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PROJEKT WAŁU |
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pola o tym kolorze należy wypełnić danymi z tematu |
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pola o tym kolorze oznaczają wartości wyliczone przez program |
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pola o tym kolorze należy wypełnić samodzielnie |
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Dane do obliczeń wału nr 2: |
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Schemat kinematyczny układu przeniesienia napędu: |
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1. Moc silnika Ms: |
11 |
[kW] |
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2. Prędkość obrotowa silnika ns: |
960 |
[1/min] |
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3. Sprawność przekładni nprz: |
99 |
[%] |
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4. Sprawność sprzęgła nsp |
99 |
[%] |
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5. Zakładana wartość przełożenia i: |
3,8 |
[ - ] |
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6. Liczba zębów koła zębatego z1: |
21 |
- |
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7. Liczba zębów koła zębatego z2: |
79 |
- |
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8. Moduł nominalny mn: |
3,5 |
[mm] |
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9. Odległość osi wałów: |
175 |
[mm] |
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10. Długość piasty koła zębatego 2 |
55 |
[mm] |
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11. Kąt przyporu alfa: |
20 |
[stopnie] |
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12. Rodzaj łożysk: |
toczne |
- |
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13: Trwalość łożysk: |
35000 |
[h] |
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14. Długość 'a' wału: |
80 |
[mm] |
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15. Długość 'b' wału: |
60 |
[mm] |
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16. Długość 'c' wału: |
80 |
[mm] |
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17. Temperatura pracy przekładni: |
75 |
[st. C] |
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18. Materiał wału: |
30H |
- |
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Wielkości wynikowe: |
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1. Rzeczywiste przełożenie przekładni: |
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i= |
3,76 |
[ - ] |
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2. Średnica podziałowa koła z1: |
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dp1= |
73,5 |
[mm] |
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3. Średnica podziałowa koła z2: |
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dp2= |
276,5 |
[mm] |
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4. Prędkość obrotowa wału 2: |
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n2= |
255,2 |
[1/min] |
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5. Moc na wale 2: |
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N2= |
10,8 |
[kW] |
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6. Moment obrotowy na wale 2: |
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M2= |
403,5 |
[Nm] |
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7. Siła obwodowa Fo: |
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Fo= |
2918,4 |
[N] |
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8. Siła promieniowa Fr: |
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Fr= |
1062,2 |
[N] |
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Obliczenia reakcji w punktach podparcia (łożyskach): |
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1. Suma sił w płaszczyźnie poziomej x-x: |
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RAX= |
1250,7 |
[N] |
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2. Suma momentów względem punktu 1: |
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RBX= |
1667,6 |
[N] |
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3. Suma sił w płaszczyźnie pionowej y-y: |
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RAY= |
455,2 |
[N] |
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4. Suma momentów względem punktu 1: |
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RBY= |
607,0 |
[N] |
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5. Wypadkowa siła reakcji w podporze A (pkt. 1): |
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RA= |
1331,0 |
[N] |
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6. Wypadkowa siła reakcji w podporze B (pkt. 3): |
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RB= |
1774,7 |
[N] |
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1. Momenty gnące pochodzące od sił poziomych x-x: |
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Przedział x1: |
0 |
- |
80 |
[mm] |
Mgx(0)= |
0 |
[Nm] |
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Mgx(80)= |
100,1 |
[Nm] |
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Przedział x2: |
80 |
- |
140 |
[mm] |
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Mgx(140)= |
0,0 |
[Nm] |
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Przedział x3: |
140 |
- |
220 |
[mm] |
Mgx(220)= |
0,0 |
[Nm] |
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2. Momenty gnące pochodzące od sił pionowych y-y: |
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Przedział y1: |
0 |
- |
80 |
[mm] |
Mgy(0)= |
0 |
[Nm] |
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Mgy(80)= |
36,4 |
[Nm] |
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Przedział y2: |
80 |
- |
140 |
[mm] |
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Mgy(140)= |
0,0 |
[Nm] |
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Przedział y3: |
140 |
- |
220 |
[mm] |
Mgy(220)= |
0,0 |
[Nm] |
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3. Momet gnący całkowity: |
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1. Moment gnący całkowity w pkt. 1 (podpora A): |
Mgc(0)= |
0 |
[Nm] |
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2. Moment gnący całkowity w pkt. 2 (koło zębate): |
Mgc(80)= |
106,5 |
[Nm] |
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3. Moment gnący całkowity w pkt. 3 (podpora B): |
Mgc(140)= |
0,0 |
[Nm] |
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4. Moment gnący całkowity w pkt. 4 (koniec wału): |
Mgc(220)= |
0,0 |
[Nm] |
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4. Moment skręcający: |
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Moment skręcający przenoszony przez wał B: |
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1. Moment skręcający w pkt.1 (podpora A): |
Ms(1)= |
0 |
[Nm] |
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2. Moment skręcający w pkt. 2 (koło zębate): |
Ms(2)= |
403,5 |
[Nm] |
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3. Moment skręcający w pkt. 3 (podpora B): |
Ms(3)= |
403,5 |
[Nm] |
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4. Momeny skręcający w pkt.4 (koniec wału): |
Ms(4)= |
403,5 |
[Nm] |
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5. Obliczenie momentu zredukowanego (dla zginania obustronnego i skręcania jednostronnego): |
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Z powyższych danych wynika, że dominuje: |
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skręcanie - wzór 2 |
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1. Dla dominującego zginania wzór na naprężenia zastępcze przyjmuje postać: |
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2. Dla dominującego skręcania wzór na naprężenia zastępcze przyjmuje postać: |
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Momenty zredukowane dla poszczególnych punktów: |
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pkt. 1 - łożysko A (środek) |
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Mz(1)= |
0,0 |
[Nm] |
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pkt. 2 - środek koła zębatego: |
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Mz(2)= |
472,5 |
[Nm] |
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pkt. 3 - łożysko B (środek): |
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Mz(3)= |
403,5 |
[Nm] |
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pkt. 4 - koniec wału: |
Mz(4)= |
403,5 |
[Nm] |
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6. Obliczanie średnic wału w poszczególnych punktach (do wyznaczenia końcowej średnicy należy uwzględnić wartość momentu zastępczego, średnicę łożysk i wymiary wpustów): |
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wprowadź wartości: |
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1. Podaj parametry materiału na wał - stal: |
30H |
: |
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1a. naprężenia dopuszczalne na obustronne zginanie dla stali |
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30H |
: |
kgo= |
110 |
[MPa] |
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1b. Naprężenia dopuszczalne na jednostronne skręcanie dla stali |
30H |
: |
ksj= |
115 |
[MPa] |
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2. Średnice w poszczególnych punktach wału: |
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Wstępnie obliczone wartości średnic z warunku na moment zastępczy: |
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2a. Średnica w punkcie 2: |
d2= |
27,4 |
[mm] |
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28,0 |
[mm] |
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2b. Średnica w punkcie 3: |
d3= |
26,0 |
[mm] |
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26,0 |
[mm] |
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2c. Średnica w punkcie 4: |
d4= |
26,0 |
[mm] |
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26,0 |
[mm] |
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7. Obliczanie parametrów łożysk: |
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Ruchową nośność dynamiczną oblicza się ze wzoru: |
gdzie: |
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Obciążenie zastępcze ruchowe Fz: |
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Obciążenie promieniowe łożyska: |
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Obciążenie osiowe łożyska - dla kół zębatych o zębach prostych nie powstają siły osiowe |
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Dla Fa=0, współczynniki: |
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Dla ruchomego wałka |
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Współczynnik nadwyżek dynamicznych: |
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fd= |
1,8 |
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Prędkość obwodowa v: |
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v= |
3,7 |
[m/s] |
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fd1= |
1,2 |
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Charakter pracy maszyny odbierającej moc: |
fd2 |
fd2= |
1,5 |
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Współczynnik trwałości: |
fh= |
4,1 |
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Dla łożysk kulkowych q=3 |
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Współczynnik prędkości obrotowej: |
fn= |
0,5 |
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Współczynnik temperaturowy: |
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ft= |
1 |
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Oznaczenie wybranego łożyska (nr katalogowy i Cr [kN]): |
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d: |
D: |
B: |
rmin: |
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7a. Nośność ruchowa łożyska w węźle A: |
CA= |
19,5 |
[kN] |
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16012 |
60 [mm] |
95[mm] |
11 [mm] |
0,6 |
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7b. Nośność ruchowa łożyska w węźle B: |
CB= |
25,9 |
[kN] |
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6306 |
30 [mm] |
72 [mm] |
19 [mm] |
1,1 |
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8. Obliczenia wpustów na wale: |
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8a. Obliczenie wpustu pod kołem zębatym dla wstępnie obliczonej średnicy d2: |
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Z normy dobierz wymiary poprzeczne bxh dla wpustu dla obliczonej wstępnie średnicy. Ponieważ rowek wpustowy osłabi wał średnica ta zostaje zwiększona o 10-15% (przy jednym wpuście): |
d2= |
33,0 |
[mm] |
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Wprowadź znormalizowaną średnicę wału d2: |
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d2= |
33,0 |
[mm] |
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b= |
10 |
[mm] |
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h= |
8 |
[mm] |
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Wprowadź wartość naprężeń dopuszczalnych na obciążenia ściskające zmienne kcj dla stali wału: |
kcj= |
145 |
[MPa] |
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Współczynnik liczbowy określający nierównomierność obciążenia - dynamiczne, nieulepszane pow. styku 0,4 - 0,6. Obciążenie równomierne 0,6 - 0,8, Koła nieprzesuwne - 1,0 |
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0,8 |
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Naprężenia dopuszczalne kd na docisk dla połączenia wpustowego dla wału: |
kd= |
116,0 |
[MPa] |
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Naprężenia dopuszczalne kd na docisk dla połączenia wpustowego dla piasty - koła zębatego: |
kd= |
200 |
[MPa] |
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Do obliczenia długości wpustu należy uwzględnić kd słabszego materiału - w tym przypadku jest to materiał: |
wału |
kd= |
116,0 |
[MPa] |
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Obliczenie długości wpustu z warunku na docisk: |
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lo= |
52,7 |
[mm] |
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Całkowita długość wpustu l: |
l= |
62,7 |
[mm] |
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Wprowadź znormalizowaną długość wpustu: |
l= |
65 |
[mm] |
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8b. Obliczenie wpustu dla sprzęgła dla średnicy d4: |
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Z normy dobierz wymiary poprzeczne bxh dla wpustu dla obliczonej wstępnie średnicy. Ponieważ rowek wpustowy osłabi wał średnica ta zostaje zwiększona o 10-15% (przy jednym wpuście): |
d4= |
30,0 |
[mm] |
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Wprowadź znormalizowaną średnicę wału d4: |
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d4= |
30,0 |
[mm] |
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b= |
8 |
[mm] |
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h= |
7 |
[mm] |
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Naprężenia dopuszczalne kd na docisk dla połączenia wpustowego dla piasty sprzęgła: |
kd= |
200 |
[MPa] |
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Do obliczenia długości wpustu należy uwzględnić kd słabszego materiału - w tym przypadku jest to materiał: |
wału |
kd= |
116,0 |
[MPa] |
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Obliczenie długości wpustu z warunku na docisk: |
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lo= |
66,2 |
[mm] |
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Całkowita długość wpustu l: |
l= |
74,2 |
[mm] |
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Wprowadź znormalizowaną długość wpustu: |
l= |
80 |
[mm] |
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9. Obliczenie rzeczywistego współczynnika bezpieczeństwa w najbardziej obciążonym przekroju: |
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Ustalenie wymiarów karbu - po uwzględnieniu średnic wynikajacych z warunku na moment zastępczy, wpust oraz łożyska: |
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Wprowadź dobrane wymiary wału: |
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D= |
33 |
[mm] |
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d= |
30 |
[mm] |
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[mm] |
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/r= |
0,03 |
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R/r= |
1,10 |
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9a. Obliczanie wartości współczynników dla zginania: |
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Wprowadź odczytaną z wykresu wartość współczynnika: |
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Wartość współczynnika kształtu: |
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alfak= |
2,18 |
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Wprowadź odczytaną z wykresu wartość współczynnika: |
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Współczynnik działania karbu: |
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Betak= |
1,1 |
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Wprowadź odczytaną z wykresu wartość współczynnika: |
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Współczynnik stanu powierzchni: |
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Betap= |
1,15 |
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Współczynnik spiętrzenia naprężeń: |
Betag= |
1,25 |
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Wprowadź odczytaną z wykresu wartość współczynnika: |
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Współczynnik wielkości przedmiotu: |
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gamma= |
1,4 |
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9b. Obliczanie wartości współczynników dla skręcania: |
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Wprowadź odczytaną z wykresu wartość współczynnika: |
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Wartość współczynnika kształtu: |
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alfak= |
2,6 |
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Wprowadź odczytaną z wykresu wartość współczynnika: |
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Współczynnik działania karbu: |
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Betak= |
1,3 |
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Wprowadź odczytaną z wykresu wartość współczynnika: |
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Współczynnik stanu powierzchni: |
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Betap= |
1,35 |
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Współczynnik spiętrzenia naprężeń: |
Betas= |
1,65 |
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Dla stali węglowych 2-5% mniej niż dla zginania, dla stali stopowych 2-5% więcej niż dla zginania: |
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Współczynnik wielkości przedmiotu: |
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gamma= |
1,5 |
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9c. Wyznaczenie naprężeń średnich oraz aplitud cyklu naprężeń: |
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Naprężenia średnie: |
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Naprężenia maksymalne: |
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0 |
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Smax= |
39,4 |
[MPa] |
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Tmax= |
74,7 |
[MPa] |
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9d. Obliczenie rzeczywistego współczynnika bezpieczeństwa delta: |
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Wprowadź parametry stali: |
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Wprowadź wartość z tabeli: |
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Wytrzymałość zmęczeniowa na obustronne zginanie: |
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Zgo= |
540 |
[MPa] |
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Wprowadź wartość z tabeli: |
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Wytrzymałość zmęczeniowa na jednostronne skręcanie: |
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Zsj= |
600 |
[MPa] |
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Rzeczywisty współczynnik bezpieczeństwa na zginanie: |
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deltag= |
7,8 |
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Rzeczywisty współczynnik bezpieczeńśtwa na skręcanie: |
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deltas= |
3,2 |
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Całkowity rzeczywisty współczynnik bezpieczeństwa: |
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delta= |
3,0 |
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Wartość współczynnika bezpieczeństwa mieści się w zalecanym przedziale |
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KONIEC |
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UWAGA: |
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Poniższe obliczenia dotyczą przekładni zębatej i służą wstępnemu określeniu wartości modułu i średnic kół zębatych. Po tak wstępnie dobraniu parametrów należy przeliczyć przekładnię, co wykracza poza zakres materiału przedmiotu. Do prześledzenia dla zainteresowanych: |
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Danymi wejściowymi do budowy przekładni są zwykle: |
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1. Przenoszona moc |
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2. Prędkość obrotowa na wyjściu lub przełożenie - wynikające ze stosunków prędkości obrotowej silnika i oczekiwanej na wyjsciowym wale reduktora |
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Powyższe dane są podane w temacie. |
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Obliczenia: |
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1. Określenie średnicy zębnika d1: |
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gdzie: |
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N |
moc przenoszona przez obliczaną parę kół [kW] |
11 |
[kW] |
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n1 |
prędkość obrotowa silnika = prędkość obrotowa koła 1 (zębnika) |
960 |
[1/min] |
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Współczynnik będący stosunkiem szerokości zębnika b do jego średnicy podziałowej d1. Dla kół o twardość <350HB i niesymetrycznie łożyskowanych przyjumuje się 0,6-1,2. |
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0,75 |
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Współczynnik określający miarę wytężenia koła. Dla przekładni ogólnego zastosowania dla stali konstrukcyjcych przyjęto wartość średnią Qu=1,5 |
1,5 |
[N/mm2] |
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ip |
Początkowa zakładana wartość przełożenia przekładni |
3,8 |
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d1= |
63 |
[mm] |
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Z warunków eksploatacyjnych przyjmuje się: |
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oraz |
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Z powyższych warunków wynika, że: |
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z1= |
21 |
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z2= |
79 |
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Moduł obliczony: |
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m= |
3,0 |
[mm] |
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Moduł znormalizowany: |
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m= |
3,50 |
[mm] |
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OSTATECZNIE: |
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z1= |
21 |
[ - ] |
liczba zębów koła zębatego 1 |
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z2= |
79 |
[ - ] |
liczba zębów koła zębatego 2 |
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i= |
3,76 |
[ - ] |
rzeczywiste przełożenie i |
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d1= |
73,5 |
[mm] |
średnica podziałowa koła 1 |
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d2= |
276,5 |
[mm] |
średnica podziałowa koła 2 |
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m= |
3,50 |
[mm] |
moduł nominalny |
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a= |
175 |
[mm] |
odległość osi wałów 1 i 2 |
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b= |
55,125 |
[mm] |
szerokość wieńca zębatego |
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