomierności |
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rozkładu obciążenia względem lini styku |
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kFb=f(HB, rozmieszczenie kół względem |
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łożysk, yBD) |
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kFb= |
1,1 |
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kFb= |
1,10 |
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7.3.Współczynnik uwzględniający |
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|
nierównomierność rozkładu obciążenia |
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między parami zębów w zazębieniu |
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kFa=f(n, klasa dokładności) |
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kFa= |
1 |
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kFa= |
1,0 |
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7.4.Jednostkowa obwodowa siła |
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obliczeniowa przy zginaniu |
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kA= |
1,1 |
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b2= |
53,83 |
[mm] |
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Ft= |
2767,7428144444 |
[N] |
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WFt= |
69,6830075769803 |
[N/mm] |
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WFt= |
69,68 |
[N/mm] |
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7.5.Ekwiwalentna liczba zębów |
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`dla zębów prostych |
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z1eq= |
19,94 |
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z2eq= |
41,87 |
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7.6.Współczynnik kształtu zębów |
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zębnika i koła zębatego |
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Obliczenia wykonuje się dla koła z pary |
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"zębnik - koło zębate", dla którego jest |
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mniejszy stosunek |
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YFS1= |
4,2 |
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YFS1= |
4,20 |
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YFS2= |
3,72 |
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YFS2= |
3,72 |
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Dane |
Obliczenia |
Wyniki |
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sFP1= |
141,75 |
[MPa] |
sFP1/YFS1= |
33,75 |
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sFP1/YFS1= |
33,75 |
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sFP2= |
133,88 |
[MPa] |
sFP2/YFS2= |
35,9879032258064 |
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sFP2/YFS2= |
35,9879032258064 |
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7.7.Naprężenia obliczeniowe gnące |
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WFt= |
69,68 |
[N/mm] |
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YFS1= |
4,20 |
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YFS2= |
3,72 |
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mn= |
4,50 |
[mm] |
sF1= |
65,037473738515 |
[MPa] |
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sF1= |
65,037473738515 |
[MPa] |
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sF2= |
57,6046195969704 |
[MPa] |
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sF2= |
57,6046195969704 |
[MPa] |
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|
gdzie: |
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Yb= |
1 |
|
Yb - współczynnik kąta pochylenia lini zęba |
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` dla zębów prostych = 1,0 |
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8.Sprawdzanie wytrzymałości zębów |
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|
przy przeciążeniach |
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|
Tmax/Tnom= |
2,8 |
|
8.1.Naprężenia stykowe przy |
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przeciążeniach |
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sH= |
497,754987951736 |
[MPa] |
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sHmax= |
832,903402854335 |
[MPa] |
|
sHmax= |
832,903402854335 |
[MPa] |
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8.2.Naprężenia gnące przy |
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|
przeciążeniach |
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|
sFPmax1= |
312,00 |
[MPa] |
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sFPmax2= |
268,00 |
[MPa] |
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sFmax1= |
182,104926467842 |
[MPa] |
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sFmax1= |
182,104926467842 |
[MPa] |
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sFmax2= |
161,292934871517 |
[MPa] |
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sFmax2= |
161,292934871517 |
[MPa] |
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9.Siły działające w zazębieniu |
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przekładni |
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9.1.Moment rzeczywisty na wale |
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wyjściowym |
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u= |
2,1 |
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urz= |
2,1 |
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T1= |
124,15 |
[Nm] |
T2rz= |
260,715 |
[Nm] |
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T2rz= |
260,715 |
[Nm] |
T2= |
260,715 |
[Nm] |
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9.2.Siły obwodowe |
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dW1= |
89,71 |
[mm] |
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dW2= |
188,40 |
[mm] |
Ft1= |
2767,7428144444 |
[N] |
|
Ft1= |
2767,7428144444 |
[N] |
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Ft2= |
2767,7428144444 |
[N] |
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Ft2= |
2767,7428144444 |
[N] |
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Dane |
Obliczenia |
Wyniki |
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9.3.Siły promieniowe |
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α= |
20 |
[°] |
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Ft1= |
2767,7428144444 |
[N] |
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|
Ft2= |
2767,7428144444 |
[N] |
Fr1= |
1007,37599982515 |
[N] |
|
Fr1= |
1007,37599982515 |
[N] |
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Fr2= |
1007,37599982515 |
[N] |
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Fr2= |
1007,37599982515 |
[N] |
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Dane |
Obliczenia |
Wyniki |
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10.Dobór materiałów na zębnik i koło |
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zębate (obróbka cieplna, naprężenia |
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dop.), stopień drugi. |
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Dla jednostkowej produkcji kół zębatych |
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dobieramy: |
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|
materiał na zębnik-C45 PN-93/H-84019 |
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|
Twardość = 215 HB |
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|
Rm3= |
720 |
[MPa] |
Rm=720 [MPa] |
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Re3= |
450 |
[MPa] |
Re=450 [MPa] |
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|
|
materiał koła zębatego-C35 PN-93/H-84019 |
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|
Twardość = 187 HB |
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Rm4= |
600 |
[MPa] |
Rm=600 [MPa] |
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Re4= |
315 |
[MPa] |
Re=315 [MPa] |
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11.Dopuszczalne naprężenia stykowe |
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11.1.Podstawa próby zmęczeniowej |
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(bazowa liczba cykli): |
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dla zębnika |
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19*10^6 |
|
NHlim3= |
19*10^6 |
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dla koła zębatego |
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16*10^6 |
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NHlim4= |
16*10^6 |
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11.2.Równoważna (ekwiwalentna) |
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liczba cykli obciążenia |
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n2= |
476,190476190476 |
[obr/min] |
NHeq3= |
155931428,571429 |
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NHeq3= |
155931428,571429 |
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n3= |
600 |
[obr/min] |
NHeq4= |
196473600 |
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NHeq4= |
196473600 |
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gdzie: |
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Lh= |
12000 |
[godz.] |
Lh - liczba godzin pracy przekładni |
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c= |
1 |
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c - liczba zazębień zęba w czasie jednego |
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obrotu(ze schematu napędu) |
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kHeq - współczynnik uwzględniający zmianę |
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obciążenia napędu w czasie |
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kHeq= |
0,4548 |
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kHeq= |
0,45 |
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mH= |
6 |
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mH - współczynnik kierunkowy nachylonego |
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odcinka na wykresie Wohlera |
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Dane |
Obliczenia |
Wyniki |
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11.3.Współczynnik trwałóści pracy |
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NHeq3= |
155931428,571429 |
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NHeq4= |
196473600 |
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ZN3= |
0,020308071924167 |
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NHlim3= |
19*10^6 |
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NHlim4= |
16*10^6 |
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ZN4= |
0,013572646231691 |
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mH= |
6 |
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dla |
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ZN3(4) = 1,0 |
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Przyjmujemy |
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ZN3= |
1 |
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ZN3= |
1,00 |
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ZN4= |
1 |
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ZN4= |
1,00 |
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11.4.Naprężenia krytyczne przy bazowej |
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liczbie cykli |
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sHlim3=2HB3+70 |
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sHlim3= |
500 |
[MPa] |
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sHlim3= |
500,00 |
[MPa] |
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sHlim4=2HB4+70 |
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sHlim4= |
444 |
[MPa] |
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sHlim4= |
444,00 |
[MPa] |
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11.5.Dopuszczalne naprężenia stykowe |
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sH3= |
409,090909090909 |
[MPa] |
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sH3= |
409,09 |
[MPa] |
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sH4= |
363,272727272727 |
[MPa] |
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sH4= |
363,27 |
[MPa] |
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gdzie: |
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SH - współczynnik bezpieczeństwa: |
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`w przypadku normalizacji, ulepszania lub |
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hartowania zebów na wskroś SH=1,1 |
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`dla hartowania powierzchniowego, |
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nawęglania, azotowania SH=1,2 |
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SH= |
1,1 |
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Przyjmujemy SH=1,1 |
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Dane |
Obliczenia |
Wyniki |
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11.6.Obliczeniowe dopuszczalne |
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naprężenia stykowe: |
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`dla przekładni walcowych z kołami o |
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zębach prostych oraz z kołami o zębach |
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skośnych przy niewielkiej różnicy twardości |
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HB3 i HB4 |
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sHP=sH4=363,27 [MPa] |
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sHP= |
363,27 |
[MPa] |
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12.Dopuszczalne naprężenia na |
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zginanie. |
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12.1.Podstawa próby zmęczeniowej |
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NFlim= |
4000000 |
[cykli] |
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12.2.Równoważna (ekwiwalentna) |
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liczba cykli obciążenia |
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c= |
1 |
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Lh= |
12000 |
[godz.] |
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n2= |
476,190476190476 |
[obr/min] |
NFeq3= |
106665462,857143 |
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NFeq3= |
106665462,857143 |
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n3= |
600 |
[obr/min] |
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mF= |
6 |
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NFeq4= |
134398483,2 |
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NFeq4= |
134398483,2 |
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gdzie: |
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KFeq= |
0,3111076 |
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KFeq= |
0,31 |
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12.3.Współczynnik trwałości |
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YN3= |
0,578547953260626 |
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YN4= |
0,556686817645271 |
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dla |
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YN3(4)= |
1,00 |
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Dane |
Obliczenia |
Wyniki |
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12.4.Naprężenia krytyczne |
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sFlim3=1,75HB3 |
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sFlim3= |
376,25 |
[MPa] |
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sFim3= |
376,25 |
[MPa] |
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sFlim4=1,75HB4 |
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sFlim4= |
327,25 |
[MPa] |
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sFlim4= |
327,25 |
[MPa] |
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12.5.Dopuszczalne naprężenia na |
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zginanie |
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sFP3= |
112,875 |
[MPa] |
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sFP3= |
112,88 |
[MPa] |
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sFP4= |
98,175 |
[MPa] |
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sFP4= |
98,18 |
[MPa] |
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gdzie: |
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YA= |
0,75 |
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YA -współczynnik uwzględniający wpływ |
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dwustronnego przekładania obciążenia na |
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ząb |
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`dla przekładni bez zmiany kierunku |
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obracania = 1,0 |
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`dla przekładni za zmianą kierunku |
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obracania = (0,7…0,8) |
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13.Graniczne naprężenia dopuszczalne |
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przy przeciążeniach |
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Dla naprężeń stykowych: |
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Rm3= |
720 |
[MPa] |
sHPmax3=2,8Re3 |
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Re3= |
450 |
[MPa] |
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Rm4= |
600 |
[MPa] |
sHPmax3= |
1260 |
[MPa] |
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sHPmax3= |
1260,00 |
[MPa] |
Re4= |
315 |
[MPa] |
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sHPmax4=2,8Re4 |
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sHPmax2= |
882 |
[MPa] |
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sHPmax4= |
882,00 |
[MPa] |
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Dla naprężeń gnących |
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sFPmax3=0,8Re3 |
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sFPmax3= |
360 |
[MPa] |
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sFPmax3= |
360,00 |
[MPa] |
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sFPmax4=0,8Re4 |
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sFPmax4= |
252 |
[MPa] |
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sFPmax4= |
252,00 |
[MPa] |
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Dane |
Obliczenia |
Wyniki |
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14.Obliczenie średnicy zębnika i dobór |
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innych parametrów przekładni. |
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14.1.Obliczeniowa średnica zębnika. |
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T2= |
260,715 |
[Nm] |
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U= |
1,67 |
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sHP= |
363,27 |
[MPa] |
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d`3= |
141,019757227413 |
[mm] |
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d`3= |
141,02 |
[mm] |
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gdzie: |
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kd= |
77 |
[MPa] |
kd=77 MPa -dla kół o zębach prostych, |
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ybd= |
0,6 |
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ybd- współczynik szerokości wieńca |
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(w stosunku do średnicy zębnika) |
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ybd=b/d3 =f(HB rozmieszczenie kół |
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względem łożysk) = 0,6 |
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kHB= |
1,06 |
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kHB - współczynnik nierównomierności |
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rozkładu obciążenia względem lini styku, |
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kHB= f(HB rozmieszczenie kół względem |
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łożysk ,ybd) = 1,06 |
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kA= |
1,1 |
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kA - współczynnik uwzględniający zew. |
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obciążenie dynamiczne = 1,1 |
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Zazębiene zewnętrzne (+) |
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14.2.Szerokość wieńca koła zębatego. |
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b4 = b = ybd*d`3 |
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b4= |
84,6118543364478 |
[mm] |
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b4= |
84,61 |
[mm] |
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Szerokość wieńca zębnika b3 = b4 + (3…5) |
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b3= |
89,6118543364478 |
[mm] |
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b3= |
89,61 |
[mm] |
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Dla przekładni o zębach prostych (b=0°) |
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z`3= |
19 |
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14.3.Przyjmując wstępnie z`3=19, |
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obliczamy moduł zazębienia |
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m`= |
6,90209248565332 |
[mm] |
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m`= |
6,90 |
[mm] |
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Moduł m` zaokrąglam do wartości zbliżonej |
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do mn ,mm zgodnej z PN-ISO 54:2001 |
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Przyjmuję mn= 6 |
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mn= |
6,00 |
[mm] |
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14.4.Liczba zębów zębnika |
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z3= |
23,5032928712355 |
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z3= |
23,50 |
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gdzie: |
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z3 - liczba całkowita (z3 ≥ 17) |
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Dane |
Obliczenia |
Wyniki |
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14.5.Liczba zębów koła |
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U2= |
1,67 |
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z3= |
23,5032928712355 |
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z4= |
39,1721547853925 |
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z4= |
39,17 |
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gdzie: |
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z4 - liczba całkowita |
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14.6.Odległość zerowa osi |
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aWO= |
188,026342969884 |
[mm] |
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aWO= |
188,03 |
[mm] |
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Zaokrąglam aWO do wielkości aW zgodnej |
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z PN-93/M-88525 |
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Przyjmuję aW= 100 [mm] |
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aW= |
100,00 |
[mm] |
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14.7.Średnice okręgów kół zębatych |
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mn= |
6 |
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` tocznych |
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z3= |
23,5032928712355 |
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z4= |
39,1721547853925 |
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dW3= |
141,019757227413 |
[mm] |
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dW3= |
141,02 |
[mm] |
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dW4= |
235,032928712355 |
[mm] |
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dW4= |
235,03 |
[mm] |
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` wierzchołków zębów |
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da3= |
153,02 |
[mm] |
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da3= |
153,02 |
[mm] |
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da4= |
247,032928712355 |
[mm] |
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da4= |
247,03 |
[mm] |
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` stóp zębów |
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df3= |
126,02 |
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df3= |
126,02 |
[mm] |
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df4= |
220,032928712355 |
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df4= |
220,03 |
[mm] |
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14.8.Przełożenie rzeczywiste przekładni |
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urz= |
1,66666666666667 |
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urz= |
1,67 |
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Dane |
Obliczenia |
Wyniki |
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15.Sprawdzanie obliczeniowych |
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naprężeń stykowych. |
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15.1.Siła obwodowa w zazębieniu |
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T3= |
260,715 |
[Nm] |
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dW3= |
141,019757227413 |
[mm] |
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Ft= |
3697,56699523404 |
[N] |
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Ft= |
3697,56699523404 |
[N] |
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15.2.Prędkość obwodowa kół |
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n2= |
476,190476190476 |
[obr/min] |
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n= |
3,51608439140204 |
[m/s] |
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n= |
3,52 |
[m/s] |
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15.3.Klasa dokładności = f(n , b) → 8 |
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15.4.Współczynnik międzyzębnego |
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obciążenia dynamicznego |
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kHn=f(n, klasa dokładności, twardość |
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zębów) |
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kHn= |
1,06 |
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kHn= |
1,06 |
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15.5.Współczynnik uwzględniający |
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nierównomierność rozkładu obciążenia |
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między parami zębów w zazębieniu |
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kHa=f(n, klasa dokładności) |
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` dla zębów prostych |
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kHa= |
1 |
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kHa= |
1,0 |
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15.6.Jednostkowa obliczeniowa siła |
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obwodowa |
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kHB= |
1,06 |
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kA= |
1,1 |
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b4= |
84,61 |
[mm] |
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WHt= |
54,0118750412595 |
[N/mm] |
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WHt= |
54,0118750412595 |
[N/mm] |
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15.7.Obliczeniowe naprężenia stykowe |
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U2= |
1,67 |
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sH= |
381,040025259604 |
[MPa] |
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sH= |
381,040025259604 |
[MPa] |
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sH ≤ sHP warunek spełniony |
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gdzie: |
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ZH - współczynnik uwzględniający kształt |
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ZH= |
1,77 |
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stykających się powierzchni zębów |
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ZM - współczynnik uwzględniający własności |
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ZM= |
275 |
[MPa1/2] |
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mechaniczne kół zębatych |
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Zε - współczynnik przyporu |
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Zε= |
1,0 |
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Dane |
Obliczenia |
Wyniki |
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16.Sprawdzanie obliczeniowych |
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naprężeń gnących |
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16.1.Współczynnik międzyzębnego |
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obciążenia dynamicznego przy zginaniu |
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zęba kFn=f(n, klasa dokładności, |
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twardość zębów) |
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kFn= |
1,12 |
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kFn= |
1,12 |
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16.2.Współczynnik nierównomierności |
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rozkładu obciążenia względem lini styku |
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kFb=f(HB, rozmieszczenie kół względem |
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łożysk, yBD) |
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kFb= |
1,1 |
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kFb= |
1,10 |
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16.3.Współczynnik uwzględniający |
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nierównomierność rozkładu obciążenia |
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między parami zębów w zazębieniu |
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kFa=f(n, klasa dokładności) |
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kFa= |
1 |
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kFa= |
1,0 |
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16.4.Jednostkowa obwodowa siła |
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obliczeniowa przy zginaniu |
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kA= |
1,1 |
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b4= |
84,61 |
[mm] |
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Ft= |
3697,56699523404 |
[N] |
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WFt= |
59,2227038544248 |
[N/mm] |
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WFt= |
59,22 |
[N/mm] |
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16.5.Ekwiwalentna liczba zębów |
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`dla zębów prostych |
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z3eq= |
23,50 |
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z4eq= |
39,17 |
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16.6.Współczynnik kształtu zębów |
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zębnika i koła zębatego |
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Obliczenia wykonuje się dla koła z pary |
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"zębnik - koło zębate", dla którego jest |
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mniejszy stosunek |
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YFS3= |
4,2 |
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YFS3= |
4,20 |
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YFS4= |
3,72 |
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YFS4= |
3,72 |
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Dane |
Obliczenia |
Wyniki |
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sFP3= |
112,88 |
[MPa] |
sFP3/YFS3= |
26,875 |
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sFP3/YFS3= |
26,875 |
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sFP4= |
98,18 |
[MPa] |
sFP4/YFS4= |
26,3911290322581 |
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sFP4/YFS4= |
26,3911290322581 |
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16.7.Naprężenia obliczeniowe gnące |
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WFt= |
59,22 |
[N/mm] |
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YFS3= |
4,20 |
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YFS4= |
3,72 |
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mn= |
6,00 |
[mm] |
sF3= |
41,4558926980974 |
[MPa] |
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sF3= |
41,4558926980974 |
[MPa] |
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sF4= |
36,7180763897434 |
[MPa] |
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sF4= |
36,7180763897434 |
[MPa] |
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gdzie: |
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Yb= |
1 |
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Yb - współczynnik kąta pochylenia lini zęba |
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` dla zębów prostych = 1,0 |
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17.Sprawdzanie wytrzymałości zębów |
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przy przeciążeniach |
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Tmax/Tnom= |
2,8 |
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17.1.Naprężenia stykowe przy |
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przeciążeniach |
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sH= |
381,040025259604 |
[MPa] |
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sHmax= |
637,601915288491 |
[MPa] |
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sHmax= |
637,601915288491 |
[MPa] |
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17.2.Naprężenia gnące przy |
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przeciążeniach |
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sFPmax3= |
360,00 |
[MPa] |
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sFPmax4= |
252,00 |
[MPa] |
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sFmax3= |
116,076499554673 |
[MPa] |
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sFmax3= |
116,076499554673 |
[MPa] |
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sFmax4= |
102,810613891281 |
[MPa] |
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sFmax4= |
102,810613891281 |
[MPa] |
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18.Siły działające w zazębieniu |
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przekładni |
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18.1.Moment rzeczywisty na wale |
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wyjściowym |
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u= |
1,67 |
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urz= |
1,66666666666667 |
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T2= |
260,72 |
[Nm] |
T3rz= |
434,525 |
[Nm] |
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T3rz= |
434,525 |
[Nm] |
T3= |
434,53 |
[Nm] |
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18.2.Siły obwodowe |
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dW3= |
141,02 |
[mm] |
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dW4= |
235,03 |
[mm] |
Ft3= |
3697,56699523404 |
[N] |
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Ft3= |
3697,56699523404 |
[N] |
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Ft4= |
3697,56699523404 |
[N] |
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Ft4= |
3697,56699523404 |
[N] |
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Dane |
Obliczenia |
Wyniki |
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18.3.Siły promieniowe |
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α= |
20 |
[°] |
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Ft3= |
3697,56699523404 |
[N] |
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Ft4= |
3697,56699523404 |
[N] |
Fr3= |
1345,80432448601 |
[N] |
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Fr3= |
1345,80432448601 |
[N] |
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Fr4= |
1345,80432448601 |
[N] |
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Fr4= |
1345,80432448601 |
[N] |
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19.Średnice wałów |
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T2= |
260,72 |
[Nm] |
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T3= |
434,53 |
[Nm] |
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ks= |
40,00 |
[MPa] |
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wał wejściowy dwał1=dse=28 [mm] |
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dwał1= |
28 |
[mm] |
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wał pośredni |
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dwał2= |
31,941747864513 |
[mm] |
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dwał2= |
31,94 |
[mm] |
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wał wyjściowy |
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dwał3= |
37,871129704332 |
[mm] |
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dwał3= |
37,87 |
[mm] |
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dwał2<dwał3 |
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20.Orientacyjne wymiary do |
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rozplanowania |
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20.1.Grubość ścianki korpusu reduktora |
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(d=>8,0 mm) |
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Dla reduktorów walcowych: |
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aWwał1= |
139,053743719052 |
[mm] |
`dwustopniowych |
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aWwał2= |
188,026342969884 |
[mm] |
d = (0,025aWwał2 + 3) mm |
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d= |
8,7006585742471 |
[mm] |
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d= |
8,70 |
[mm] |
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20.2.Minimalna odległość od wew. pow. |
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ścianki reduktora do: |
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`bocznej powierzchni obracającej się |
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części e = (1,0…1,2)d mm; e = 1,1d |
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e= |
9,57072443167181 |
[mm] |
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e= |
9,57 |
[mm] |
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`bocznej powierzchni łożyska tocznego |
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e1 = (0…5) mm |
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e1= |
2,00 |
[mm] |
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Dane |
Obliczenia |
Wyniki |
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d= |
8,70 |
[mm] |
20.3.Minimalna odległość w kierunku |
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osiowy między obracającymi się |
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częściami osadzonymi na: |
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` różnych wałach e3 = (0,5…1,0)d mm |
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e3 = 0,75d |
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e3= |
6,52549393068533 |
[mm] |
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e3= |
6,53 |
[mm] |
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20.4.Minimalna odległość w kierunku |
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promieniowym między kołem zębatym |
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jednego stopnia a wałem drugiego st. |
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e4 = (5,0…7,0) mm |
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e4= |
6,00 |
[mm] |
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20.5.Minimalna odległość w kierunku |
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promieniowym od wierzchołków kół |
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zbatych do: |
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` wewnętrznej powierzchni ścianki korpusu |
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e5 = 1,2d mm |
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e5= |
10,4407902890965 |
[mm] |
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e5= |
10,44 |
[mm] |
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` wew. dolnej pow. ścianki korpusu |
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e6 = (5…10)m mm |
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m= |
6 |
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e6= |
42 |
[mm] |
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e6= |
42 |
[mm] |
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20.6.Minimalna odległość od bocznych |
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pow. części obracających się razem z |
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wałem do nieruchomych części |
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zew. reduktora |
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e7 = (5,0…8,0) mm |
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e7= |
6,50 |
[mm] |
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e8 = (0,7…1,0)d mm |
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e8= |
7,39555978811004 |
[mm] |
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e8= |
7,40 |
[mm] |
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20.7.Szerokość kołnierzy s łączonych |
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śrubą o średnicy dśr2 = 1,2d |
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dśr2= |
10,44 |
[mm] |
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z uwzględnieniem grubości ścianki d |
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s = k+d+4 |
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s= |
40,70 |
[mm] |
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s= |
40,70 |
[mm] |
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k = f(dśr) |
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k= |
28 |
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20.8.Grubość kołnierza pokrywy bocznej |
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h1 = f(D) |
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h1= |
8,00 |
[mm] |
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20.9.Wysokość łba śruby |
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h = 0,8 h1 |
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h= |
6,40 |
[mm] |
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Dane |
Obliczenia |
Wyniki |
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20.10.Grubość tulei |
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h3 = f(D) |
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h3= |
8,00 |
[mm] |
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20.11.Grubość kołnierza tulei |
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h2 = h1 |
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h2= |
8,00 |
[mm] |
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20.12.Odległość od bocznej pow.łożyska |
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do bocznej pow. nakładanej pokrywy |
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h4 dobiera się konstrukcyjnie |
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h4= |
5,00 |
[mm] |
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20.13.Odległość między bocznymi pow. |
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łożysk montowanych parami |
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h5 = (0…5) mm |
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h5= |
3,00 |
[mm] |
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20.14.Dobór pozostałych wymiarów |
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dwał2= |
31,94 |
[mm] |
Dobieram Lp ≈ Dp ≈ (1,6…1,8)dwał |
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dwał3= |
37,87 |
[mm] |
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Przedłużam piasty do wym: |
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Lp2≈Dp2 ≈ |
57,4951461561234 |
[mm] |
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Lp2≈Dp2 ≈ |
73,5 |
[mm] |
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Lp3≈Dp3 ≈ |
68,1680334677976 |
[mm] |
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Lp3≈Dp3 ≈ |
94,7 |
[mm] |
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Dane |
Obliczenia |
Wyniki |
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21.Obliczenia wałów |
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Wał wejściowy |
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Dane: |
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Ms1= |
124,15 |
[Nm] |
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Ms2=Ms3= |
260,72 |
[Nm] |
Po1= |
2767,7428144444 |
[N] |
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Po1= |
2768 |
[N] |
Ms4= |
434,53 |
[Nm] |
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n1= |
1000 |
[obr/min] |
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n2= |
476,19 |
[obr/min] |
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n3= |
600,00 |
[obr/min] |
Pr1= |
1007,37599982515 |
[N] |
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Pr1= |
1007 |
[N] |
d1= |
89,7120927219693 |
[mm] |
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d2= |
188,40 |
[mm] |
Wał pośredni |
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d3= |
141,02 |
[mm] |
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d4= |
235,032928712355 |
[mm] |
Po2= |
2767,7428144444 |
[N] |
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Po2= |
2768 |
[N] |
tgαo= |
0,36 |
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N= |
13 |
[kW] |
Pr2= |
1007,37599982515 |
[N] |
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Pr2= |
1007 |
[N] |
Nwyj= |
27,3 |
[kW] |
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r1= |
44,83 |
[mm] |
Po3= |
3697,56699523404 |
[N] |
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Po3= |
3698 |
[N] |
r2= |
94,18 |
[mm] |
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r3= |
70,5 |
[mm] |
Pr3= |
1345,80432448601 |
[N] |
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Pr3= |
1346 |
[N] |
r4= |
117,5 |
[mm] |
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Wał napędzany |
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Po4= |
3697,56699523404 |
[N] |
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Po4= |
3698 |
[N] |
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Pr4= |
1345,80432448601 |
[N] |
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Pr4= |
1346 |
[N] |
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22.Wyniki otrzymane z programu WALCAD |
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Wał 1 |
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Reakcje podpór: |
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Podpora stała: |
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Fx= |
653,67 |
[N] |
Fy= |
-237,92 |
[N] |
Fz = 0 [N] |
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Podpora ruchoma: |
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Fx= |
2114,07 |
[N] |
Fy= |
-769,74 |
[N] |
Fz = 0 [N] |
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Płaszczyzna z-y |
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Wykres sił Fy [N] |
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Wykres sił Fz [N] |
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Wykres momentów gnących [N*mm] |
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Płaszczyzna z-x |
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Wykres sił Fx [N] |
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Wykres momentów gnących [N*mm] |
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Wykres momentów skręcających [N*mm] |
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Wstępne wyniki obliczeń |
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Wykres momentów gnących wypadkowych [N*mm] |
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Wykres momentów zastępczych [N*mm] |
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Wyliczone średnice wałów w [mm] |
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Naprężenia w przedziałach |
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Dobór łożysk z programu ŁożyskaCAD |
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Dobieram łożyska kulkowe zwykłe: |
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Lewe: |
6006 |
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Prawe: |
6008 |
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d = |
30 [mm] |
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d = |
40 [mm] |
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D = |
55 [mm] |
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D = |
68 [mm] |
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B = |
13 [mm] |
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B = |
15 [mm] |
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C = |
13300 [N] |
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C = |
17000 [N] |
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Co = |
8000 [N] |
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Co = |
12400 [N] |
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Wał 2 |
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Reakcje podpór: |
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Podpora stała: |
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Fx= |
-1943,75 |
[N] |
Fy= |
-707,47 |
[N] |
Fz = 0 [N] |
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Podpora ruchoma: |
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Fx= |
1013,92 |
[N] |
Fy = |
369,04 |
[N] |
Fz = 0 [N] |
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Płaszczyzna z-y |
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Wykres sił Fy [N] |
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Wykres sił Fz [N] |
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Wykres momentów gnących [N*mm] |
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Płaszczyzna z-x |
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Wykres sił Fx [N] |
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Wykres momentów gnących [N*mm] |
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Wykres momentów skręcających [N*mm] |
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Wstępne wyniki obliczeń |
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Wykres momentów gnących wypadkowych [N*mm] |
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Wykres momentów zastępczych [N*mm] |
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Wyliczone średnice wałów w [mm] |
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Naprężenia w przedziałach |
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Dobór łożysk z programu ŁożyskaCAD |
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Dobieram łożyska kulkowe zwykłe: |
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Lewe: |
6006 |
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Prawe: |
6007 |
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d = |
30 [mm] |
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d = |
35 [mm] |
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D = |
55 [mm] |
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D = |
62 [mm] |
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B = |
15 [mm] |
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B = |
14 [mm] |
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C = |
13300 [N] |
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C = |
15900 [N] |
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Co = |
8000 [N] |
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Co = |
10000 [N] |
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Wał 3 |
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Reakcje podpór: |
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Podpora stała: |
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Fx= |
-1176,59 |
[N] |
Fy= |
428,24 |
[N] |
Fz = 0 [N] |
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Podpora ruchoma: |
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Fx= |
-2520,98 |
[N] |
Fy= |
917,56 |
[N] |
Fz = 0 [N] |
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Płaszczyzna z-y |
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Wykres sił Fy [N] |
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Wykres sił Fz [N] |
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Wykres momentów gnących [N*mm] |
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Płaszczyzna z-x |
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Wykres sił Fx [N] |
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Wykres momentów gnących [N*mm] |
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Wykres momentów skręcających [N*mm] |
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Wstępne wyniki obliczeń |
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Wykres momentów gnących wypadkowych [N*mm] |
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Wykres momentów zastępczych [N*mm] |
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Wyliczone średnice wałów w [mm] |
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Naprężenia w przedziałach |
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Dobór łożysk z programu ŁożyskaCAD |
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Dobieram łożyska kulkowe zwykłe: |
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Lewe: |
6108 |
|
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Prawe: |
6008 |
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|
d = |
44 [mm] |
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d = |
40 [mm] |
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D = |
68 [mm] |
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D = |
68 [mm] |
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B = |
15 [mm] |
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B = |
15 [mm] |
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C = |
16800 [N] |
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C = |
16800 [N] |
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Co = |
11600 [N] |
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Co = |
11600 [N] |
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Dane |
Obliczenia |
Wyniki |
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23.Dobór wpustów |
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T1= |
124,15 |
[Nm] |
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T2= |
260,72 |
[Nm] |
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T3= |
434,53 |
[Nm] |
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dwał2= |
31,94 |
[mm] |
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dwał3= |
37,87 |
[mm] |
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dwał1= |
28 |
[mm] |
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Lp1≈ |
57,83 |
[mm] |
Wpust 1 |
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Lp2≈ |
73,5 |
[mm] |
h= |
7 |
[mm] |
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Lp4≈ |
94,7 |
[mm] |
b= |
8 |
[mm] |
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|
Lp3≈ |
89,61 |
[mm] |
Naciski powierzchniowe |
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pdop= |
85 |
[MPa] |
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kA= |
1,1 |
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kT= |
95 |
[MPa] |
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Naprężenia ścinające |
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L≥ |
32,7887154861945 |
[mm] |
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L≥ |
12,8350563909774 |
[mm] |
|
L= |
45,00 |
[mm] |
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Wpust 2 |
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h= |
8 |
[mm] |
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b= |
10 |
[mm] |
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L≥ |
52,814248579018 |
[mm] |
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L≥ |
18,9019415967012 |
[mm] |
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L= |
56,00 |
[mm] |
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Wpust 3 |
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h= |
8 |
[mm] |
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b= |
10 |
[mm] |
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L≥ |
52,814248579018 |
[mm] |
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L≥ |
18,9019415967012 |
[mm] |
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L= |
70,00 |
[mm] |
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Wpust 4 |
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h= |
8 |
[mm] |
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b= |
10 |
[mm] |
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L≥ |
74,2421040748504 |
[mm] |
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L≥ |
26,5708583004728 |
[mm] |
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L= |
80,00 |
[mm] |
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Dane |
Obliczenia |
Wyniki |
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Dobierm śruby |
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M4x20-5,6 |
|
(PN-74/M-82106) |
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M6x20-5,6 |
|
(PN-74/M-82106) |
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M10x20-5,6 |
|
(PN-74/M-82106) |
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M10x20-5,6 |
|
(PN-74/M-82106) |
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Nakrętki |
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M10-6 |
|
(PN-86/M-82155) |
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Podkładki zwykłe |
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4,2 |
(PN-EN ISO 10673:2002 |
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6,2 |
(PN-EN ISO 10673:2003 |
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10,2 |
(PN-EN ISO 10673:2004 |
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Podkładki sprężyste |
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6,1 |
(PN-77/M-82008) |
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10,2 |
(PN-77/M-82009) |
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Pierścień uszczelniający |
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A 30x40x7 (PN-72/M-86964) |
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A 40x50x10 (PN-72/M-86964) |
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